problem solving involving linear equation in two variables

$The equations are 2 x plus y equals 7 and x minus 2y equals 6. Step 1 is to write both equations in standard form. Both equations are in standard form, Ax plus By equals C. If any coefficients are fractions, clear them. There are no fractions.$

Try It 4.17

Solve the system by elimination: { 3 x + y = 5 2 x − 3 y = 7 . { 3 x + y = 5 2 x − 3 y = 7 .

Try It 4.18

Solve the system by elimination: { 4 x + y = − 5 − 2 x − 2 y = − 2 . { 4 x + y = − 5 − 2 x − 2 y = − 2 .

The steps are listed here for easy reference.

Solve a system of equations by elimination.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example 4.10

Solve the system by elimination: { 4 x − 3 y = 9 7 x + 2 y = −6 . { 4 x − 3 y = 9 7 x + 2 y = −6 .

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

Try It 4.19

Solve the system by elimination: { 3 x − 4 y = − 9 5 x + 3 y = 14 . { 3 x − 4 y = − 9 5 x + 3 y = 14 .

Try It 4.20

Solve each system by elimination: { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 .

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

Example 4.11

Solve the system by elimination: { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 . { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 .

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

Try It 4.21

Solve each system by elimination: { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 . { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 .

Try It 4.22

Solve each system by elimination: { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 . { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 .

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Example 4.12

Solve the system by elimination: { 3 x + 4 y = 12 y = 3 − 3 4 x . { 3 x + 4 y = 12 y = 3 − 3 4 x .

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 4.23

Solve the system by elimination: { 5 x − 3 y = 15 y = − 5 + 5 3 x . { 5 x − 3 y = 15 y = − 5 + 5 3 x .

Try It 4.24

Solve the system by elimination: { x + 2 y = 6 y = − 1 2 x + 3 . { x + 2 y = 6 y = − 1 2 x + 3 .

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example 4.13

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

Since both equations are in standard form, using elimination will be most convenient.

Since one equation is already solved for y , using substitution will be most convenient.

Try It 4.25

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = −7 { x = 2 y − 1 3 x − 5 y = −7

Try It 4.26

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

Section 4.1 Exercises

Practice makes perfect.

In the following exercises, determine if the following points are solutions to the given system of equations.

{ 2 x − 6 y = 0 3 x − 4 y = 5 { 2 x − 6 y = 0 3 x − 4 y = 5

ⓐ ( 3 , 1 ) ( 3 , 1 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

{ − 3 x + y = 8 − x + 2 y = −9 { − 3 x + y = 8 − x + 2 y = −9

ⓐ ( −5 , −7 ) ( −5 , −7 ) ⓑ ( −5 , 7 ) ( −5 , 7 )

{ x + y = 2 y = 3 4 x { x + y = 2 y = 3 4 x

ⓐ ( 8 7 , 6 7 ) ( 8 7 , 6 7 ) ⓑ ( 1 , 3 4 ) ( 1 , 3 4 )

{ 2 x + 3 y = 6 y = 2 3 x + 2 { 2 x + 3 y = 6 y = 2 3 x + 2 ⓐ ( −6 , 2 ) ( −6 , 2 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

In the following exercises, solve the following systems of equations by graphing.

{ 3 x + y = −3 2 x + 3 y = 5 { 3 x + y = −3 2 x + 3 y = 5

{ − x + y = 2 2 x + y = −4 { − x + y = 2 2 x + y = −4

{ y = x + 2 y = −2 x + 2 { y = x + 2 y = −2 x + 2

{ y = x − 2 y = −3 x + 2 { y = x − 2 y = −3 x + 2

{ y = 3 2 x + 1 y = − 1 2 x + 5 { y = 3 2 x + 1 y = − 1 2 x + 5

{ y = 2 3 x − 2 y = − 1 3 x − 5 { y = 2 3 x − 2 y = − 1 3 x − 5

{ x + y = −4 − x + 2 y = −2 { x + y = −4 − x + 2 y = −2

{ − x + 3 y = 3 x + 3 y = 3 { − x + 3 y = 3 x + 3 y = 3

{ − 2 x + 3 y = 3 x + 3 y = 12 { − 2 x + 3 y = 3 x + 3 y = 12

{ 2 x − y = 4 2 x + 3 y = 12 { 2 x − y = 4 2 x + 3 y = 12

{ x + 3 y = −6 y = − 4 3 x + 4 { x + 3 y = −6 y = − 4 3 x + 4

{ − x + 2 y = −6 y = − 1 2 x − 1 { − x + 2 y = −6 y = − 1 2 x − 1

{ − 2 x + 4 y = 4 y = 1 2 x { − 2 x + 4 y = 4 y = 1 2 x

{ 3 x + 5 y = 10 y = − 3 5 x + 1 { 3 x + 5 y = 10 y = − 3 5 x + 1

{ 4 x − 3 y = 8 8 x − 6 y = 14 { 4 x − 3 y = 8 8 x − 6 y = 14

{ x + 3 y = 4 − 2 x − 6 y = 3 { x + 3 y = 4 − 2 x − 6 y = 3

{ x = −3 y + 4 2 x + 6 y = 8 { x = −3 y + 4 2 x + 6 y = 8

{ 4 x = 3 y + 7 8 x − 6 y = 14 { 4 x = 3 y + 7 8 x − 6 y = 14

{ 2 x + y = 6 − 8 x − 4 y = −24 { 2 x + y = 6 − 8 x − 4 y = −24

{ 5 x + 2 y = 7 − 10 x − 4 y = −14 { 5 x + 2 y = 7 − 10 x − 4 y = −14

{ y = 2 3 x + 1 − 2 x + 3 y = 5 { y = 2 3 x + 1 − 2 x + 3 y = 5

{ y = 3 2 x + 1 2 x − 3 y = 7 { y = 3 2 x + 1 2 x − 3 y = 7

{ 5 x + 3 y = 4 2 x − 3 y = 5 { 5 x + 3 y = 4 2 x − 3 y = 5

{ y = − 1 2 x + 5 x + 2 y = 10 { y = − 1 2 x + 5 x + 2 y = 10

{ 5 x − 2 y = 10 y = 5 2 x − 5 { 5 x − 2 y = 10 y = 5 2 x − 5

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ − 2 x + 2 y = 6 y = −3 x + 1 { − 2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x − 16 y = 8 − x − 8 y = −4 { 2 x − 16 y = 8 − x − 8 y = −4

{ y = 7 8 x + 4 − 7 x + 8 y = 6 { y = 7 8 x + 4 − 7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 − 3 x − y = 0 { 5 x + 2 y = 2 − 3 x − y = 0

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 3 x + 1 2 y = 1 { x + 1 3 y = −1 1 3 x + 1 2 y = 1

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 − 3 x + 12 y = 3 { x − 4 y = −1 − 3 x + 12 y = 3

{ − 3 x − y = 8 6 x + 2 y = −16 { − 3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Solve the system of equations by substitution and explain all your steps in words: { 3 x + y = 12 x = y − 8 . { 3 x + y = 12 x = y − 8 .

Solve the system of equations by elimination and explain all your steps in words: { 5 x + 4 y = 10 2 x = 3 y + 27 . { 5 x + 4 y = 10 2 x = 3 y + 27 .

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing ⓑ by substitution ⓒ Which method do you prefer? Why?

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns 5 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: determine whether an ordered pair is a solution of a system of equations, solve a system of linear equations by graphing, solve a system of equations by substitution, solve a system of equations by elimination, choose the most convenient method to solve a system of linear equations. The remaining columns are blank.

If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Linear Equations in Two Variables

A linear equation in two variables is an equation in which two variables have the exponent 1. A system of equations with two variables has a unique solution, no solutions, or infinitely many solutions. A linear system of equations may have 'n' number of variables. An important thing to keep in mind while solving linear equations with n number of variables is that there must be n equations to solve and determine the value of variables.

Linear equations in two variables are the algebraic equations which are of the form (or can be converted to the form) y = mx + b, where m is the slope and b is the y-intercept . They are the equations of the first order. For example, y = 2x + 3 and 2y = 4x + 9 are two-variable linear equations.

What are Linear Equations in Two Variables?

The linear equations in two variables are the equations in which each of the two variables is of the highest order ( exponent ) of 1 and may have one, none, or infinitely many solutions. The standard form of a two-variable linear equation is ax + by + c = 0 where x and y are the two variables. The solutions can also be written in ordered pairs like (x, y). The graphical representation of the pairs of linear equations in two variables includes two straight lines which could be:

Forms of Linear Equations in Two Variables

A linear equation in two variables can be in different forms like standard form , intercept form and point-slope form . For example, the same equation 2x + 3y=9 can be represented in each of the forms like 2x + 3y - 9=0 (standard form), y = (-2/3)x + 3 ( slope-intercept form ), and y - 5/3 = -2/3(x + (-2)) (point-slope form). Look at the image given below showing all these three forms of representing linear equations in two variables with examples.

Linear Equations in Two Variables are of three forms standard form, slope intercept form and the point slope form. Each form is provided with an example

The system of equations means the collection of equations and they are also referred to as simultaneous linear equations . We will learn how to solve pair of linear equations in two variables using different methods.

Solving Pairs of Linear Equations in Two Variables

There are five methods to solve pairs of linear equations in two variables as shown below:

Determinant Method

Graphical method for solving linear equations in two variables.

The steps to solve linear equations in two variables graphically are given below:

Example: Find the solution of the following system of equations graphically.

Solution: We will graph them and see whether they intersect at a point. As you can see below, both lines meet at (1, 2). Thus, the solution of the given system of linear equations is x=1 and y=2.

Graph of pair of Linear Equations in Two Variables has two lines that meet at the point with coordinates 1 comma 2.

But both lines may not intersect always. Sometimes they may be parallel. In that case, the pairs of linear equations in two variables have no solution. In some other cases, both lines coincide with each other. In that case, each point on that line is a solution of the given system and hence the given system has an infinite number of solutions.

Consistent and Inconsistent System of Linear Equations:

Independent and Dependent System of Linear Equations:

Consider a system of two linear equations: a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0. Here we can understand when a linear system with two variables is consistent/inconsistent and independent/dependent.

Types of Solutions of linear equations in two variables, consistent and inconsistent, dependent and independent

Method of Substitution

To solve a system of two linear equations in two variables using the substitution method , we have to use the steps given below:

Example: Solve the following system of equations using the substitution method. x+2y-7=0 2x-5y+13=0

Solution: Let us solve the equation, x+2y-7=0 for y: x+2y-7=0 ⇒2y=7-x ⇒ y=(7-x)/2

Substitute this in the equation, 2x-5y+13=0:

2x-5y+13=0 ⇒ 2x-5((7-x)/2)+13=0 ⇒ 2x-(35/2)+(5x/2)+13=0 ⇒ 2x + (5x/2) = 35/2 - 13 ⇒ 9x/2 = 9/2 ⇒ x=1

Substitute x=1 this in the equation y=(7-x)/2:

y=(7-1)/2 = 3

Therefore, the solution of the given system is x=1 and y=3.

Consider a system of linear equations: a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0.

To solve this using the cross multiplication method , we first write the coefficients of each of x and y and constants as follows:

Cross Multiplication Method to Solve Linear Equations in Two Variables

Here, the arrows indicate that those coefficients have to be multiplied. Now we write the following equation by cross-multiplying and subtracting the products. $\dfrac{x}{b_{1} c_{2}-b_{2} c_{1}}=\dfrac{y}{c_{1} a_{2}-c_{2} a_{1}}=\dfrac{1}{a_{1} b_{2}-a_{2} b_{1}}$

From this equation, we get two equations:

$\begin{align} \dfrac{x}{b_{1} c_{2}-b_{2} c_{1}}&=\dfrac{1}{a_{1} b_{2}-a_{2} b_{1}} \\[0.2cm] \dfrac{y}{c_{1} a_{2}-c_{2} a_{1}}&=\dfrac{1}{a_{1} b_{2}-a_{2} b_{1}} \end{align}$

Solving each of these for x and y, the solution of the given system is:

$\begin{align} x&=\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\[0.2cm] y&=\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}} \end{align}$

Method of Elimination

To solve a system of linear equations in two variables using the elimination method , we will use the steps given below:

Example: Solve the following system of equations using the elimination method. 2x+3y-11=0 3x+2y-9=0

Adding or subtracting these two equations would not result in the cancellation of any variable. Let us aim at the cancellation of x. The coefficients of x in both equations are 2 and 3. Their LCM is 6. We will make the coefficients of x in both equations 6 and -6 such that the x terms get canceled when we add the equations.

3 × (2x+3y-11=0) ⇒ 6x+9y-33=0 -2 × (3x+2y-9=0) ⇒ -6x-4y+18=0

Now we will add these two equations: 6x+9y-33=0 -6x-4y+18=0

On adding both the above equations we get, ⇒ 5y-15=0 ⇒ 5y=15 ⇒ y=3

Substitute this in one of the given two equations and solve the resultant variable for x. 2x+3y-11=0 ⇒ 2x+3(3)-11=0 ⇒ 2x+9-11=0 ⇒ 2x=2 ⇒ x=1

Therefore, the solution of the given system of equations is x=1 and y=3.

The determinant of a 2 × 2 matrix is obtained by cross-multiplying elements starting from the top left corner and subtracting the products.

Determinant Method to Solve Linear Equations in Two Variables

Consider a system of linear equations in two variables: a 1 x + b 1 y = c 1 and a 2 x + b 2 y = c 2 . To solve them using the determinants method (which is also known as Crammer's Rule ), follow the steps given below:

Now, the solution of the given system of linear equations is obtained by the formulas:

Important Points on Linear Equations with Two Variables:

Tricks and Tips:

While solving the equations using either the substitution method or the elimination method:

☛Related Topics:

Linear Equations in Two Variables Examples

Example 1: The sum of the digits of a two-digit number is 8. When the digits are reversed, the number is increased by 18. Find the number.

Solution: Let us assume that x and y are the tens digit and the ones digit of the required number. Then the number is 10x+y.

And the number when the digits are reversed is 10y+x.

The question says, "The sum of the digits of a two-digit number is 8".

So from this, we get a linear equation in two variables: x+y=8.

⇒ y=8-x

Also, when the digits are reversed, the number is increased by 18.

So, the equation is 10y+x =10x+y+18

⇒ 10(8-x)+x =10x+(8-x)+18 (by substituting the value of y) ⇒ 80-10x+x =10x+8-x+18 ⇒ 80-9x=9x+26 ⇒ 18x = 54 ⇒ x=3

Substituting x=3 in y=8-x, we get, ⇒ y = 8-3 = 5 ⇒ 10x+y=10(3)+5 =35 Answer: The required number is 35.

Example 2: Jake's piggy bank has 11 coins (only quarters or dimes) that have a total value of $1.85. How many dimes and quarters does the piggy bank has?

Solution: Let us assume that the number of dimes be x and the number of quarters be y in the piggy bank. Let us form linear equations in two variables based on the given information.

Since there are 11 coins in total, x+y=11 ⇒ y=11-x. We know that, 1 dime = 10 cents and 1 quarter = 25 cents. The total value of the money in the piggy bank is $1.85 (185 cents).

Thus we get the equation 10x + 25y = 185 ⇒ 10x + 25(11-x) = 185 (as y = 11-x) ⇒ 10x + 275 - 25x =185 ⇒ -15x +275 =185 ⇒ -15x=-90 ⇒ x = 6

Substitute this value of x in x+y=11. ⇒ y=11-6=5

Answer: Therefore, the number of dimes is 6 and the number of quarters is 5.

Example 3: In a river, a boat can travel 30 miles upstream in 2 hours. The same boat can travel 51 miles downstream in 3 hours. Find,

Solution: Let us assume that:

During upstream, the current pulls back the boat's speed and the speed of the boat upstream = (x-y). During downstream, the current's speed adds to the boat's speed and the speed of the boat downstream = (x+y).

Using the last two columns of the table, we can form a pair of linear equations in two variables: x-y=15 x+y=17

Adding both equations we get: 2x = 32 ⇒ x=16

Substitute x=16 in x+y=17 16+y= 17 y=1

Answer: Therefore, the speed of the boat is 16 miles per hour and the speed of the current is 1 mile per hour.

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problem solving involving linear equation in two variables

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Practice Questions on Linear Equations in Two Variables

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FAQs on Linear Equations in Two Variables

What is meant by linear equation in two variables.

A linear equation is an equation with degree 1. A linear equation in two variables is a type of linear equation in which there are 2 variables present. For example, 2x - y = 45, x+y =35, a-b = 45 etc.

How do you Identify Linear Equations in Two Variables?

We can identify a linear equation in two variables if it can be expressed in the form ax+by+ c = 0, consisting of two variables x and y and the highest degree of the given equation is 1.

Can You Solve a Pair of Linear Equations in Two Variables?

Yes, we can solve pair of linear equations in two variables using different methods and ensure there are two equations present in the given system of equations so as to obtain the values of variables. If there is one solution it means that the given lines are intersecting, if there is no solution possible, then it means that the given equations are of parallel lines. If there are infinitely many solutions possible, it means that the given equations are forming coincidental lines.

How to Graphically Represent a Pair of Linear Equations in Two Variables?

We can represent linear equations in two variables graphically using the steps given below:

How Does One Solve the System of Linear Equations in Two Variables?

We have different methods to solve the system of linear equations:

Determinant or Matrix Method

How Many Solutions Does a Linear Equation with Two Variables Have?

Suppose we have a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0. The solutions of a linear equation with two variables are:

How is a Linear Inequality in Two Variables like a Linear Equation in Two Variables?

A linear inequality in two variables and a linear equation in two variables have the following things in common:

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Course: SAT > Unit 6

Systems of linear equations word problems | Lesson

What are systems of linear equations word problems, and how frequently do they appear on the test.

How do I solve systems of linear equations word problems?

Systems of equations examples, how do i write systems of linear equations.

Let's look at an example!

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IMAGES

Word Problem
Elimination Method of solving a Pair of linear equations in two variables/Class 10/CBSE
Word Problems based on linear equations in two variables: part II
Solving Systems of Linear Equations in Two Variables by Elimination (Addition)
Solving Problems Involving Linear Equations In Two Variables Worksheet
Solving a system of two linear equations with two variables no solution

VIDEO

Solving Linear Equations
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Substitution Involving Linear Equation
class-10।। Linear equation two variables ।।words problems ।।by vinod godiyal Sir
Methods to Solve Linear Equations
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COMMENTS

4.1 Solve Systems of Linear Equations with Two Variables
Solve a system of equations by elimination. Step 1. Write both equations in standard form. If any coefficients are fractions, clear them. Step 2. Make the coefficients of one variable opposites. Step 3. Add the equations resulting from Step 2 to eliminate one variable. Step 4. Solve for the ...
Solutions to 2-variable equations (video)
To isolate y, first subtract 2x from both sides (4y = -2x + 100), then divide by 4 (y = -1/2 x + 25). This is the more common usage because this is a linear function in slope intercept form - y in terms of x or y dependent on x. To solve for x, subtract 4y from both sides (2x = - 4y + 100), then divide by 2 (x = - 2y + 50).
4.1: Solve Systems of Linear Equations with Two Variables
An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations. {2x + y = 7 x − 2y = 6. A linear equation in two variables, such as 2x + y = 7, has an infinite number of solutions. Its graph is a line.
Systems of linear equations word problems
To solve a system of linear equations word problem: Select variables to represent the unknown quantities. Using the given information, write a system of two linear equations relating the two variables. Solve the system of linear equations using either substitution or elimination.