- 4.1 Solve Systems of Linear Equations with Two Variables
- Introduction
- 1.1 Use the Language of Algebra
- 1.2 Integers
- 1.3 Fractions
- 1.4 Decimals
- 1.5 Properties of Real Numbers
- Key Concepts
- Review Exercises
- Practice Test
- 2.1 Use a General Strategy to Solve Linear Equations
- 2.2 Use a Problem Solving Strategy
- 2.3 Solve a Formula for a Specific Variable
- 2.4 Solve Mixture and Uniform Motion Applications
- 2.5 Solve Linear Inequalities
- 2.6 Solve Compound Inequalities
- 2.7 Solve Absolute Value Inequalities
- 3.1 Graph Linear Equations in Two Variables
- 3.2 Slope of a Line
- 3.3 Find the Equation of a Line
- 3.4 Graph Linear Inequalities in Two Variables
- 3.5 Relations and Functions
- 3.6 Graphs of Functions
- 4.2 Solve Applications with Systems of Equations
- 4.3 Solve Mixture Applications with Systems of Equations
- 4.4 Solve Systems of Equations with Three Variables
- 4.5 Solve Systems of Equations Using Matrices
- 4.6 Solve Systems of Equations Using Determinants
- 4.7 Graphing Systems of Linear Inequalities
- 5.1 Add and Subtract Polynomials
- 5.2 Properties of Exponents and Scientific Notation
- 5.3 Multiply Polynomials
- 5.4 Dividing Polynomials
- Introduction to Factoring
- 6.1 Greatest Common Factor and Factor by Grouping
- 6.2 Factor Trinomials
- 6.3 Factor Special Products
- 6.4 General Strategy for Factoring Polynomials
- 6.5 Polynomial Equations
- 7.1 Multiply and Divide Rational Expressions
- 7.2 Add and Subtract Rational Expressions
- 7.3 Simplify Complex Rational Expressions
- 7.4 Solve Rational Equations
- 7.5 Solve Applications with Rational Equations
- 7.6 Solve Rational Inequalities
- 8.1 Simplify Expressions with Roots
- 8.2 Simplify Radical Expressions
- 8.3 Simplify Rational Exponents
- 8.4 Add, Subtract, and Multiply Radical Expressions
- 8.5 Divide Radical Expressions
- 8.6 Solve Radical Equations
- 8.7 Use Radicals in Functions
- 8.8 Use the Complex Number System
- 9.1 Solve Quadratic Equations Using the Square Root Property
- 9.2 Solve Quadratic Equations by Completing the Square
- 9.3 Solve Quadratic Equations Using the Quadratic Formula
- 9.4 Solve Equations in Quadratic Form
- 9.5 Solve Applications of Quadratic Equations
- 9.6 Graph Quadratic Functions Using Properties
- 9.7 Graph Quadratic Functions Using Transformations
- 9.8 Solve Quadratic Inequalities
- 10.1 Finding Composite and Inverse Functions
- 10.2 Evaluate and Graph Exponential Functions
- 10.3 Evaluate and Graph Logarithmic Functions
- 10.4 Use the Properties of Logarithms
- 10.5 Solve Exponential and Logarithmic Equations
- 11.1 Distance and Midpoint Formulas; Circles
- 11.2 Parabolas
- 11.3 Ellipses
- 11.4 Hyperbolas
- 11.5 Solve Systems of Nonlinear Equations
- 12.1 Sequences
- 12.2 Arithmetic Sequences
- 12.3 Geometric Sequences and Series
- 12.4 Binomial Theorem

## Learning Objectives

By the end of this section, you will be able to:

- 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

## Be Prepared 4.1

Before you get started, take this readiness quiz.

## Be Prepared 4.2

## Be Prepared 4.3

Determine Whether an Ordered Pair is a Solution of a System of Equations

## System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations .

## Solutions of a System of Equations

## Example 4.1

ⓐ ( −2 , −1 ) ( −2 , −1 ) ⓑ ( −4 , −3 ) ( −4 , −3 )

ⓐ ( 1 , −3 ) ( 1 , −3 ) ⓑ ( 0 , 0 ) ( 0 , 0 )

ⓐ ( 2 , −2 ) ( 2 , −2 ) ⓑ ( −2 , 2 ) ( −2 , 2 )

Solve a System of Linear Equations by Graphing

## Example 4.2

How to solve a system of equations by graphing.

Solve the system by graphing { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

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

Solve the system by graphing: { − x + y = 1 3 x + 2 y = 12 . { − x + y = 1 3 x + 2 y = 12 .

The steps to use to solve a system of linear equations by graphing are shown here.

## Solve a system of linear equations by graphing.

- Step 1. Graph the first equation.
- Step 2. Graph the second equation on the same rectangular coordinate system.
- Step 3. Determine whether the lines intersect, are parallel, or are the same line.
- If the lines intersect, identify the point of intersection. This is the solution to the system.
- If the lines are parallel, the system has no solution.
- If the lines are the same, the system has an infinite number of solutions.
- Step 5. Check the solution in both equations.

## Example 4.3

Solve the system by graphing: { 3 x + y = − 1 2 x + y = 0 . { 3 x + y = − 1 2 x + y = 0 .

Solve the system by graphing: { − x + y = 1 2 x + y = 10 . { − x + y = 1 2 x + y = 10 .

Solve the system by graphing: { 2 x + y = 6 x + y = 1 . { 2 x + y = 6 x + y = 1 .

## Example 4.4

Solve the system by graphing: { y = 1 2 x − 3 x − 2 y = 4 . { y = 1 2 x − 3 x − 2 y = 4 .

Solve the system by graphing: { y = − 1 4 x + 2 x + 4 y = − 8 . { y = − 1 4 x + 2 x + 4 y = − 8 .

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

## Example 4.5

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

Solve the system by graphing: { y = − 3 x − 6 6 x + 2 y = − 12 . { y = − 3 x − 6 6 x + 2 y = − 12 .

## Try It 4.10

Solve the system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16 . { y = 1 2 x − 4 2 x − 4 y = 16 .

## Coincident Lines

Coincident lines have the same slope and same y- intercept.

## Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

Let’s sum this up by looking at the graphs of the three types of systems. See below and Table 4.1 .

## Example 4.6

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ We will compare the slopes and intercepts of the two lines.

ⓑ We will compare the slope and intercepts of the two lines.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

## Try It 4.11

## Try It 4.12

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

## Example 4.7

How to solve a system of equations by substitution.

Solve the system by substitution: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

## Try It 4.13

Solve the system by substitution: { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 .

## Try It 4.14

Solve the system by substitution: { 2 x + y = −1 4 x + 3 y = 3 . { 2 x + y = −1 4 x + 3 y = 3 .

## Solve a system of equations by substitution.

- Step 1. Solve one of the equations for either variable.
- Step 2. Substitute the expression from Step 1 into the other equation.
- Step 3. Solve the resulting equation.
- Step 4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
- Step 5. Write the solution as an ordered pair.
- Step 6. Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

## Example 4.8

Solve the system by substitution: { 4 x + 2 y = 4 6 x − y = 8 . { 4 x + 2 y = 4 6 x − y = 8 .

We need to solve one equation for one variable. We will solve the first equation for y .

## Try It 4.15

Solve the system by substitution: { x − 4 y = −4 − 3 x + 4 y = 0 . { x − 4 y = −4 − 3 x + 4 y = 0 .

## Try It 4.16

Solve the system by substitution: { 4 x − y = 0 2 x − 3 y = 5 . { 4 x − y = 0 2 x − 3 y = 5 .

Solve a System of Equations by Elimination

For any expressions a, b, c, and d .

Notice how that works when we add these two equations together:

The y ’s add to zero and we have one equation with one variable.

This time we don’t see a variable that can be immediately eliminated if we add the equations.

Then rewrite the system of equations.

## Example 4.9

How to solve a system of equations by elimination.

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

## 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

The steps are listed here for easy reference.

## Solve a system of equations by elimination.

- Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
- Decide which variable you will eliminate.
- Multiply one or both equations so that the coefficients of that variable are opposites.
- Step 3. Add the equations resulting from Step 2 to eliminate one variable.
- Step 4. Solve for the remaining variable.
- Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
- Step 6. Write the solution as an ordered pair.
- Step 7. Check that the ordered pair is a solution to both original equations.

## 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 .

## Try It 4.19

## 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 .

## Example 4.11

## Try It 4.21

## Try It 4.22

## 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 .

## Try It 4.23

## 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

## Example 4.13

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

## Try It 4.26

## Section 4.1 Exercises

{ 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 )

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

## Writing Exercises

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?

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Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

- Authors: Lynn Marecek, Andrea Honeycutt Mathis
- Publisher/website: OpenStax
- Book title: Intermediate Algebra 2e
- Publication date: May 6, 2020
- Location: Houston, Texas
- Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
- Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/4-1-solve-systems-of-linear-equations-with-two-variables

## Linear Equations in Two Variables

## What are Linear Equations in Two Variables?

## Forms of Linear Equations in Two Variables

## 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:

- Step 1 : To solve a system of two equations in two variables graphically , we graph each equation. To know how, click here or follow steps 2 and 3 below.
- Step 2 : To graph an equation manually, first convert it to the form y = mx+b by solving the equation for y.
- Step 3 : Start putting the values of x as 0, 1, 2, and so on and find the corresponding values of y, or vice-versa.
- Step 4 : Identify the point where both lines meet.
- Step 5 : The point of intersection is the solution of the given system.

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

Consistent and Inconsistent System of Linear Equations:

- If the system has a solution, then it is said to be consistent;
- otherwise, it is said to be inconsistent.

Independent and Dependent System of Linear Equations:

- If the system has a unique solution, then it is independent.
- If it has an infinite number of solutions, then it is dependent. It means that one variable depends on the other.

## Method of Substitution

- Step 1: Solve one of the equations for one variable.
- Step 2: Substitute this in the other equation to get an equation in terms of a single variable.
- Step 3: Solve it for the variable.
- Step 4: Substitute it in any of the equations to get the value of another variable.

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:

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.

From this equation, we get two equations:

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

## Method of Elimination

- Step 1: Arrange the equations in the standard form: ax+by+c=0 or ax+by=c.
- Step 2: Check if adding or subtracting the equations would result in the cancellation of a variable.
- Step 3: If not, multiply one or both equations by either the coefficient of x or y such that their addition or subtraction would result in the cancellation of any one of the variables.
- Step 4: Solve the resulting single variable equation.
- Step 5: Substitute it in any of the given equations to get the value of another variable.

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

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

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

- Step 1: We first find the determinant formed by the coefficients of x and y and label it Δ. Δ = \(\left|\begin{array}{ll}a_1 & b_1 \\a_2 & b_2\end{array}\right| = a_1 b_2 - a_2b_1\)
- Step 2: Then we find the determinant Δ x which is obtained by replacing the first column of Δ with constants. Δ x = \(\left|\begin{array}{ll}c_1 & b_1 \\c_2 & b_2\end{array}\right| = c_1 b_2 - c_2b_1\)
- Step 3: We then find the determinant Δ y which is obtained by replacing the second column of Δ with constants. Δ y = \(\left|\begin{array}{ll}a_1 & c_1 \\a_2 & c_2\end{array}\right| = a_1 c_2 - a_2c_1\)

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

Important Points on Linear Equations with Two Variables:

- A linear equation in two variables is of the form ax + by + c = 0, where x and y are variables; and a, b, and c are real numbers.
- A pair of linear equations are of the form a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 and its solution is a pair of values (x, y) that satisfy both equations.
- To solve linear equations in two variables, we must have at least two equations.
- A linear equation in two variables has infinitely many solutions.

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

- If we get an equation that is true (i.e., something like 0 = 0, -1 = -1, etc), then it means that the system has an infinite number of solutions.
- If we get an equation that is false (i.e., something like 0 = 2, 3 = -1, etc), then it means that the system has no solution.

- Solving Linear Equations Calculator
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## Linear Equations in Two Variables Examples

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.

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

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

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.

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

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

go to slide go to slide go to slide

## Practice Questions on Linear Equations in Two Variables

## FAQs on Linear Equations in Two Variables

What is meant by linear equation in two variables.

## How do you Identify Linear Equations in Two Variables?

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

## 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:

- Step 1: A system of two equations in two variables can be solved graphically by graphing each equation by converting it to the form y=mx+b by solving the equation for y.
- Step 2: The points where both lines meet are identified.
- Step 3: The point of intersection is the solution of the given pair of linear equations in two variables.

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

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

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

- One and unique if a 1 /a 2 ≠ b 1 /b 2
- None if a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2
- Infinitely many if a 1 /a 2 = b 1 /b 2 = c 1 /c 2

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

- The degree of a linear equation and linear inequality is always 1.
- Both of them can be solved graphically.
- The way to solve a linear inequality is the same as linear equations except that it is separated by an inequality symbol. But note that the inequality rules should be taken care of.

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## Course: SAT > Unit 6

- Solving linear equations and linear inequalities | Lesson
- Understanding linear relationships | Lesson
- Linear inequality word problems | Lesson
- Graphing linear equations | Lesson
- Systems of linear inequalities word problems | Lesson
- Solving systems of linear equations | Lesson

## 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.

- 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.

## Let's look at an example!

- (Choice A) c − 1 = b 4 b + 6 c = 31 \begin{aligned} c-1 &= b \\ \\ 4b+6c &=31 \end{aligned} c − 1 4 b + 6 c = b = 3 1 A c − 1 = b 4 b + 6 c = 31 \begin{aligned} c-1 &= b \\ \\ 4b+6c &=31 \end{aligned} c − 1 4 b + 6 c = b = 3 1
- (Choice B) c + 1 = b 4 b + 6 c = 31 \begin{aligned} c+1 &= b \\ \\ 4b+6c &=31 \end{aligned} c + 1 4 b + 6 c = b = 3 1 B c + 1 = b 4 b + 6 c = 31 \begin{aligned} c+1 &= b \\ \\ 4b+6c &=31 \end{aligned} c + 1 4 b + 6 c = b = 3 1
- (Choice C) c − 1 = b 6 b + 4 c = 31 \begin{aligned} c-1 &= b \\ \\ 6b+4c &=31 \end{aligned} c − 1 6 b + 4 c = b = 3 1 C c − 1 = b 6 b + 4 c = 31 \begin{aligned} c-1 &= b \\ \\ 6b+4c &=31 \end{aligned} c − 1 6 b + 4 c = b = 3 1
- (Choice D) c + 1 = b 6 b + 4 c = 31 \begin{aligned} c+1 &= b \\ \\ 6b+4c &=31 \end{aligned} c + 1 6 b + 4 c = b = 3 1 D c + 1 = b 6 b + 4 c = 31 \begin{aligned} c+1 &= b \\ \\ 6b+4c &=31 \end{aligned} c + 1 6 b + 4 c = b = 3 1
- (Choice A) 1 1 1 1 A 1 1 1 1
- (Choice B) 2 2 2 2 B 2 2 2 2
- (Choice C) 3 3 3 3 C 3 3 3 3
- (Choice D) 4 4 4 4 D 4 4 4 4
- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 pi 12\ \text{pi} 1 2 pi 12, space, start text, p, i, end text or 2 / 3 pi 2/3\ \text{pi} 2 / 3 pi 2, slash, 3, space, start text, p, i, end text

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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 ...

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).

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.

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.