1. 5.3 Solving Systems of Linear Equations by Elimination

2. Solving Systems of Linear Equations by Elimination

3. Example 1: Solving a System of Linear Equations by Elimination
Solve the system of linear equations by elimination. 3x + 2y = 4 Equation 1 3x – 2y = -4 Equation 2 Step 1: Because the coefficients of the y terms are opposites, you do not need to multiply either equation by a constant. Step 2: Add the equations. 3x + 2y = 4 3x – 2y = -4 6x = 0

4. Step 3: Solve for x. 6x = 0 x = 0 Step 4: Substitute 0 in for x in one of the original equations and solve for y. 3(0) + 2y = 4 Equation 1 2y = 4 y = 2
3x + 2y = 4 Equation 1
3x – 2y = -4 Equation 2
The solution is (0,2).

5. You try! Solve the system of linear equations by elimination.
3x + 2y = Equation 1
-3x + 4y = Equation 2
Step 1: Because the coefficients of the x terms
are opposites, you do not need to multiply either
equation by a constant.
Step 2: Add the equations.
3x + 2y = 7
-3x + 4y = 5
6y = 12
Step 3: Solve for y.
6y = 12
y = 2
The solution is (1,2).
Step 4: Substitute 2 in for y in one
of the original equations and solve
for x.
3x + 2(2) = 7 Equation 1
3x + 4 = 7
3x = 3
x = 1

6. Example 2: Solving a System of Linear Equations by Elimination
Solve the system of linear equations by elimination. -10x + 3y = 1 Equation 1 -5x – 6y = 23 Equation 2 Step 1: Multiply equation 2 by -2 so that the x-terms are opposites. -10x + 3y = 1 -10x + 3y = 1 Equation 1 -5x – 6y = 23 10x + 12y = -46 Equation 2 Step 2: Add the equations. -10x + 3y = 1 10x + 12y = y = -45

7. Step 4: Substitute -3 in for y in one
Step 3: Solve for y.
15y = -45
y = -3
Step 4: Substitute -3 in for y in one
of the original equations and solve
for x.
-5x – 6(-3) = 23
-5x + 18 = 23
-5x = 5
x = -1
-10x + 3y = 1 Equation 1
-5x -6y = 23 Equation 2
The solution is (-1,-3).

8. You try! Solve the system of linear equations by elimination.
2) x – 3y = Equation 1
3x + y = Equation 2
Step 1: Multiply equation 2 by 3 so that the coefficients
on the y-terms are opposites.
Step 2: Add the equations.
x – 3y = 24
9x + 3y = 36
10x = 60
x – 3y = 24
9x + 3y = 36
Step 3: Solve for y.
10x = 60
x = 6
The solution is (6,-6).
Step 4: Substitute 6 in for x in one
of the original equations and solve
for y.
3(6) + y = Equation 2
18 + y = 12
y = -6

9. Example 3: Solving Real-Life Problems Modeling with MathmaticsMathematics
A business with two locations buys seven large delivery vans and five small delivery vans. Location A receives five large vans and two small vans for a total of $235,000. Location B receives two large vans and three small vans for a total of $160,000. What is the cost of each van? 5 • cost of large van + 2 • cost of small van = 235,000 2 • cost of large van + 3 • cost of small van = 160,000 g = cost of large van s = cost of small van 5g + 2s = 235,000 Equation 1 2g + 3s = 160,000 Equation 2

10. 5g + 2s = 235,000 Equation 1 2g + 3s = 160,000 Equation 2 Step 1: Multiply equation 1 by 3 and equation 2 by -2 to make the s- terms be opposites. 5g + 2s = 235,000 15g + 6s = 705,000 Equation 1 2g + 3s = 160,000 -4g – 6s = -320,000 Equation 2 Step 2: Add equations together 15g + 6s = 705,000 -4g – 6s = -320,000 11g = 385,000

11. Step 3: Solve for g.
11g = 385,000
g = 35,000
Step 4: Substitute 35,000 for g in one of the original equations and solve for s.
5g + 2s = 235, Equation 1
2g + 3s = 160, Equation 2
2(35,000) + 3s = 160,000
70, s = 160,000
3s = 90,000
s = 30,000
The cost of a large van is $35,000.
The cost of a small van is $30,000

12. You try!
How could you check your two solutions? Substitute your values back into both of the original equations to be sure they work!