Name: ___________________________
Date: ___________________
Algebra 2
Solving Linear Systems: Substitution and Elimination
Solving Linear Systems by Substitution and Graphing
Remember:
A linear system consists of two-or-more linear equations involving two-or-more
variables.
A solution is a set of numerical values for the variables that, when plugged into the
equations, makes all of the equations true. For today’s lesson, the systems we solve will
have two equations involving x and y, so the solutions can be thought of as points (x, y).
I. Substitution Method
Step 1: Solve one of the equations for either x or y (your choice of which equation
and which variable — try to choose whichever solving is going to be easiest).
Step 2: Take the result of the solving, and substitute into the other equation.
Step 3: You should now have a one-variable equation. Solve for that variable.
Step 4: Use one of the original equations, put in the just-found number for one
variable, and solve for the other variable.
Step 5: Check your answer.
Example:
Solve the linear system:

x  y 1
2x  3y 12
Name: ___________________________
Date: ___________________
Algebra 2
Solving Linear Systems: Substitution and Elimination
II. Elimination Method
Step 1: Write both equations in the standard form Ax + By = C, which lines up the like
terms.
Step 2: Pick either x or y to eliminate. Think of a common multiple for the terms having
that variable. Multiply in one or both equations to get two equations that have that
common multiple.
Step 3: If the equations have opposite terms (such as 6x and –6x), add the equations.
OR: If the equations have identical terms (such as –6x and –6x), subtract the
equations.
Step 4: One of the variables should have been eliminated. Solve for the remaining
variable.
Step 5: Substitute the value you’ve found into one of the original equations. Solve for the other
variable.
Step 6: Write the solution. Check your answer.
Example:
Solve the linear system:

4x  3y  11
3x  2y  13
Name: ___________________________
Date: ___________________
Algebra 2
Solving Linear Systems: Substitution and Elimination
Practice Problems:
Directions: Solve each system of equations using the substitution method. Check your
solutions in both of the original equations.
1. 3x – y = –15, 2x + y = 0
Solve
Check
2. –2x + 4y = 6, 3x – y = 1
Solve
Check
3. x – 5y = 18, 2x – 3y = –13
Solve
4.
1
3
Solve
Check
x + y = 9, –2x + 2y = –6
Check
Name: ___________________________
Date: ___________________
5. –3x = 9,
Algebra 2
Solving Linear Systems: Substitution and Elimination
2x + y = –13
Solve
Check
6. A large family holds a potluck dinner. Each person attending brings either 1 dish or 2
dishes. In total, 14 people come to the dinner, and there are 23 dishes. How many
people brought 1 dish and how many people brought 2 dishes?
Variables
x=
y=
System
Solve
Check
Answer
7. A caterer is planning a party for 64 people. The customer has $150 to spend. A $39
pan of pasta feeds 14 people and a $12 sandwich tray feeds 6 people. How many pans
of pasta and how many sandwich trays should the caterer make?
Name: ___________________________
Date: ___________________
Algebra 2
Solving Linear Systems: Substitution and Elimination
8. Xena and Yvonne are sisters. Xena’s age and Yvonne’s age add up to 30.
Yvonne is four years older than Xena. How old are the sisters?
Directions: Solve each system of equations using the elimination method. Check your
solutions in both of the original equations.
9. 2x + 3y = 11
5x + 4y = 17
10. 5x + 4y = 22
7x – 6y = –4
11. 3x = 16 + 2y
5y = –2x –2
Name: ___________________________
Date: ___________________
12.
1
5
1
3
x  53 y  2
x  43 y  4
13. 0.2x + 0.6y = 0.4
0.3x – 0.7y = 0.6
14. y  9  32 x
y  4  14 x
Algebra 2
Solving Linear Systems: Substitution and Elimination
Name: ___________________________
Date: ___________________
Algebra 2
Solving Linear Systems: Substitution and Elimination
Systems that do not have solutions
Sometimes a linear system will not have any solutions. Here’s a system that has no
solutions:
x+ y= 4
2x + 2y = 10
What happens in our system solving methods, if the system actually has no solution?
Substitution Method:
Here’s what happens for the example problem.
First equation solved for y:
Substitute into second equation:
Distribute and simplify:
y=4–x
2x + 2(4– x) = 10
2x + 8 – 2x = 10
8 = 10
We end up with an always-false equation, which tells us that there’s no solution.
Elimination Method:
Here’s what happens for the example problem.
First equation multiplied by 2:
Second equation:
Subtract:
2x + 2y = 8
2x + 2y = 10
0x + 0y = 2
0 =2
We end up with an always-false equation, which tells us that there’s no solution.
In general, using either of the methods, the indicator that a system has no solutions is that
you reach an equation that is always false, such as a number equaling a different number.
Since the equation is false no matter what values x and y have, that means the system has
no solution.
You try it
15. a. Solve by substitution:
2x + 4y = 20
3x + 6y = 50
b. Solve by elimination:
2x + 4y = 20
3x + 6y = 50
Name: ___________________________
Date: ___________________
Algebra 2
Solving Linear Systems: Substitution and Elimination
Systems that have infinitely many solutions
Sometimes a linear system will have infinitely many solutions. Here’s such a system:
x+ y= 4
2x + 2y = 8
What happens in our system solving methods, if the system has infinitely many
solutions?
Substitution Method:
Here’s what happens for the example problem.
First equation solved for y:
Substitute into second equation:
Distribute and simplify:
y=4–x
2x + 2(4– x) = 8
2x + 8 – 2x = 8
8=8
We get an always-true equation, which tells that there are infinitely many
solutions.
Elimination Method:
Here’s what happens for the example problem.
First equation multiplied by 2:
Second equation:
Subtract:
2x + 2y = 8
2x + 2y = 8
0x + 0y = 0
0 =0
We get an always-true equation, which tells that there are infinitely many
solutions.
In general, using either of the methods, the indicator that a system has infinitely many
solutions is that you reach an equation that is always true, such as a number equaling the
same number. When that happens, any of the infinitely many (x, y) pairs that makes one
of the equations true also makes the other equation true.
You try it
16. a. Solve by substitution:
2x + 4y = 20
3x + 6y = 30
b. Solve by elimination:
2x + 4y = 20
3x + 6y = 30
Name: ___________________________
Date: ___________________
Algebra 2
Solving Linear Systems: Substitution and Elimination
More Problems:
Directions: Solve these systems using your choice of substitution or elimination.
17. 4x – 2y = 5
–8x + 4y = –10
18. –3x – y = 6
6x + 2y = 12
19.
2x + 5y = 20
–4x + 10y = 40