Solve for the variable
1. 5x – 4 = 2x + 8
2. -4(x + 2) + 3x = 2
x = -2
2x + y = 1
4
2
Wait! It’s not in slopeintercept form!
y = -2x + 1
Solution:
(-2, 5)
.
.
.. .
.. .
. .
6
-6
-4
-2
2
4
-2
-4
-6
.
6
x = -2
2x + y = 1
Do you notice anything
about one of the
equations? 
Do you know the value of x?
Do you know the value of y?
x = -2
2(-2) + y = 1
 -4 + y = 1
+4
=+4
y=5
(-2 , 5)
x = -2
2x + y = 1
y = -2x + 1
4
2
-6
Solution:
(-2, 5)
.
.
.. .
.. .
. .
6
-4
-2
2
4
-2
-4
-6
.
6

Substitution





1) One equation will be ISOLATED (it will
have either x or y by itself) or can be
solved for x or y easily.
2) SUBSTITUTE the expression from Step 1
into the other equation and solve for
the other variable.
3) SUBSTITUTE the value from Step 2 into
the equation from Step 1 and solve.
4) Your SOLUTION is the ordered pair
formed by x & y.
5) CHECK the solution in each of the
original equations.
x = -4
3(-4) + 2y = 20
-12 + 2y = 20
+12
=+12
2y
__ = 32
__
2
2
y = 16
Why would
x = -4
substitution be a
3x + 2y = 20
good method to
use?
3x + 2y = 20
3x + 2(16) = 20
3x + 32 = 20
- 32 = -32
3x = -12
__
__
3
3
x = -4
(-4, 16)
x y
x = -4
3x + 2y = 20
(-4, 16)
3x + 2y = 20
3(-4) + 2(16) = 20
3(-4) + 2(16) = 20
We are
correct!
-12 + 32 = 20
20 = 20
y=x–1
x+y=3
y=x–1
x+x–1=3
2x – 1 = 3
+ 1 = +1
Why would
substitution be a
good method to
use?
y=x–1
x+y=3
y=2–1
2+y=3
y= 1
2x
__ = __
4
2
2
x =2
(2, 1 )
x y
y= 1
y=x-1
y=x–1
x+y=3
(2, 1)
x y
1=2-1
1=1
x+y=3
The values work
in both
equations, so we
are correct!
2+1=3
3=3
3x + 2y = -12
y=x-1
y=x-1
y=x-1
3x + 2(x – 1) = -12
y = -2 - 1
3x + 2x – 2 = -12
y = -3
Does it matter
which equation
we use to
substitute x?
5x – 2 = -12
+2 = +2
5x = ____
-10
___
5
5
x = -2
Why would we want to use
y = x – 1 instead of
3x + 2y = -12?
(-2 , -3)
x y
3x + 2y = -12
y=x–1
(-2, -3)
x y
3x + 2y = -12
3 (-2) + 2 (-3) = -12
-6 + -6 = -12
-12 = -12
The values work
in both
equations, so we
are correct!
y=x-1
-3 = -2 - 1
-3 = -3
x = 1/2y – 3
4x – y = 10
x = 1/2y – 3
4(1/2y – 3) – y = 10
2y – 12 – y = 10
y – 12 = 10
x = ½(22) – 3
x = 11– 3
x=8
+12 = +12
y = 22
(8, 22)
xy
Does it matter
which equation
we use to
substitute y?
x = 1/2y – 3 (8, 22)
4x – y = 10
x y
x=½y-3
8 = ½ (22) - 3
8 = 11 - 3
8=8
The values work
in both
equations, so we
are correct!
4x – y = 10
4(8) – 22 = 10
32 – 22 = 10
10 = 10
x = -5y + 4
3x + 15y = -1
x = -5y + 4
3(-5y + 4) + 15y = -1
-15y + 12 + 15y = -1
What is our
answer?
-15y + 12 + 15y = -1
12 = -1
Does 12 = -1?
No Solution
x = -5y + 4
No solution
3x + 15y = -1
There aren’t any
values to substitute
and check.
So, what do we do?
x = -5y + 4
3x + 15y = -1
x = -5y + 4
3(-5y + 4) + 15y = -1
-15y + 12 + 15y = -1
Anytime the
answer is No
Solution….
-15y + 12 + 15y = -1
12 = -1
Does 12 = -1?
Just go back and
check your work
and make sure
there is no solution.
2x – 5y = 29
x = -4y + 8
x = -4y + 8
2(-4y + 8) – 5y = 29
- 8y + 16 – 5y = 29
x = -4y + 8
x = -4(-1) + 8
x=4+8
- 13y + 16 = 29
x = 12
-16 = -16
-13y
13
___ = ___
-13 -13
y = -1
(12, -1)
x y
Does it matter
which equation
we use to
substitute y?
2x – 5y = 29
x = -4y + 8
x = -4y + 8
12 = -4 (-1) + 8
12 = 4 + 8
12 = 12
(12, -1)
x y
2x – 5y = 29
2 (12) – 5 (-1) = 29
24 – (-5) = 29
24 + 5 = 29
29 = 29
The values work
in both
equations, so we
are correct!
1) One equation will be ISOLATED (it will
have either x or y by itself) or can be
solved for x or y easily.
2) SUBSTITUTE the expression from Step 1
into the other equation and solve for
the other variable.
3) SUBSTITUTE the value from Step 2 into
the equation from Step 1 and solve.
4) Your SOLUTION is the ordered pair
formed by x & y.
5) CHECK the solution in each of the
original equations.