Answer:
−1, 2, −1)
Explanation:
Eliminate one variable.
The coefficients of z are opposites in the first and third equations.
x + 2y + z = 2
+(2x − 3y − z= −7)
3x − y − z = −5 Add equations 1 and 3.
Use equations 1 and 2 to create a second equation in x and y.
2(x + 2y + z = 2)→2x + 4y + 2z= 4 Multiply equation 1 by 2.
− (3x + y + 2z = −3)
−x + 3y + 2z = 7 Subtract.
Write the 2-by-2 system.
3x − y = −5
−x + 3y = 7
Eliminate y, and solve for x.
3(3x − y = −5) → 9x − 3y= −15Multiply the first equation in the 2-by-2 system by 3.
+ (−x + 3y = 7)
8x + 3y = −8 Add.
x = −1 Divide both sides by 8.
Use one of the equations in the 2-by-2 system to solve for y.
3x − y = −5
3(−1) − y = −5 Substitute −1 for x.
−3 − y = −5 Multiply.
−y = −2 Add 3 to both sides.
y = 2 Divide both sides by −1.
Substitute for x and y in one of the original equations to solve for z.
x + 2y + z = 2
(−1) + 2(2) + z = 2 Substitute −1 for x and 2 for y.
3 + z = 2 Simplify.
z = −1 Subtract 3 from both sides.
Therefore, the solution is (−1, 2, −1).