Algebra 1 · Unit 4
Decide whether a system of two linear equations has one solution, no solution, or infinitely many solutions.
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Different slopes → one
Same slope, different intercept → none
Same line → infinitely many
One solution
The lines have different slopes, so they cross once. The ordered pair is the intersection.
No solution
The lines have the same slope and different y-intercepts. They are parallel and never meet. After elimination you get a false number sentence such as 0 = 5.
Infinitely many solutions
The two equations are the same line. Same slope and same intercept. After elimination you get 0 = 0.
Rewrite first
Put both equations in y = mx + b, or use elimination, before you decide.
Both equations are already y = mx + b. Compare slope and intercept.
Both slopes are 2.
The y-intercepts are 1 and −4, which are different.
Parallel distinct lines, so no solution.
Answer: no solution
Slopes 3 and −1 are different.
The lines cross once.
Answer: one solution
Divide 2y = 2x + 4 by 2.
y = x + 2
This is the same as the first equation.
Infinitely many solutions.
Answer: infinitely many solutions
The equations are identical.
Every point on the line is a solution.
Answer: infinitely many solutions
Two different horizontal lines.
Same slope 0, different intercepts.
No solution.
Answer: no solution
Slopes 12 and 2 are different.
One solution.
Answer: one solution
Rewrite into slope-intercept form, or eliminate and read 0 = 0 or 0 = 5.
Divide the first equation by 2.
x + y = 4
The second equation is x + y = 5.
Same slope, different totals: 4 ≠ 5.
No solution.
Answer: no solution
Multiply the first equation by 2.
6x − 2y = 8
This matches the second equation.
Infinitely many solutions.
Answer: infinitely many solutions
Solve the first for y.
2y = −4x + 10
y = −2x + 5
Compare with y = −2x + 1.
Same slope, different intercepts.
No solution.
Answer: no solution
Solve the second for y.
y = −2x + 5
From the first, x = 3y + 6, which is not a constant multiple of the second.
Slopes: first line y = 13x − 2.
13 ≠ −2, so one solution.
Answer: one solution
Divide the first equation by 5.
x + 2y = 3
Same as the second equation.
Infinitely many solutions.
Answer: infinitely many solutions
Multiply the first equation by 2.
4x + 2y = 14
The second says 4x + 2y = 9.
14 = 9 is false.
No solution.
Answer: no solution
Find a value of k, or classify after several steps.
Multiply the first equation by 2.
2x + 4y = 12
This must match 2x + 4y = k.
So k = 12.
If k ≠ 12 the lines are parallel and there is no solution.
Answer: k = 12
Multiply the first equation by 2.
6x − 2y = 8
For no solution, the left sides match but the right sides do not.
k ≠ 8
If k = 8 there are infinitely many solutions.
Answer: k ≠ 8
First: x + y = 5
Second: x + y = 5
Same equation twice.
Infinitely many solutions.
Answer: infinitely many solutions
Subtract: (k − 2)x = 0.
If k ≠ 2, then x = 0 and y = 4: exactly one solution.
If k = 2, the equations are the same: infinitely many solutions.
There is no k that gives no solution, because the intercepts already match.
Answer: every k ≠ 2
Multiply the first equation by 2.
x + 2y = 6
The second is x + 2y = 7.
6 = 7 is false.
No solution.
Answer: no solution
No solution means the same slope and different intercepts.
m = 3
The intercepts 2 and −1 are already different, so the lines are parallel when m = 3.
Answer: m = 3
Multiply the first equation by 2: 2x + 2ky = 10.
Compare with 2x + 6y = 10.
If 2k = 6, then k = 3 and the equations match: infinitely many solutions.
The first line has slope −1k when k ≠ 0.
Equal slopes with a different intercept cannot happen here: matching slopes force k = 3, which is the same line.
If k = 0, the first equation is x = 5, a vertical line, which meets 2x + 6y = 10 once.
(a) k = 3 (b) none (c) all other k
Answer: (a) k = 3 (b) none (c) all other k
If the plans cost the same for every hour, the two cost equations are the same line: infinitely many solutions.
If elimination produces 0 = 5, the plans never cost the same: no solution.
If they cross once, there is one break-even time.
Answer: same line: always equal; 0 = 5: never equal
The second equation is exactly twice the first, so it is the same line.
Infinitely many solutions: every (x, y) on ax + by = c.
Replace 2c with any other number, such as 2c + 1.
Then 2ax + 2by = 2c + 1 is parallel and distinct: no solution.
Answer: infinitely many; no solution if the second total is not 2c
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