Algebra 1 · Unit 4
Solve a system of three linear equations in two variables and verify the solution.
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Solve two · check the third
All three, or no common point
Three lines, one pair (x, y)
A solution of all three equations is a point that lies on every line. Substitute into all three to verify.
Find two, check the third
Solve any two equations, then plug the ordered pair into the third. If it works, all three meet at that point. If it fails, there is no common solution.
Pairs can still intersect
Each pair of lines may cross even when all three do not share a single point. The three pairwise intersections form a triangle.
Parallel among the three
If two of the lines are parallel and distinct, they never meet, so all three cannot meet at one point.
Solve two equations, then check the third. Graph only in the answer.
Solve y = x + 1 and y = −2x + 7.
x + 1 = −2x + 7
3x + 1 = 7
3x = 6
x = 2
y = 2 + 1 = 3
Check in x + y = 5.
2 + 3 = 5, true.
All three pass through (2, 3).
Answer: (2, 3)
Solve y = x and y = −1x + 4.
x = −1x + 4
2x = 4
x = 2, y = 2
Check y = 1: 2 = 1 is false.
The first two meet at (2, 2), which is not on y = 1.
No common solution.
Answer: no common solution
2(1) + 2 = 4, true.
−1(1) + 5 = 4, true.
1 + 4 = 5, true.
(1, 4) is on all three lines.
Answer: yes; (1, 4)
x = 2 and y = 3 give the candidate (2, 3).
Check y = x + 1: 3 = 2 + 1, true.
All three meet at (2, 3).
Answer: (2, 3)
The first two have the same slope 2 and different intercepts, so they are parallel.
Parallel distinct lines never meet.
No third line can pass through a common point of the first two, because that point does not exist.
No common solution.
Answer: no common solution
x + (x − 1) = 7
2x − 1 = 7
2x = 8
x = 4, y = 3
2(4) − 3 = 8 − 3 = 5, true.
All three meet at (4, 3).
Answer: (4, 3)
Rewrite first. Some triples form a triangle instead of one point.
x + y = 6 → y = −x + 6
Set −x + 6 = −2x + 3.
x + 6 = 3
x = −3
y = 3 + 6 = 9
Check 2x − y = 0.
2(−3) − 9 = −6 − 9 = −15
−15 ≠ 0.
The first two meet at (−3, 9), which is not on the third line.
No common solution.
Answer: no common solution
From x − y = 2, x = y + 2.
(y + 2) + 2y = 8
3y + 2 = 8
3y = 6
y = 2, x = 4
3(4) + 2 = 14, true.
All three meet at (4, 2).
Answer: (4, 2)
y = x and y = 4: (4, 4)
y = x and x = 1: (1, 1)
y = 4 and x = 1: (1, 4)
No single point lies on all three lines.
The system of all three has no solution.
Answer: (4, 4), (1, 1), (1, 4); no common solution
2(3) + (−1) = 6 − 1 = 5, true.
3 − 3(−1) = 3 + 3 = 6, true.
−1 = 3 − 4, true.
Yes. All three contain (3, −1).
Answer: yes
Plug (1, 5) into 2x + y = k.
2(1) + 5 = k
k = 7
If k ≠ 7, the third line misses (1, 5).
Answer: k = 7
The first two are the same line: x + y = 5.
A solution of all three must lie on x + y = 5 and on y = x + 2.
x + (x + 2) = 5
2x = 3
x = 32, y = 72
Check: 32 + 72 = 5.
Yes. The first two are the same line, and they meet the third at (32, 72).
Answer: (32, 72)
Parameters, three pairwise points, and a careful check.
The first two meet at (2, 3), from x + 1 = −2x + 7.
3x = 6, x = 2, y = 3.
That point must lie on x + y = k.
2 + 3 = k
k = 5
Answer: k = 5
x = 3 and y = −1x + 5 give y = −3 + 5 = 2.
The candidate is (3, 2).
It must lie on y = mx + 1.
2 = 3m + 1
m = 13
The point is (3, 2).
Answer: m = 13; (3, 2)
The first two: multiply 2x − y = 3 by 2 to get 4x − 2y = 6.
The second equation says 4x − 2y = 10.
6 = 10 is false, so the first two are parallel and distinct.
They never meet, so all three cannot meet.
The problem is the first pair, not the third line.
Answer: no common solution; first two are parallel
From 3x + y = 11, y = 11 − 3x.
x − 2(11 − 3x) = −1
x − 22 + 6x = −1
7x = 21
x = 3
y = 11 − 9 = 2
2(3) + 3(2) = 6 + 6 = 12, true.
All three meet at (3, 2).
Answer: (3, 2)
Concurrent means one point on all three lines.
If the pairwise intersections are three different points, each pair meets somewhere the third line does not.
Three distinct pairwise intersections cannot be the same point.
The lines are not concurrent.
Answer: no
Solve the first two.
Add: 3x = 9, x = 3
3 + y = 4, y = 1
The candidate is (3, 1).
3(3) + k(1) = 11
9 + k = 11
k = 2
Only k = 2 places the third line through (3, 1).
Answer: k = 2
y = 4 and y = x + 2 meet at (2, 4).
For that point to lie on y = mx + 2: 4 = 2m + 2, so m = 1.
When m = 1, y = mx + 2 is the same as y = x + 2, so the first two are the same line, and they meet y = 4 at (2, 4).
If m ≠ 1, y = x + 2 and y = mx + 2 are distinct lines through (0, 2). They only meet at (0, 2), and (0, 2) is not on y = 4.
The only concurrent case is m = 1, at (2, 4).
Answer: m = 1; (2, 4)
Example: y = −3
x = 2
y = −1x − 1, because −3 = −2 − 1.
Check: (2, −3) has y = −3, x = 2, and −1(2) − 1 = −3.
Any three lines through the same point work.
Answer: any three lines through (2, −3)
Right: any equation true at (5, 1), such as x + y = 6.
5 + 1 = 6.
Wrong: any equation false at (5, 1), such as x + y = 8.
5 + 1 = 6 ≠ 8.
You must check the third equation every time.
Answer: e.g. x + y = 6 works; x + y = 8 fails
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