3.2 Solving Systems of
Equations Algebraically
Substitution Method
Elimination Method
Substitution Method
Here you replace one variable with an expression.
x + 4y = 26
x – 5y = - 10
Solve for a variable,
x = 26 – 4y
Replace “x” in the other equation
(26 – 4y) – 5y = -10
Solve for y
Solve for y
(26 – 4y) – 5y = -10
26 – 4y – 5y = - 10
Remove parentheses by
multiplying by 1
26 – 9y = - 10
Add like terms
-9y = - 36
y=4
Subtract 26 from both sides
Divide by - 9
Solve for x
x = 26 – 4y
x = 26 – 4(4)
Substitute for y
x = 26 – 16
x = 10
The order pair is (10, 4).
This is where the lines cross.
The Elimination Method
Here we add the equations together when
the coefficients are different signs.
x + 2y = 10
x+y=6
Here both lead coefficients are 1.
We can change the coefficient to – 1, by
multiplying by – 1.
x + 2y = 10
x+y=6
Multiply the bottom equation by – 1.
x + 2y = 10
-x-y =-6
y=4
When adding the
equations together,
x go to zero.
Find x by replace it back in either equation.
x + 2(4) = 10;
x + 8 = 10;
x=2
So the order pair (2, 4)
works in both equations.
2 + 2(4) = 10
2+4=6
We have two way to solve the systems,
Substitution and Elimination; which way is
better depends on the problem.
What about this problem
2x + 3y = 12
5x – 2y = 11
Here we have to multiply both equations
If we wanted to remove the “x”, then we
have to find the Least common multiple
(L.C.M.) of 2 and 5.
If we wanted to remove the “y”, then we
have to find the least common multiple of
3 and -2.
Lets get rid of the “y”
The L.C.M of 2 and 3 is 6. Since we want
the coefficients to be opposite, - 2 will help
in the equation.
we multiply the top equation by 2.
2x + 3y = 12
4x + 6y = 24
The bottom equation by 3
5x – 2y = 11
15x – 6y = 33
Add the new equations together
4x + 6y = 24
15x – 6y = 33
19x
= 57
Divide by 19
x=3
Replace in original equation and solve for y
2(3) + 3y = 12
6 + 3y = 12
3y = 6
y= 2
What about inconsistent systems?
y – x = 5 Multiply the top equation by – 2,
2y – 2x = 8
2y – 2x = -10
then add the bottom.
2y – 2x = 8
0=-2
This shows no solutions.
What if it is dependent (Many solutions)
1.6y = 0.4x + 1
0.4y = 0.1x + 0.25
Multiply the top and bottom equation by 100
to remove decimals.
160y = 40x + 100
40y = 10x + 25
Then multiply the bottom equation by -4
-160y = -40x – 100
Add the new equations together
160y = 40x + 100
-160y = -40x – 100
0=0
This is a system with many solutions.
Solve this system
a–b=2
-2a + 3b = 3
How about this system
y = 3x – 4
y=4+x
Homework
Page 120
# 13 – 35 odd
Homework
Page 120
# 14 – 34 even, 37