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Algebra 2
Section 10.2 B
Substitution Method
Solve using the substitution method.
1. 2x + y = 5
-2x
-2x
y = -2x + 5
substitute into other equation for y
solve one equation for one variable
get one to read x = or y =
4x + 3y = 7
4x + 3(-2x + 5) = 7
4x – 6x + 15 = 7
-2x + 15 = 7
solve for x
distributive property
combine like terms
-15 -15
-2x = -8
x=4
substitute into equation
that reads y =
y = -2x + 5
y = -2(4) + 5
y = -8 + 5
y = -3
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2. 4x + 2y = 2
solve one equation for one variable
get one to read x = or y =
x + 3y = 13
-3y
-3y
x = -3y + 13
substitute into other
equation for x
4(-3y + 13) + 2y = 2
-12y + 52 + 2y = 2
-10y + 52 = 2
solve for y
distributive property
combine like terms
-52 -52
-10y = -50
y=5
x = -3y + 13
x = -3(5) + 13
x = -15 + 13
x = -2
substitute into equation
that reads x =
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3. 6x – 2y = - 4
get one equation to
read x = or y =
-8x + y = -3
+8x
+8x
y = 8x - 3
substitute into other
equation for y
6x – 2(8x – 3) = - 4
solve for x
6x – 16x + 6 = - 4
-10x + 6 = - 4
-6
distributive property
-6
-10x = -10
x=1
substitute into equation
that reads y =
y = 8x - 3
y = 8(1) – 3
y=8–3
y=5
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4. x – 3y = 18
x = 3y + 18
substitute into other equation for x
+3y +3y get one equation to
-x + 3y = 12
read x = or y =
-(3y + 18) + 3y = 12
-3y – 18 + 3y = 12
-18 = 12
solve for y
distributive property
combine like terms
since –18 = 12 is never true,
there is no solution
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5. 2x  5y  1
get one equation to
read x = or y =
5
1
x y
2
2
-5/2y
-5/2y
5
1
x   y
2
2
substitute into other
equation for x
1
 5
2  y    5y  1
2
 2
-5y + 1 + 5y = 1
1=1
distributive property
combine like terms
since 1 =1 is always true, there are
infinitely many solutions
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6. x – 3y = 8
+3y +3y
x = 3y + 8
substitute into other equation for x
get one equation to
read x = or y =
-2x + 3y = -7
-2(3y + 8) + 3y = -7 solve for y
-6y – 16 + 3y = -7 distributive property
-3y – 16 = -7 combine like terms
+16
+16
-3y = 9
y = -3
substitute into equation
that reads x =
x = 3y + 8
x = 3(-3) + 8
x = -9 + 8
x = -1
Mr. Sims
Solve using the substitution method.
Algebra 2
Section 10.2 B
Substitution Method
1.) 2x – 5y = 9
-3x + y = -7
2.) -3x + 4y = 1
x – 2y = 1
3.) 6x + 2y = 11
4x + y = 6
4.) x – 2y = -1
5x – 7y = 4
5.) 4x + 3y = 3
2x + y = -3
6.) 10x – 16y = 17
x+y=3
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