The document provides 12 systems of 3 equations each that can be solved by substitution. For each system, the steps to solve by substitution are shown, along with the unique solution. The solutions are provided in the form (x, y, z) where x, y, z are the values of the variables that satisfy all 3 equations simultaneously. System 8 is noted as having no unique solution.

1. Kuta Software - Infinite Algebra 2 Name___________________________________
Solving Systems of Three Equations w/ Substitution Date________________ Period____
Solve each system by substitution.
1) x = 4 y − 11 2) x = −4 y + 4z + 4
−3 x + 4z = −7 z = 5 x − 25
y = −5 x + 2z + 25 −2 x − 5z = 17
3) z = −4 x + 4 y + 13 4) z = −2 x + 5 y + 24
x + 2 y − 2z = 10 x = 3 y − 3z + 21
x = 2z + 10 5 y − 3z = −24
5) −5 x − 3 y + z = −4 6) −4 x + 2z = 14
−2 x − 2 y + 2z = 4 y = x + z + 12
z=x+5 −2 x − 4z = 22
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2. 7) 3 x − 3 y = −6 8) x = −5 y + 4z + 1
z = −3 x − 3 y + 9 x − 2 y + 3z = 1
−4 x + 5 y + z = 8 2x + 3 y − z = 2
9) −2r − 2s + 2t = −4 10) −6 x + y − 4z = 3
−4r + 2t = −16 5 x − 3 y = −8
r + s + 6t = −12 − x − 5 y = −4
11) −4 y + 5z = −19 12) −4r + 2s = 12
5 x − 5 y − 6z = 8 4r − 4s − 2t = −4
2 x + z = −5 −4r + 3t = −10
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3. Kuta Software - Infinite Algebra 2 Name___________________________________
Solving Systems of Three Equations w/ Substitution Date________________ Period____
Solve each system by substitution.
1) x = 4 y − 11 2) x = −4 y + 4z + 4
−3 x + 4z = −7 z = 5 x − 25
y = −5 x + 2z + 25 −2 x − 5z = 17
(5, 4, 2) (4, −5, −5)
3) z = −4 x + 4 y + 13 4) z = −2 x + 5 y + 24
x + 2 y − 2z = 10 x = 3 y − 3z + 21
x = 2z + 10 5 y − 3z = −24
(4, 0, −3) (3, −3, 3)
5) −5 x − 3 y + z = −4 6) −4 x + 2z = 14
−2 x − 2 y + 2z = 4 y = x + z + 12
z=x+5 −2 x − 4z = 22
(0, 3, 5) (−5, 4, −3)
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4. 7) 3 x − 3 y = −6 8) x = −5 y + 4z + 1
z = −3 x − 3 y + 9 x − 2 y + 3z = 1
−4 x + 5 y + z = 8 2x + 3 y − z = 2
(1, 3, −3) No unique solution
9) −2r − 2s + 2t = −4 10) −6 x + y − 4z = 3
−4r + 2t = −16 5 x − 3 y = −8
r + s + 6t = −12 − x − 5 y = −4
(3, −3, −2) (−1, 1, 1)
11) −4 y + 5z = −19 12) −4r + 2s = 12
5 x − 5 y − 6z = 8 4r − 4s − 2t = −4
2 x + z = −5 −4r + 3t = −10
(−1, 1, −3) (−2, 2, −6)
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