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Transcript
```Mr. Sims
Solve using the substitution method.
1. x + 3y = 0
Algebra 1
Sect. 7.4 A
Ch #7 Review for Test
x = -3y
-3y -3y
4x – 2y = -14
get one equation to
read x = or y =
4(-3y) – 2y = -14
-12y – 2y = -14
-14y = -14
y=1
x = -3y
x = -3(1)
x = -3
Mr. Sims
2. 2 8x + 3y = 15
3 5x – 2y = -10
multiply to cancel
16x + 6y = 30
15x – 6y = -30
31x = 0
x=0
8x + 3y = 15
8(0) + 3y = 15
3y = 15
y=5
Mr. Sims
Solve with the substitution method.
1
y  1 x  2
xy2
3
3. 3
-1/3x
-1/3x
2x  6y  12
get one equation to
read x = or y =
 1

2x  6  x  2   12
 3

2x – 2x + 12 = 12
0=0
infinite solutions
Mr. Sims
4.
2 7x – 3y = -1
4x + 6y = -16
multiply to cancel out
14x – 6y = -2
4x + 6y = -16
18x = -18
x = -1
4x + 6y = -16
4(-1) + 6y = -16
- 4 + 6y = -16
+4
+4
6y = -12
y = -2
Mr. Sims
Solve with the substitution method.
5.
4x – 3y = -2
4x + y = 4
-4x
-4x
y = - 4x + 4
4x – 3(- 4x + 4) = -2
4x + 12x – 12 = - 2
16x – 12 = -2
+12 +12
16x = 10
5
x
8
5
y 4
2
3
y
2
y = - 4x + 4
5
y  4   4
8
Mr. Sims
6.
2 - 4x – 5y = 7
3x + 10y = -24
-8x – 10y = 14
3x + 10y = -24
-5x = -10
x=2
3x + 10y = -24
3(2) + 10y = -24
6 + 10y = -24
-6
-6
10y = -30
y = -3
Mr. Sims
Algebra 1
Sect. 7.4 A
Assignment
Use substitution to solve the linear system.
1.) 2x – 5y = -12
- 4x + y = 6
2.) -x + 3y = 4
2x – 6y = -8
3.)
6x + y = 12
- 4x – 2y = 0
6.)
4x + 7y = -11
14x – 12y = -2
Use the addition method to solve the linear system.
4.) 8x – 4y = -11
6x + 2y = -2
5.)
1
3
x  y  2
4
4
2x – 6y = -16
Mr. Sims
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