1. Solve Linear Systems By Multiplying First
February 4, 2014
Pages

2. 7.4 SOLVE SYSTEMS BY MULTIPLYING Students will be able to:
Solve special systems of linear equations in two variables.
Classify systems of linear equations and determine the number of solutions.

3. Solve the linear system by multiplying one equation, then add.
6x +5y = 19
Equation 1
2x +3y = 5
Equation 2
SOLUTION
STEP 1
Multiply: Equation 2 by –3 so that the coefficients of x are opposites.
6x + 5y = 19
6x + 5y = 19
2x + 3y = 5
–6x – 9y = –15
STEP 2
Add: the equations.
–4y = 4
STEP 3
Solve: for y.
y = –1

4. STEP 4
Substitute: –1 for y in either of the original equations and solve for x.
2x + 3y = 5
Write Equation 2.
2x + 3(–1) = 5
Substitute –1 for y.
2x + (–3) = 5
Multiply.
2x = 8
Add 3 to each side.
x = 4
Divide each side by 2.
ANSWER
The solution is (4, –1).

5. Solve the linear system by multiplying both equations, then subtract.
4x + 5y = 35
Equation 1
2y = 3x – 9
Equation 2
SOLUTION
STEP 1
Arrange: the equations so that like terms are in columns.
4x + 5y = 35
Write Equation 1.
–3x + 2y = –9
Rewrite Equation 2.

6. STEP 2
Multiply: Equation 1 by 2 and Equation 2 by 5 so that the coefficient of y in each equation is the least common multiple of 5 and 2, or 10.
4x + 5y = 35
8x + 10y = 70
–3x + 2y = –9
–15x +10y = –45
STEP 3
Multiply Equation 2 by - 1
15x – 10 y= 45
STEP 4
Solve: for x.
23x = 115
x = 5

7. STEP 5
Substitute: 5 for x in either of the original equations and solve for y.
4x + 5y = 35
Write Equation 1.
4(5) + 5y = 35
Substitute 5 for x.
y = 3
Solve for y.
ANSWER
The solution is (5, 3).

8. 3. Solve the linear system using elimination.
6x – 2y = 1
Equation 1
–2x + 3y = –5
Equation 2
SOLUTION
STEP 1
Multiply equation 2 by 3 so that the coefficient of x are opposites.
6x – 2y = 1
6x – 2y = 1
–2x + 3y = –5
× (3)
–6x + 9y = –15
STEP 2
Solve for y
7y = –14
y = –2

9. STEP 3
Substitute –2 for y in either of the original equations and solve for x.
6x – 2y = 1
Write Equation 1.
6x –2(–2) = 1
Substitute –2 for y.
6x –2(–2) = 1
Multiply.
6x = –3
Substitute 4 form each side.
x = –0.5
Divide each by 3.
ANSWER
The solution is (–0.5, –2).

10. 4. Solve the linear system using elimination.
2x + 5y = 3
Equation 1
3x + 10y = –3
Equation 2
SOLUTION
STEP 1
Multiply equation 1 by (–2) so that the coefficients of y are opposites.
2x + 5y = 3
4x – 10y = 6
× (–2)
3x + 10y = –3
3x + 10y = –3
STEP 2
Add the equation
–x = –9
STEP 3
Solve for x
x = 9

11. STEP 4
Substitute 9 for x in either of the original equations and solve for y.
2x + 5y = 3
Write Equation 1.
2(9) + 5y = 3
Substitute 9 for x.
18 + 5y = 3
Multiply.
5y = –15
Substitute 18 form each side.
y = –3
ANSWER
The solution is (9, –3).

12. 5. Solve the linear system using elimination.
3x – 7y = 5
Equation 1
9y = 5x +5
Equation 2
SOLUTION
STEP 1
Arrange the equation so that like terms are in columns.
3x – 7y = 5
Write Equation 1
–5x + 9y = 5
Rewrite Equation 2

13. STEP 2
Multiply equation 1 by 5 and equation 2 by 3 so that the coefficient of y in each equation is the least common multiple of 5 and 3.
15x – 35y = 25
× (5)
3x – 7y = 5
–5x + 9y = 5
15x + 27y = 15
× (3)
–8y = 40
STEP 3
Add the equations.
STEP 4
Solve for x.
y = –5

14. STEP 5
Substitute: –5 for x in either of the original equations and solve for x.
3x – 7y = 5
Write Equation 1.
3x – 7(– 5) = 5
Substitute 5 for x.
x = –10
Solve for y.
ANSWER
The solution is (–10, –5).

15. Pages 454-455, Problems #4-16, even, #22-26, even
HOMEWORK
Pages , Problems #4-16, even, #22-26, even