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Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

• For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
• For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

• Add or Subtract the same value from both sides
• Clear out any fractions by Multiplying every term by the bottom parts
• Divide every term by the same nonzero value
• Combine Like Terms
• Expanding (the opposite of factoring) may also help
• Recognizing a pattern, such as the difference of squares
• Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

• solve Quadratic Equations
• solve Radical Equations
• solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

• Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
• Show all the steps , so it can be checked later (by you or someone else)

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

Common Core State Standards

How does this relate to 8 th grade and high school math?

• Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
• High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Solving Equations Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of solving equations. 15 questions with answers to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

• Combine like terms .
• Simplify the equation by using the opposite operation to both sides.
• Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

• Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

• Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
• Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
• Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
• Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

• Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
• Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
• Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
• Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
• Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Solve equations with fractions
• Linear equations
• Writing linear equations
• Substitution
• Identity math
• One step equation

Practice solving equations questions

1. Solve 4x-2=14.

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

• Inequalities
• Types of graph
• Coordinate plane

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Solving Equations

Solving equations involves finding the value of the unknown variables in the given equation. The condition that the two expressions are equal is satisfied by the value of the variable. Solving a linear equation in one variable results in a unique solution, solving a linear equation involving two variables gives two results. Solving a quadratic equation gives two roots. There are many methods and procedures followed in solving an equation. Let us discuss the techniques in solving an equation one by one, in detail.

What is the Meaning of Solving Equations?

Solving equations is computing the value of the unknown variable still balancing the equation on both sides. An equation is a condition on a variable such that two expressions in the variable have equal value. The value of the variable for which the equation is satisfied is said to be the solution of the equation. An equation remains the same if the LHS and the RHS are interchanged. The variable for which the value is to be found is isolated and the solution is obtained. Solving an equation depends on what type of equation that we are dealing with. The equations can be linear equations, quadratic equations, rational equations, or radical equations.

Steps in Solving an Equation

The aim of solving an equation is to find the value of the variable that satisfies the condition of the equation true. To isolate the variable, the following operations are performed still balancing the equation on both sides. By doing so LHS remains equal to RHS, and eventually, the balance remains undisturbed throughout.

• Addition property of equality : Add the same number to both the sides. If a = b, then a + c = b + c
• Subtraction property of equality : Subtract the same number from both sides. If a = b, then a - c = b - c
• Multiplication property of equality: Multiply the same number on both sides. If a = b, then ac = bc
• Division property of equality : Divide by the same number on both sides. If a = b, then a/c = b/c (where c ≠ 0)

After performing this systematic balancing method of solving an equation by a series of identical arithmetical operations on both sides of the equation, we separate the variable on one of the sides and the ultimate step is the solution of the equation.

Solving Equations of One Variable

A linear equation of one variable is of the form ax + b = 0, where a, b, c are real numbers. The following steps are followed while solving an equation that is linear.

• Remove the parenthesis and use the distributive property if required.
• Simplify both sides of the equation by combining like terms.
• If there are fractions , multiply both sides of the equation by the LCD (Least common denominator) of all the fractions.
• If there are decimals , multiply both sides of the equation by the lowest power of 10 to convert them into whole numbers.
• Bring the variable terms to one side of the equation and the constant terms to the other side using the addition and subtraction properties of equality.
• Make the coefficient of the variable as 1, using the multiplication or division properties of equality.
• isolate the variable and get the solution.

Consider this example: 3(x + 4) = 24 + x

We simplify the LHS using the distributive property.

3x + 12 = 24 + x

Group the like terms together using the transposing method. This becomes 3x - x = 24-12

Simplify further ⇒ 2x = 12

Use the division property of equality, 2x/2 = 12/2

isolate the variable x. x = 6 is the solution of the equation.

Use any one of the following techniques to simplify the linear equation and solve for the unknown variable. The trial and error method, balancing method and the transposing method are used to isolate the variable.

Solving an Equation by Trial And Error Method

Consider 12x = 60. To find x, we intuitively try to find that 12 times what number is 60. We find that 5 is the required number. Solving equations by trial and error method is not always easy.

Solving an Equation by Balancing Method

We need to isolate the variable x for solving an equation. Let us use the separation of variables method or the balancing method to solve it. Consider an equation 2x + 3 = 17.

We first eliminate 3 in the first step. To keep the balance while solving the equation, we subtract 3 from either side of the equation.

Thus 2x + 3 - 3 = 17 - 3

We have 2x = 14

Now to isolate x, we divide by 2 on both sides. (Division property of equality)

2x/2 = 14/2

Thus, we isolate the variable using the properties of equality while solving an equation in the balancing method.

Solving an Equation by Transposing Method

While solving an equation, we change the sides of the numbers. This process is called transposing. While transposing a number, we change its sign or reverse the operation. Consider 5y + 2 = 22.

We need to find y, so isolate it. Hence we transpose the number 2 to the other side. The equation becomes,

Now taking 5 to the other side, we reverse the operation of multiplication to division. y = 20/5 = 4

Solving an Equation That is Quadratic

There are equations that yield more than one solution. Quadratic polynomials are of degree two and the zeroes of a quadratic polynomial represent the quadratic equation.

Consider (x+3) (x+2)= 0. This is quadratic in nature. We just equate each of the expressions in the LHS to 0.

Either x+3 = 0 or x+2 =0.

We arrive at x = -3 and x = -2.

A quadratic equation is of the form ax 2 + bx + c = 0. Solving an equation that is quadratic, results in two roots : α and β.

Steps involved in solving a quadratic equation are:

By Completing The Squares Method

By factorization method, by formula method.

Solving an equation of quadratic type by completing the squares method is quite easy as we apply our knowledge of algebraic identity: (a+b) 2

• Write the equation in the standard form ax 2 + bx + c = 0.
• Divide both the sides of the equation by a.
• Move the constant term to the other side
• Add the square of one-half of the coefficient of x on both sides.
• Complete the left-hand side as a square and simplify the right-hand side.
• Take the square root on both sides and solve for x.

For more information about solving equations (quadratic) by completing the squares, click here .

Solving an equation of quadratic type using the factorization method , follow the steps discussed here. Write the given equation in the standard form and by splitting the middle terms, factorize the equation. Rewrite the equation obtained as a product of two linear factors. Equate each linear factor to zero and solve for x. Consider 2x 2 + 19x + 30 =0. This is of the standard form ax 2 + bx + c = 0.

Split the middle term in such a way that the product of the terms should equal the product of the coefficient of x 2 and c and the sum of the terms should be b. Here the product of the terms should be 60 and the sum should be 19. Thus, split 19x as 4x and 15x (as the sum of 4 and 15 is 19 and their product is 60).

2x 2 + 4x + 15x + 30 = 0

Take the common factor out of the first two terms and the common factors out of the last two terms.

2x(x + 2) + 15(x + 2) = 0

Factoring (x+2) again, we get

(x + 2)(2x + 15) = 0

x = -2 and x = -15/2

Solving an equation that is quadratic involves such steps while splitting the middle terms on factorization.

Solving an equation of quadratic type using the formula

x = [-b ± √[(b 2 -4ac)]/2a helps us find the roots of the quadratic equation ax 2 + bx + c = 0. Plugging in the values of a, b, and c in the formula, we arrive at the solution.

Consider the example: 9x 2 -12 x + 4 = 0

a= 9, b = -12 and c = 4

x = [-b ± √[(b 2 -4ac)]/2a

= [12 ± √[((-12) 2 -4×9×4)] / (2 × 9)

= [12 ± √(144 - 144)] / 18

= (12 ± 0)/18

x = 12/18 = 2/3

Solving an Equation That is Rational

An equation with at least one polynomial expression in its denominator is known as the rational equation. Solving an equation that is rational involves the following steps. Reduce the fractions to a common denominator and then solve the equation of the numerators .

Consider x/(x-1) = 5/3

On cross-multiplication, we get

3x = 5(x-1)

3x = 5x - 5

3x - 5x = - 5

Solving an Equation That is Radical

An equation in which the variable is under a radical is termed the radical equation. Solving an equation that is a radical involves a few steps. Express the given radical equation in terms of the index of the radical and balance the equation. Solve for the variable.

Consider √(x+1) = 4

Now square both the sides to balance it. [ √(x+1)] 2 = 4 2

Thus x = 16-1 =15

Important Notes on Solving Equations:

• Solving an equation is finding the value of the variable in the equation.
• The solution of an equation satisfies the condition of the given equation.
• Solving an equation of linear type can be also done  graphically .
• If the right side part of an equation is zero, then for solving equation, just graph the left side of the equation and the x-intercept (s) of the graph would be the solution(s).

☛ Related Articles:

• Solving Equations Calculator
• Simultaneous linear equations
• One variable linear equations and inequations
• Simple equations and their applications

Examples of Solving an Equation

Example 1. Use the balancing method of solving equations: (x-2) / 5 - (x-4) / 2 = 2

The given equation is (x-2) / 5 - (x-4) / 2 = 2.

Solving an equation that is rational involves the following steps.

Simplify the LHS. Take the LCD of the denominators. LCD is 10.

[2(x-2) -5(x-4)]/10 = 2

Use distributive property and simplify the numerator.

We get [2x- 4 -5x+20]/10 = 2

Use the multiplication property of equality to get rid of the denominator.

10 × [2x- 4 -5x+20]/10 = 10 × 2

Simplifying we get - 3x + 16 = 20

isolate the term with the variable using the addition property of equality

-3x + 16 - 16 = 20 - 16

isolate the variable using the division property of equality

-3x/3 = 4/3

Answer: Thus solving (x-2)/ 5 - (x-4)/2 = 2, x = -4/3

Example 2. Use the transposing method of solving an equation 0.4(a+10)= 2 - 0.6a

Solving equations that are linear involving decimals involves the following steps.

Given 0.4(a+10)= 2 - 0.6a

0.4 a + 0.4 × 10 = 2- 0.6a

0.4 a + 4 = 2- 0.6a

0.4 a + 0.6a = 2-4

Answer: The solution is a = -2

Example 3. What is the value of p on solving an equation: 4 (p - 3) - (p - 5) = 4?

Given: 4 (p - 3) - (p - 5) = 4

Let us use the transposing method in solving equations.

4p - 12 - p + 5 = 4 (distributive property)

Answer: The value of p = 11/3

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Practice Questions on Solving an Equation

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FAQs on Solving Equations

What is solving an equation.

Solving an equation is finding the value of the unknown variables in the given equation. The process of solving an equation depends on the type of the equation .

What are The Steps in Solving Equations?

Identify the type of equation: linear, quadratic, logarithmic, exponential, radical or rational.

• Remove the brackets, if any in the given equation. Apply the distributive property.
• Add the same number to both the sides
• Subtract the same number from both the sides
• Multiply the same number on both the sides
• Divide by the same number on both sides.

What are The Golden Rule in Solving an Equation?

The type of the equation is identified. If it is a linear equation, separating the variables method or transposing method is used. If it is a quadratic equation, completing the squares, splitting the middle terms using factorization is used or by formula method.

How Do You use 3 Steps in Solving an Equation?

The 3 steps in solving an equation are to

• remove the brackets, if any using the distributive property,
• simplify the equation by adding or subtracting the like terms ,
• isolating the variable and solving it.

How Do You Solve Linear Equations?

While solving an equation that is linear, we isolate the variable whose value is to be found. We either use transposing method or the balancing method.

How Do You Solve Quadratic Equations?

While solving an equation that is quadratic, we write the equation in the standard form ax 2 + bx + c = 0, and then solve using the formula method or factorization method or completing the squares method.

How Do You Solve Radical Equations?

While solving an equation that is radical , we remove the radical sign, by raising both the sides of the equation to the index of the radical, isolate the variable and solve for x.

How Do You Solve Rational Equations?

While solving an equation that is rational, we simplify the expression on each side of the equation, cross multiply , combine the like terms and then isolate the variable to solve for x.

• Solve equations and inequalities
• Simplify expressions
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The equations section of QuickMath allows you to solve and plot virtually any equation or system of equations. In most cases, you can find exact solutions to your equations. Even when this is not possible, QuickMath may be able to give you approximate solutions to almost any level of accuracy you require. It also contains a number of special commands for dealing with quadratic equations.

The Solve command can be uses to solve either a single equation for a single unknown from the basic solve page or to simultaneously solve a system of many equations in many unknowns from the advanced solve page . The advanced command allows you to specify whether you want approximate numerical answers as well as exact ones, and how many digits of accuracy (up to 16) you require. It also allows you to eliminate certain variables from the equations.

Go to the Solve page

The Plot command, from the Graphs section, will plot any function of two variables. In order to plot a single function of x, go to the basic equation plotting page , where you can enter the equation and specify the upper and lower limits on x that you want the graph to be plotted for. The advanced plotting page allows you to plot up to 6 equations on the one graph, each with their own color. It also gives you control over such things as whether or not to show the axes, where the axes should be located, what the aspect ratio of the plot should be and what the range of the dependent variable should be. All equations can be given in the explicit y = f(x) form or the implicit g(x,y) = c form.

Go to the Equation Plotting page

The Quadratics page contains 13 separate commands for dealing with the most common questions concerning quadratics. It allows you to : factor a quadratic function (by two different methods); solve a quadratic equation by factoring the quadratic, using the quadratic formula or by completing the square; rewrite a quadratic function in a different form by completing the square; calculate the concavity, x-intercepts, y-intercept, axis of symmetry and vertex of a parabola; plot a parabola; calculate the discriminant of a quadratic equation and use the discriminant to find the number of roots of a quadratic equation. Each command generates a complete and detailed custom-made explanation of all the steps needed to solve the problem.

Go to the Quadratics page

Introduction to Equations

By an equation we mean a mathematical sentence that states that two algebraic expressions are equal. For example, a (b + c) =ab + ac, ab = ba, and x 2 -1 = (x-1)(x+1) are all equations that we have been using. We recall that we defined a variable as a letter that may be replaced by numbers out of a given set, during a given discussion. This specified set of numbers is sometimes called the replacement set. In this chapter we will deal with equations involving variables where the replacement set, unless otherwise specified, is the set of all real numbers for which all the expressions in the equation are defined.

If an equation is true after the variable has been replaced by a specific number, then the number is called a solution of the equation and is said to satisfy it. Obviously, every solution is a member of the replacement set. The real number 3 is a solution of the equation 2x-1 = x+2, since 2*3-1=3+2. while 1 is a solution of the equation (x-1)(x+2) = 0. The set of all solutions of an equation is called the solution set of the equation.

In the first equation above {3} is the solution set, while in the second example {-2,1} is the solution set. We can verify by substitution that each of these numbers is a solution of its respective equation, and we will see later that these are the only solutions.

A conditional equation is an equation that is satisfied by some numbers from its replacement set and not satisfied by others. An identity is an equation that is satisfied by all numbers from its replacement set.

Example 1 Consider the equation 2x-1 = x+2

The replacement set here is the set of all real numbers. The equation is conditional since, for example, 1 is a member of the replacement set but not of the solution set.

Example 2 Consider the equation (x-1)(x+1) =x 2 -1 The replacement set is the set of all real numbers. From our laws of real numbers if a is any real number, then (a-1)(a+1) = a 2 -1 Therefore, every member of the replacement set is also a member of the solution set. Consequently this equation is an identity.  

The replacement set for this equation is the set of real numbers except 0, since 1/x and (1- x)/x are not defined for x = 0. If a is any real number in the replacement set, then

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Chapter 1: Solving Equations and Inequalities

Problem solving, learning objectives.

• Translate words into algebraic expressions and equations
• Define a process for solving word problems
• Apply the steps for solving word problems to distance, rate, and time problems
• Apply the steps for solving word problems to interest rate problems
• Evaluate a formula using substitution
• Rearrange formulas to isolate specific variables
• Identify an unknown given a formula
• Apply the steps for solving word problems to geometry problems
• Use the formula for converting between Fahrenheit and Celsius

Define a Process for Problem Solving

Word problems can be tricky. Often it takes a bit of practice to convert an English sentence into a mathematical sentence, which is one of the first steps to solving word problems. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it’s good to see what mathematical operators are associated with them.

Some examples follow:

• [latex]x\text{ is }5[/latex]  becomes [latex]x=5[/latex]
• Three more than a number becomes [latex]x+3[/latex]
• Four less than a number becomes [latex]x-4[/latex]
• Double the cost becomes [latex]2\cdot\text{ cost }[/latex]
• Groceries and gas together for the week cost $250 means [latex]\text{ groceries }+\text{ gas }=250[/latex]
• The difference of 9 and a number becomes [latex]9-x[/latex]. Notice how 9 is first in the sentence and the expression

Let’s practice translating a few more English phrases into algebraic expressions.

Translate the table into algebraic expressions:

In this example video, we show how to translate more words into mathematical expressions.

The power of algebra is how it can help you model real situations in order to answer questions about them.

Here are some steps to translate problem situations into algebraic equations you can solve. Not every word problem fits perfectly into these steps, but they will help you get started.

• Read and understand the problem.
• Determine the constants and variables in the problem.
• Translate words into algebraic expressions and equations.
• Write an equation to represent the problem.
• Solve the equation.
• Check and interpret your answer. Sometimes writing a sentence helps.

Twenty-eight less than five times a certain number is 232. What is the number?

Following the steps provided:

• Read and understand: we are looking for a number.
• Constants and variables: 28 and 232 are constants, “a certain number” is our variable because we don’t know its value, and we are asked to find it. We will call it x.
• Translate:  five times a certain number translates to [latex]5x[/latex] Twenty-eight less than five times a certain number translates to [latex]5x-28[/latex] because subtraction is built backward. is 232 translates to [latex]=232[/latex] because “is” is associated with equals.
• Write an equation:  [latex]5x-28=232[/latex]

[latex]\begin{array}{r}5x-28=232\\5x=260\\x=52\,\,\,\end{array}[/latex]

[latex]\begin{array}{r}5\left(52\right)-28=232\\5\left(52\right)=260\\260=260\end{array}[/latex].

In the video that follows, we show another example of how to translate a sentence into a mathematical expression using a problem solving method.

Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation.

For example, let’s say I want to know the next consecutive integer after 4. In mathematical terms, we would add 1 to 4 to get 5. We can generalize this idea as follows: the consecutive integer of any number, x , is [latex]x+1[/latex]. If we continue this pattern we can define any number of consecutive integers from any starting point. The following table shows how to describe four consecutive integers using algebraic notation.

We apply the idea of consecutive integers to solving a word problem in the following example.

The sum of three consecutive integers is 93. What are the integers?

• Read and understand:  We are looking for three numbers, and we know they are consecutive integers.
• Constants and Variables:  93 is a constant. The first integer we will call x . Second: [latex]x+1[/latex] Third: [latex]x+2[/latex]
• Translate:  The sum of three consecutive integers translates to [latex]x+\left(x+1\right)+\left(x+2\right)[/latex], based on how we defined the first, second, and third integers. Notice how we placed parentheses around the second and third integers. This is just to make each integer more distinct. is 93 translates to [latex]=93[/latex] because is is associated with equals.
• Write an equation:  [latex]x+\left(x+1\right)+\left(x+2\right)=93[/latex]

[latex]x+x+1+x+2=93[/latex]

Combine like terms, simplify, and solve.

[latex]\begin{array}{r}x+x+1+x+2=93\\3x+3 = 93\\\underline{-3\,\,\,\,\,-3}\\3x=90\\\frac{3x}{3}=\frac{90}{3}\\x=30\end{array}[/latex]

• Check and Interpret: Okay, we have found a value for x . We were asked to find the value of three consecutive integers, so we need to do a couple more steps. Remember how we defined our variables: The first integer we will call [latex]x[/latex], [latex]x=30[/latex] Second: [latex]x+1[/latex] so [latex]30+1=31[/latex] Third: [latex]x+2[/latex] so [latex]30+2=32[/latex] The three consecutive integers whose sum is [latex]93[/latex] are [latex]30\text{, }31\text{, and }32[/latex]

There is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would want to know how long it would take you to reach your destination. [latex]d=rt[/latex] is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes.

Distance, Rate, and Time

If you know two of the quantities in the relationship [latex]d=rt[/latex], you can easily find the third using methods for solving linear equations. For example, if you know that you will be traveling on a road with a speed limit of [latex]30\frac{\text{ miles }}{\text{ hour }}[/latex] for 2 hours, you can find the distance you would travel by multiplying rate times time or [latex]\left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }[/latex].

We can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the [latex]d=rt[/latex] equation for t using division:

[latex]d=rt\\\frac{d}{r}=t[/latex]

Likewise, if we want to find rate, we can isolate r using division:

[latex]d=rt\\\frac{d}{t}=r[/latex]

In the following examples you will see how this formula is applied to answer questions about ultra marathon running.

Ultra marathon running (defined as anything longer than 26.2 miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River 50-mile Endurance Run which begins in Sacramento, California, and ends in Auburn, California. [1] In 1993 Trason finished the run with a time of 6:09:08.  The men’s record for the same course was set in 1994 by Tom Johnson who finished the course with a time of 5:33:21. [2]

In the next examples we will use the [latex]d=rt[/latex] formula to answer the following questions about the two runners.

What was each runner’s rate for their record-setting runs?

By the time Johnson had finished, how many more miles did Trason have to run?

How much further could Johnson have run if he had run as long as Trason?

What was each runner’s time for running one mile?

To make answering the questions easier, we will round the two runners’ times to 6 hours and 5.5 hours.

Read and Understand:  We are looking for rate and we know distance and time, so we can use the idea: [latex]d=rt\\\frac{d}{t}=r[/latex]

Define and Translate: Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of what how all the terms are related to each other.

Write and Solve:

Trason’s rate:

[latex]\begin{array}{c}d=rt\\\\50\text{ miles }=\text{r}\left(6\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{8.33\text{ miles }}{\text{ hour }}\end{array}[/latex].

(rounded to two decimal places)

Johnson’s rate:

[latex]\begin{array}{c}d=rt\\\\,\,\,\,\,\,\,50\text{ miles }=\text{r}\left(5.5\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{9.1\text{ miles }}{\text{ hour }}\end{array}[/latex]

Check and Interpret:

We can fill in our table with this information.

Now that we know each runner’s rate we can answer the second question.

Here is the table we created for reference:

Read and Understand:  We are looking for how many miles Trason still had on the trail when Johnson had finished after 5.5 hours. This is a distance, and we know rate and time.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , and the time Trason had run is 5.5 hours.

[latex]\begin{array}{l}d=rt\\\\d=8.33\frac{\text{ miles }}{\text{ hour }}\left(5.5\text{ hours }\right)\\\\d=45.82\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles she had to run after 5.5 hours.  What we have found is how long she had run after 5.5 hours. We need to subtract [latex]d=45.82\text{ miles }[/latex] from the total distance of the course.

[latex]50\text{ miles }-45.82\text{ miles }=1.48\text{ miles }[/latex]

The third question is similar to the second. Now that we know each runner’s rate, we can answer questions about individual distances or times.

Read and Understand:  The word further implies we are looking for a distance.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , the time is 6 hours, and Johnson’s rate is [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex]

[latex]\begin{array}{l}d=rt\\\\d=9.1\frac{\text{ miles }}{\text{ hour }}\left(6\text{ hours }\right)\\\\d=54.6\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles Johnson would have run if he had run at his rate of [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex] for 6 hours.

Johnson would have run 54.6 miles, so that’s 4.6 more miles than than he ran for the race.

Now we will tackle the last question where we are asked to find a time for each runner.

Read and Understand:  we are looking for time, and this time our distance has changed from 50 miles to 1 mile, so we can use

Define and Translate: we can use the formula [latex]d=rt[/latex] again. This time the unknown is t , the distance is 1 mile, and we know each runner’s rate. It may help to create a new table:

We will need to divide to isolate time.

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=8.33\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{8.33\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.12\text{ hours }=t\end{array}[/latex].

0.12 hours is about 7.2 minutes, so Trason’s time for running one mile was about 7.2 minutes. WOW! She did that for 6 hours!

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=9.1\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{9.1\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.11\text{ hours }=t\end{array}[/latex].

0.11 hours is about 6.6 minutes, so Johnson’s time for running one mile was about 6.6 minutes. WOW! He did that for 5.5 hours!

Have we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole 50-mile course.  Yes, we answered our question.

Trason’s mile time was [latex]7.2\frac{\text{minutes}}{\text{mile}}[/latex] and Johnsons’ mile time was [latex]6.6\frac{\text{minutes}}{\text{mile}}[/latex]

In the following video, we show another example of answering many rate questions given distance and time.

Simple Interest

In order to entice customers to invest their money, many banks will offer interest-bearing accounts. The accounts work like this: a customer deposits a certain amount of money (called the Principal, or P ), which then grows slowly according to the interest rate ( R , measured in percent) and the length of time ( T , usually measured in months) that the money stays in the account. The amount earned over time is called the interest ( I ), which is then given to the customer.

The simplest way to calculate interest earned on an account is through the formula [latex]\displaystyle I=P\,\cdot \,R\,\cdot \,T[/latex].

If we know any of the three amounts related to this equation, we can find the fourth. For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for T using division:

[latex]\displaystyle\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,I=P\,\cdot \,R\,\cdot \,T\\\\ \frac{I}{{P}\,\cdot \,R}=\frac{P\cdot\,R\,\cdot \,T}{\,P\,\cdot \,R}\\\\\,\,\,\,\,\,\,\,\,\,\,{T}=\frac{I}{\,R\,\cdot \,T}\end{array}[/latex]

Below is a table showing the result of solving for each individual variable in the formula.

In the next examples, we will show how to substitute given values into the simple interest formula, and decipher which variable to solve for.

If a customer deposits a principal of $2000 at a monthly rate of 0.7%, what is the total amount that she has after 24 months?

Substitute in the values given for the Principal, Rate, and Time.

[latex]\displaystyle\begin{array}{l}I=P\,\cdot \,R\,\cdot \,T\\I=2000\cdot 0.7\%\cdot 24\end{array}[/latex]

Rewrite 0.7% as the decimal 0.007, then multiply.

[latex]\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}[/latex]

Add the interest and the original principal amount to get the total amount in her account.

[latex] \displaystyle 2000+336=2336[/latex]

She has $2336 after 24 months.

The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate.

In the following example you will see why it is important to make sure the units of the interest rate match the units of time when using the simple interest formula.

Alex invests $600 at 3.25% monthly interest for 3 years. What amount of interest has Alex earned?

Read and Understand: The question asks for an amount, so we can substitute what we are given into the simple interest formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex]

Define and Translate:  we know P, R, and T so we can use substitution. R  = 0.0325, P = $600, and T = 3 years. We have to be careful! R is in months, and T is in years.  We need to change T into months because we can’t change the rate—it is set by the bank.

[latex]{T}=3\text{ years }\cdot12\frac{\text{ months }}{ year }=36\text{ months }[/latex]

Substitute the given values into the formula.

[latex]\begin{array}{l} I=P\,\cdot \,R\,\cdot \,T\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}[/latex]

We were asked what amount Alex earned, which is the amount provided by the formula. In the previous example we were asked the total amount in the account, which included the principal and interest earned.

Alex has earned $756.

After 10 years, Jodi’s account balance has earned $1080 in interest. The rate on the account is 0.09% monthly. What was the original amount she invested in the account?

Read and Understand: The question asks for the original amount invested, the principal. We are given a length of time in years, and an interest rate in months, and the amount of interest earned.

Define and Translate:  we know I = $1080, R = 0.009, and T = 10 years so we can use [latex]{P}=\frac{I}{{R}\,\cdot \,T}[/latex]

We also need to make sure the units on the interest rate and the length of time match, and they do not. We need to change time into months again.

[latex]{T}=10\text{ years }\cdot12\frac{\text{ months }}{ year }=120\text{ months }[/latex]

Substitute the given values into the formula

[latex]\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}[/latex]

We were asked to find the principal given the amount of interest earned on an account.  If we substitute P = $1000 into the formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex] we get

[latex]I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080[/latex]

Our solution checks out. Jodi invested $1000.

Further Applications of Linear Equations

Formulas come up in many different areas of life. We have seen the formula that relates distance, rate, and time and the formula for simple interest on an investment. In this section we will look further at formulas and see examples of formulas for dimensions of geometric shapes as well as the formula for converting temperature between Fahrenheit and Celsius.

There are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.

Perimeter is the distance around an object. For example, consider a rectangle with a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[/latex]. Since there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using the formula [latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex] where

In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.

You want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there you realize you didn’t write down the width dimension—only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.

Here is what you have written down:

Perimeter = 16.4 feet Length = 4.7 feet

Can you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?

Read and Understand:  We know perimeter = 16.4 feet and length = 4.7 feet, and we want to find width.

Define and Translate:

Define the known and unknown dimensions:

First we will substitute the dimensions we know into the formula for perimeter:

[latex]\begin{array}{l}\,\,\,\,\,P=2{W}+2{L}\\\\16.4=2\left(w\right)+2\left(4.7\right)\end{array}[/latex]

Then we will isolate w to find the unknown width.

[latex]\begin{array}{l}16.4=2\left(w\right)+2\left(4.7\right)\\16.4=2{w}+9.4\\\underline{-9.4\,\,\,\,\,\,\,\,\,\,\,\,\,-9.4}\\\,\,\,\,\,\,\,7=2\left(w\right)\\\,\,\,\,\,\,\,\frac{7}{2}=\frac{2\left(w\right)}{2}\\\,\,\,\,3.5=w\end{array}[/latex]

Write the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won’t get the right size of board!

If we replace the width we found, [latex]w=3.5\text{ feet }[/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:

[latex]\begin{array}{l}\,\,\,\,\,{P}=2\left({L}\right)+2\left({W}\right)\\\\{16.4}=2\left({4.7}\right)+2\left({3.5}\right)\\\\{16.4}=9.4+7\\\\{16.4}=16.4\end{array}[/latex]

Our calculation for width checks out. We need to ask for 2 boards cut to 3.5 feet and 2 boards cut to 4.7 feet so we can make the new garden box.

This video shows a similar garden box problem.

We could have isolated the w in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle :  

[latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex].

First, isolate the term with  w by subtracting 2 l from both sides of the equation.

[latex] \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]

Next, clear the coefficient of w by dividing both sides of the equation by 2.

[latex]\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\frac{p-2l}{2}\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.

[latex]w=\frac{p-2l}{2}[/latex]

The area of a triangle is given by [latex] A=\frac{1}{2}bh[/latex] where

A = area b = the length of the base h = the height of the triangle

Remember that when two variables or a number and a variable are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough that it has been adopted as convention.

In the next example we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.

Find the base ( b) of a triangle with an area ( A ) of 20 square feet and a height ( h) of 8 feet.

Use the formula for the area of a triangle, [latex] {A}=\frac{{1}}{{2}}{bh}[/latex] .

Substitute the given lengths into the formula and solve for  b.

[latex]\displaystyle \begin{array}{l}\,\,A=\frac{1}{2}bh\\\\20=\frac{1}{2}b\cdot 8\\\\20=\frac{8}{2}b\\\\20=4b\\\\\frac{20}{4}=\frac{4b}{4}\\\\ \,\,\,5=b\end{array}[/latex]

The base of the triangle measures 5 feet.

We can rewrite the formula in terms of b or h as we did with perimeter previously. This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just x .

Use the multiplication and division properties of equality to isolate the variable b .

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

Use the multiplication and division properties of equality to isolate the variable h .

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]

Temperature

Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

[latex]C=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].

Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:

[latex]12=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Isolate the variable F to obtain the equivalent temperature.

[latex]\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\frac{108}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.

To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].

[latex]\begin{array}{l}\\\,\,\,\,\left(\frac{9}{5}\right)C=\left(F-32\right)\left(\frac{5}{9}\right)\left(\frac{9}{5}\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\frac{9}{5}C=F-32\end{array}[/latex]

Add 32 to both sides.

[latex]\begin{array}{l}\frac{9}{5}\,C+32=F-32+32\\\\\frac{9}{5}\,C+32=F\end{array}[/latex]

[latex]F=\frac{9}{5}C+32[/latex]

Think About It

Express the formula for the surface area of a cylinder, [latex]s=2\pi rh+2\pi r^{2}[/latex], in terms of the height, h .

In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h .

[latex]\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Next, isolate the variable h by dividing both sides of the equation by [latex]2\pi r[/latex].

[latex]\begin{array}{r}\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\ \frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

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Mathematics > Analysis of PDEs

Title: energy solutions of the cauchy-dirichlet problem for fractional nonlinear diffusion equations.

Abstract: The present paper is concerned with the Cauchy-Dirichlet problem for fractional (and non-fractional) nonlinear diffusion equations posed in bounded domains. Main results consist of well-posedness in an energy class with no sign restriction and convergence of such (possibly sign-changing) energy solutions to asymptotic profiles after a proper rescaling. They will be proved in a variational scheme only, without any use of semigroup theories nor classical quasilinear parabolic theories. Proofs are self-contained and performed in a totally unified fashion for both fractional and non-fractional cases as well as for both porous medium and fast diffusion cases.

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