Solve the following system of linear equations: x − 2y − 2z = 0, 2x + 3y + z = 1, 3x − y − 3z = 3.

Question

ScanSolve AI Tutor · Accepted Answer

Here's how to solve the system of linear equations:

The given system of equations is:

Eliminate $z$ from equations (1) and (2).
Multiply equation (2) by 2:
$2(2x + 3y + z) = 2(1)$ $4x + 6y + 2z = 2 \quad \text{(Equation 4)}$
Add equation (1) and equation (4):
$(x - 2y - 2z) + (4x + 6y + 2z) = 0 + 2$ $5x + 4y = 2 \quad \text{(Equation A)}$
Eliminate $z$ from equations (2) and (3).
Multiply equation (2) by 3:
$3(2x + 3y + z) = 3(1)$ $6x + 9y + 3z = 3 \quad \text{(Equation 5)}$
Add equation (3) and equation (5):
$(3x - y - 3z) + (6x + 9y + 3z) = 3 + 3$ $9x + 8y = 6 \quad \text{(Equation B)}$
Solve the system of two equations (A and B) for $x$ and $y$ .
Equation A: $5x + 4y = 2$
Equation B: $9x + 8y = 6$
Multiply equation (A) by 2:
$2(5x + 4y) = 2(2)$ $10x + 8y = 4 \quad \text{(Equation C)}$
Subtract equation (C) from equation (B):
$(9x + 8y) - (10x + 8y) = 6 - 4$ $-x = 2$ $x = -2$
Substitute the value of $x$ into equation (A) to find $y$ .
$5(-2) + 4y = 2$ $-10 + 4y = 2$ $4y = 2 + 10$ $4y = 12$ $y = \frac{12}{4}$ $y = 3$
Substitute the values of $x$ and $y$ into equation (1) to find $z$ .
$(-2) - 2(3) - 2z = 0$ $-2 - 6 - 2z = 0$ $-8 - 2z = 0$ $-2z = 8$ $z = \frac{8}{-2}$ $z = -4$
Verify the solution with the original equations.
For (1): $(-2) - 2(3) - 2(-4) = -2 - 6 + 8 = 0$ (Correct)
For (2): $2(-2) + 3(3) + (-4) = -4 + 9 - 4 = 1$ (Correct)
For (3): $3(-2) - (3) - 3(-4) = -6 - 3 + 12 = 3$ (Correct)

The solution is $x=-2$ , $y=3$ , and $z=-4$ .
$x=-2, y=3, z=-4$

Answer