Systems of Linear Equations A system of linear equations consists of two or more linear equations. We will focus on only two equations at a time. The solution.

Systems of Linear Equations

A system of linear equations consists of two or more linear equations. We will focus on only two equations at a time.

The solution of a system of linear equations in two variables is any ordered pair that solves both of the linear equations. The solution to the system is the point that satisfies ALL of the equations. This point will be an ordered pair.

When graphing, you will encounter three possibilities.

NUMBER OF SOLUTIONS OF A LINEAR SYSTEM

IDENTIFYING THE NUMBER OF SOLUTIONS

y

x

y

x

Lines intersectone solution

Lines are parallelno solution

y

x

Lines coincideinfinitely many solutions

Parallel Lines

These lines never intersect!

Since the lines never cross, there is NO SOLUTION!

Parallel lines have the same slope with different y-intercepts.

2Slope = = 21

y-intercept = 2y-intercept = -1

Same Lines

These lines are the same!

Since the lines are on top of each other, there are INFINITELY MANY SOLUTIONS!

Coinciding lines have the same slope and y-intercepts.

2Slope = = 21

y-intercept = -1

What is the solution of the system graphed below?

1. (2, -2)

2. (-2, 2)

3. No solution

4. Infinitely many solutions

Name the Solution

Solving a system of equations by graphing.Let's summarize! There are 3 steps to solving a system

using a graph.

Step 1: Graph both equations.

Step 2: Do the graphs intersect?

Step 3: Check your solution.

Graph using slope and y – intercept or x- and y-intercepts. Be sure to use a ruler and graph paper!

This is the solution! LABEL the solution!

Substitute the x and y values into both equations to verify the point is a solution to both equations.

Determine whether the given point is a solution of the following system.

point: (– 3, 1)system: x – y = – 4 and 2x + 10y = 4

•Plug the values into the equations.First equation: – 3 – 1 = – 4 trueSecond equation: 2(– 3) + 10(1) = – 6 + 10 = 4 true

•Since the point (– 3, 1) produces a true statement in both equations, it is a solution.

Solution of a SystemExample

1) Find the solution to the following system:

2x + y = 4

x - y = 2

Graph both equations. I will graph using x- and y-intercepts (plug in zeros).

Graph the ordered pairs.

2x + y = 4(0, 4) and (2, 0)

x – y = 2(0, -2) and (2, 0)

Graph the equations.

2x + y = 4

(0, 4) and (2, 0)

x - y = 2(0, -2) and (2, 0)

Where do the lines intersect?(2, 0)

2x + y = 4

x – y = 2

Check your answer!

To check your answer, plug the point back into both equations.

2x + y = 4

2(2) + (0) = 4

x - y = 2(2) – (0) = 2 Nice job…let’s try another!

2) Find the solution to the following system:

y = 2x – 3

-2x + y = 1

Graph both equations. Put both equations in slope-intercept or standard form. I’ll do slope-intercept form on this one!

y = 2x – 3

y = 2x + 1

Graph using slope and y-intercept

Graph the equations.y = 2x – 3

m = 2 and b = -3

y = 2x + 1

m = 2 and b = 1

Where do the lines intersect?No solution!

Notice that the slopes are the same with different y-intercepts. If you recognize this early, you don’t

have to graph them!

Practice – Solving by Graphing

y – x = 1 (0,1) and (-1,0)

y + x = 3 (0,3) and (3,0)

Solution is probably (1,2) …

Check it:

2 – 1 = 1 true

2 + 1 = 3 true

therefore, (1,2) is the solution

(1,2)

Practice – Solving by Graphing

Inconsistent: no solutions

y = -3x + 5 (0,5) and (3,-4)

y = -3x – 2 (0,-2) and (-2,4)

They look parallel: No solution

Check it:

m1 = m2 = -3

Slopes are equal

therefore it’s an inconsistent system

Consistent: infinite sol’s

3y – 2x = 6 (0,2) and (-3,0)

-12y + 8x = -24 (0,2) and (-3,0)

Looks like a dependant system …

Check it:

divide all terms in the 2nd equation by -4

and it becomes identical to the 1st equation

therefore, consistent, dependant system

(1,2)

Graph the system of equations. Determine whether the system has one solution, no solution, or infinitely many solutions. If the system has one solution, determine the solution.

1 3 33 9 9

.

x yx y

2 35

4

5 3

. y x

y x

3 32 6

. x yx y

Solving a System of Linear Equations by the Substitution Method

1) Solve one of the equations for a variable.2) Substitute the expression from step 1 into the

other equation.3) Solve the new equation.4) Substitute the value found in step 3 into

either equation containing both variables.5) Check the proposed solution in the original

equations.

The Substitution Method

Solve the following system using the substitution method.3x – y = 6 and – 4x + 2y = –8

Solving the first equation for y, 3x – y = 6

–y = –3x + 6 (subtract 3x from both sides) y = 3x – 6 (multiply both sides by – 1)

Substitute this value for y in the second equation. –4x + 2y = –8 –4x + 2(3x – 6) = –8 (replace y with result from first equation) –4x + 6x – 12 = –8 (use the distributive property) 2x – 12 = –8 (simplify the left side) 2x = 4 (add 12 to both sides) x = 2 (divide both sides by 2)

The Substitution Method

Continued.

Substitute x = 2 into the first equation solved for y.y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0

Our computations have produced the point (2, 0).Check the point in the original equations.

First equation, 3x – y = 6 3(2) – 0 = 6 trueSecond equation, –4x + 2y = –8 –4(2) + 2(0) = –8 true

The solution of the system is (2, 0).

The Substitution Method

Solving a System of Linear Equations by the Addition or Elimination Method

1) Rewrite each equation in standard form, eliminating fraction coefficients.

2) If necessary, multiply one or both equations by a number so that the coefficients of a chosen variable are opposites.

3) Add the equations.4) Find the value of one variable by solving equation

from step 3.5) Find the value of the second variable by substituting

the value found in step 4 into either original equation.6) Check the proposed solution in the original equations.

The Elimination Method

Solve the following system of equations using the elimination method.

6x – 3y = –3 and 4x + 5y = –9

Multiply both sides of the first equation by 5 and the second equation by 3.

First equation,5(6x – 3y) = 5(–3) 30x – 15y = –15 (use the distributive property)Second equation,3(4x + 5y) = 3(–9) 12x + 15y = –27 (use the distributive property)

The Elimination Method

Continued.

Combine the two resulting equations (eliminating the variable y).

30x – 15y = –15 12x + 15y = –27 42x = –42 x = –1 (divide both sides by 42)

The Elimination Method

Continued.

Substitute the value for x into one of the original equations.

6x – 3y = –3 6(–1) – 3y = –3 (replace the x value in the first equation) –6 – 3y = –3 (simplify the left side) –3y = –3 + 6 = 3 (add 6 to both sides and simplify) y = –1 (divide both sides by –3)

Our computations have produced the point (–1, –1).

The Elimination Method

Continued.

Check the point in the original equations.First equation,

6x – 3y = –3 6(–1) – 3(–1) = –3 true

Second equation, 4x + 5y = –94(–1) + 5(–1) = –9 true

The solution of the system is (–1, –1).

The Elimination Method