Algebra 1 · Unit 4
Solve a system of three linear equations in three variables using reasoning and algebra, then verify.
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Three letters · three equations
Eliminate · substitute · check all three
Three letters, three equations
A solution is an ordered triple (x, y, z) that makes every equation true. You need three independent equations for a unique triple.
Reasoning first
If one equation is already solved, or two equations share a pair such as x + y, subtract to find the leftover variable.
Elimination still works
Add or subtract two equations to cancel one variable. Repeat until one letter is left, then substitute back.
Verify in all three
Plug x, y, and z into every original equation. All three must be true.
One variable is already known, or two equations share a sum you can subtract.
y = 2 and z = 3 are given.
Substitute into x + y + z = 8.
x + 2 + 3 = 8
x + 5 = 8
x = 3
Check.
3 + 2 + 3 = 8
Answer: (3, 2, 3)
x = 2
2 + y = 6, so y = 4
2 + 4 + z = 10
z = 4
Check.
2 + 4 + 4 = 10
Answer: (2, 4, 4)
z = 1
y + 1 = 5, so y = 4
x + 4 + 1 = 9
x = 4
Check.
4 + 4 + 1 = 9
Answer: (4, 4, 1)
2x + 1 + 2 = 9
2x + 3 = 9
2x = 6
x = 3
Check.
2(3) + 1 + 2 = 9
Answer: (3, 1, 2)
z = 4
y + 4 = 7, so y = 3
x + 3 = 5, so x = 2
Check.
3 + 4 = 7 and 2 + 3 = 5
Answer: (2, 3, 4)
5 + (−1) + z = 12
4 + z = 12
z = 8
Check.
5 − 1 + 8 = 12
Answer: (5, −1, 8)
Subtract shared sums, or eliminate one variable from a pair of equations.
From x + y + z = 9 and x + y = 5, subtract.
z = 4
y + 4 = 6, so y = 2
x + 2 = 5, so x = 3
Check.
3 + 2 + 4 = 9
Answer: (3, 2, 4)
z = 1, so x + y + 1 = 6
x + y = 5
2x + y = 7
Subtract: x = 2
2 + y = 5, so y = 3
Check.
2(2) + 3 = 7 and 2 + 3 + 1 = 6
Answer: (2, 3, 1)
z = 2, so x − y + 2 = 4
x − y = 2
x + y = 6
Add: 2x = 8, x = 4
4 + y = 6, y = 2
Check.
4 − 2 + 2 = 4
Answer: (4, 2, 2)
2x + 2 + z = 11
2x + z = 9
x − 2 + z = 3
x + z = 5
Subtract: (2x + z) − (x + z) = 9 − 5
x = 4
4 + z = 5, z = 1
Check.
2(4) + 2 + 1 = 11
Answer: (4, 2, 1)
From x + y + z = 0 and x + y = 3, z = −3.
x − (−3) = 5
x + 3 = 5
x = 2
2 + y = 3, y = 1
Check.
2 + 1 + (−3) = 0
Answer: (2, 1, −3)
x + 2(3) + z = 10
x + z + 6 = 10
x + z = 4
This matches the second equation.
y = 3 and x + z = 4.
Infinitely many solutions: every triple with y = 3 and z = 4 − x.
Example: (1, 3, 3), because 1 + 6 + 3 = 10 and 1 + 3 = 4.
Answer: infinitely many; y = 3 and x + z = 4
Eliminate twice. Watch for no solution or infinitely many.
Add the first and the third.
2x + 3y = 13
Add the second and the third.
3x + y = 9
y = 9 − 3x
2x + 3(9 − 3x) = 13
2x + 27 − 9x = 13
−7x = −14
x = 2
y = 9 − 6 = 3
2 + 3 + z = 6
z = 1
Check the second equation.
2(2) − 3 + 1 = 4 − 3 + 1 = 2
Answer: (2, 3, 1)
Add the first two.
2x + 2z = 6
x + z = 3
z = 3 − x
Subtract the second from the first.
2y = −2
y = −1
2x + (−1) − (3 − x) = −1
2x − 1 − 3 + x = −1
3x − 4 = −1
3x = 3
x = 1
z = 2
Check.
1 + (−1) + 2 = 2
Answer: (1, −1, 2)
The first two say 4 = 5, which is false.
No values of x, y, z can make both true.
No solution.
Answer: no solution
Add the first two.
2x + 2z = 4
x + z = 2
Subtract the second from the first.
2y = 0, so y = 0
2x + 0 − z = 7
2x − z = 7
From x + z = 2, z = 2 − x
2x − (2 − x) = 7
2x − 2 + x = 7
3x = 9
x = 3
z = −1
Check.
3 + 0 + (−1) = 2
2(3) + 0 − (−1) = 6 + 1 = 7
Answer: (3, 0, −1)
The second equation is twice the first, so it is the same fact.
x = 2 and y + z = 4.
Infinitely many solutions: (2, y, 4 − y) for any y.
Check an example: (2, 1, 3) gives 2 + 1 + 3 = 6.
Answer: infinitely many; x = 2 and y + z = 4
Add the second and the third.
3x + 2z = 10
Add the first and the third.
3x + y − z + 2x − y + z = 5 + 3
5x = 8
x = 85
y = 6 − 285 = 305 − 165 = 145
From x + y + z = 7,
85 + 145 + z = 7
225 + z = 355
z = 135
Check 2x − y + z.
285 − 145 + 135 = 165 − 145 + 135 = 155 = 3
Answer: (85, 145, 135)
The second equation is twice the first when k = 12.
If k = 12, you have x = 1 and y + z = 5: infinitely many triples.
If k ≠ 12, the first two contradict: no solution.
The x terms never isolate a unique (x, y, z) because the first two equations only constrain x + y + z.
It never has exactly one solution.
Answer: infinitely many when k = 12; none otherwise; never exactly one
From x + y + z = 10 and x + y = 7, the leftover is z = 3.
From y + z = 8, y + 3 = 8, so y = 5.
From x + y = 7, x + 5 = 7, so x = 2.
Check: 2 + 5 + 3 = 10.
Answer: (2, 5, 3)
x + y + z = 9
z = 2
Subtract the second from the first.
x + y = 7
Subtraction property of equality: subtract equal amounts from both sides, here the two sides of z = 2.
Any triple with z = 2 and x + y = 7 works, such as (3, 4, 2).
Answer: x + y = 7; subtraction property of equality
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