Solving Systems
Of Equations by
Substitution
Algebra 1
Glencoe McGraw-Hill
Malinda Young
Substitution is an algebraic method of
solving systems of equations. It’s more
accurate than the graphing system. Any
small inaccuracies that happen during
graphing can make it impossible to
determine what the solution to a system
really is!
If one of the equations in a
system can be easily solved for
one of its variables, the
substitution method is easy to
do. Just follow these steps:
2. Substitute this
expression into the
other equation and
solve for the
OTHER variable.
3. Substitute this
value into the revised
first equation and
solve.
1. Solve one of
the equations
for one of its
variables.
4. Check the
solution pair in
each of the
original equations.
Use substitution to solve the system of equations.
x  4y
4x  y  75
Since x = 4y, substitute 4y for x in the 2nd equation.
4x – y = 75
4(4y) – y = 75
16y – y = 75
15y = 75
y =5
Solution:
(
, )
Substitute y = 5 in the first equation to find x.
x = 4y
x = 4(5)
x = 20
20
Check to verify that (20, 5) is a solution
for both equations.
x  4y
(20)  4(5)
20  20
4x  y  75
4(20)  (5)  75
80  5  75
75  75
(20, 5) is the solution for the system of
equations. You found it algebraically
instead of by graphing.
If the system had been graphed, where
would the two lines intersect?
(20, 5)
Use substitution to solve the system of equations.
4x  y  12
 2x  3y  14
Solve the first equation for y since the coefficient of y is 1.
4x + y = 12
y = -4x + 12
y = -8
-8
y = -4(5) + 12
Substitute -4x + 12 in the 2nd equation in place of y.
-2x – 3y = 14
-2x – 3(-4x + 12) = 14
-2x + 12x - 36 = 14
10x - 36 = 14
10x = 50
x=5
Solution:
( ,
)
Check to verify that (5, -8) is a solution
for both equations.
4x  y  12
4(5)  ( 8)  12
12  12
 2x  3y  14
 2(5)  3( 8)  14
 10  ( 24)  14
14  14
(5, -8) is the solution for the system of
equations. You found it algebraically
instead of by graphing.
If the system had been graphed, where
would the two lines intersect?
(5, -8)
Use the substitution method to solve: 2x  2y  3
x  4 y  1
Solve the 2nd
equation for x since
the coefficient is 1.
x – 4y = -1
x = 4y - 1
x = 4(½) - 1
2x + 2y = 3
2(4y – 1) + 2y = 3
8y – 2 + 2y = 3
10y – 2 = 3
10y = 5
1
y=
2
x= 1
Solution:
1
(1 , )
2
Use the substitution method to solve: 7 x  3y  43
Solve the 2nd equation for y
since the coefficient is 1.
2x – y = -10
-y = -2x - 10
y = 2x + 10
2x  y  10
y = 2(1) + 10
y = 12
7x + 3y = 43
7x + 3(2x + 10) = 43
7x + 6x + 30 = 43
13x + 30 = 43
13x = 13
x=1
Solution:
( 1 , 12 )
Use the substitution method to solve:
 x  2y  12
 3x  2y
x  6y  20
x  5y  17
(-4, 4)
(2, 3)
Don’t be overconfident!
Always check the apparent
solution by substituting it
back in to both original
equations. That’s the only
way to catch an error.
How about a system of
equations that has
no solution or infinitely
many solutions? How
does that work when
using the substitution
method?
Use the substitution method to solve:
2x  2y  8
x  y  2
2x + 2y = 8
2(-y – 2) + 2y = 8
-2y – 4 + 2y = 8
-4 = 8
If you solve a linear system
and the result is
a false statement
(such as -4 = 8), there is
no solution
for the system.
A system with no
solution...
Parallel lines!
Use the substitution method to solve:
3x  2y  3
2

3 y  1   2y  3
3

2y  3  2y  3
33
Why? The equations
represent the
same line!
3x  2y  3
2
x  y1
3
If you solve a linear
system and the
result is
a true statement
(such as 3 = 3), there
are infinitely many
solutions
for the system.