Algebra 1 · Unit 4
Use substitution, elimination, or a graph to solve a system of two linear equations.
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Substitute · Eliminate · Graph
Check in both equations
Substitution
Solve one equation for a variable, then replace that variable in the other equation. After you find one value, substitute back to find the other.
Elimination
Add or subtract the equations so one variable cancels. Multiply first if the coefficients are not opposites.
Graphing
Write both equations as y = mx + b. Graph both lines. The intersection is the solution. Check it in both original equations.
Always check
Plug the ordered pair into both original equations. Both must be true.
Substitution when y is already solved, elimination by adding, or graph two slope-intercept lines.
Substitute y = x + 3 into x + y = 11.
x + (x + 3) = 11
2x + 3 = 11
2x = 8
x = 4
y = 4 + 3 = 7
Check.
4 + 7 = 11
Answer: (4, 7)
2x + 5 = 13
2x = 8
x = 4
y = 5
Check.
2(4) + 5 = 13
Answer: (4, 5)
Add the equations.
2x = 12
x = 6
6 + y = 9
y = 3
Check.
6 − 3 = 3
Answer: (6, 3)
y = x has slope 1 and y-intercept (0, 0).
y = −x + 6 has slope −1 and y-intercept (0, 6).
The lines meet at (3, 3).
Check.
3 = 3 and 3 = −3 + 6
Answer: (3, 3)
Substitute x = 2y into x + 3y = 15.
2y + 3y = 15
5y = 15
y = 3
x = 2(3) = 6
Check.
6 + 3(3) = 15
Answer: (6, 3)
Add the equations.
3x = 12
x = 4
4 − y = 2
−y = −2
y = 2
Check.
2(4) + 2 = 10
Answer: (4, 2)
Solve one equation for a variable first, or multiply one equation before you add.
Add the equations.
3x = 9
x = 3
3 − y = 0
y = 3
Check.
2(3) + 3 = 9
Answer: (3, 3)
Subtract the second from the first.
2x = 6
x = 3
3 + 2y = 10
2y = 7
y = 72
Check.
3(3) + 272 = 9 + 7 = 16
Answer: (3, 72)
Solve the second for x.
x = y + 3
Substitute into 2x + 3y = 11.
2(y + 3) + 3y = 11
2y + 6 + 3y = 11
5y + 6 = 11
5y = 5
y = 1
x = 1 + 3 = 4
Check.
2(4) + 3(1) = 11
Answer: (4, 1)
Multiply the second equation by 3.
12x − 3y = 15
Add to 2x + 3y = 12.
14x = 27
x = 2714
42714 − y = 5
10814 − y = 5
547 − y = 357
− y = 357 − 547
− y = −197
y = 197
Check.
22714 + 3197 = 277 + 577 = 847 = 12
Answer: (2714, 197)
First line: slope 2, y-intercept (0, −1).
Second line: slope −1, y-intercept (0, 5).
The lines meet at (2, 3).
Check.
2(2) − 1 = 3 and −2 + 5 = 3
Answer: (2, 3)
Subtract the second from the first.
x = 5
2(5) + y = 9
10 + y = 9
y = −1
Check.
3(5) + (−1) = 15 − 1 = 14
Answer: (5, −1)
Multiply both equations, or expect a fractional solution. Graph after rewriting in slope-intercept form.
Multiply the first equation by 5 and the second by 3.
10x + 15y = 65
9x + 15y = 63
Subtract.
x = 2
2(2) + 3y = 13
4 + 3y = 13
3y = 9
y = 3
Check.
3(2) + 5(3) = 6 + 15 = 21
Answer: (2, 3)
Multiply the first equation by 2.
6x + 4y = 16
Subtract 2x + 4y = 9.
4x = 7
x = 74
274 + 4y = 9
72 + 4y = 9
4y = 9 − 72
4y = 182 − 72
4y = 112
y = 112 × 14
y = 118
Check.
374 + 2118 = 214 + 114 = 324 = 8
Answer: (74, 118)
Multiply the second equation by 3.
12x + 3y = 33
Add to 2x − 3y = 5.
14x = 38
x = 3814
x = 197
4197 + y = 11
767 + y = 777
y = 17
Check.
2197 − 317 = 387 − 37 = 357 = 5
Answer: (197, 17)
Rewrite x + y = 6 as y = −x + 6.
Graph y = −x + 6 and y = −2x + 3.
The lines meet at (−3, 9).
Check.
−3 + 9 = 6
−2(−3) + 3 = 6 + 3 = 9
Answer: (−3, 9)
Multiply the first equation by 2.
10x − 4y = 18
Add to 3x + 4y = 1.
13x = 19
x = 1913
31913 + 4y = 1
5713 + 4y = 1313
4y = 1313 − 5713
4y = −4413
y = −1113
Check.
51913 − 2−1113 = 9513 + 2213 = 11713 = 9
Answer: (1913, −1113)
Substitute y = 12x + 1 into 2x − 4y = −4.
2x − 412x − 4(1) = −4
2x − 2x − 4 = −4
−4 = −4
The equation is always true, so the lines are the same.
Infinitely many solutions. Every point on y = 12x + 1.
Answer: infinitely many solutions
Multiply the first equation by 6 to clear the fractions.
3x + 2y = 24
From x − y = 2, x = y + 2.
3(y + 2) + 2y = 24
3y + 6 + 2y = 24
5y = 18
y = 185
x = 185 + 2 = 285
Check.
285 − 185 = 2
Answer: (285, 185)
Multiply the first equation by 2: 4x + 2ky = 16.
Compare with 4x + 6y = 16.
If 2k = 6, then k = 3 and the equations are the same: infinitely many solutions.
For exactly one solution, k ≠ 3.
When k = 1: 2x + y = 8 and 4x + 6y = 16.
Multiply the first by 2: 4x + 2y = 16.
Subtract from 4x + 6y = 16.
4y = 0, so y = 0
2x = 8, so x = 4
The solution is (4, 0).
Answer: k ≠ 3; when k = 1, (4, 0)
2x + 2y = 10
x + y = 5
y = −x + 5
Set −x + 5 = −x + 8.
5 = 8
5 = 8 is false, so there is no intersection.
The lines are parallel: both have slope −1 and different intercepts 5 and 8.
Answer: no solution
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