Question

solve the following system of equations for all three variables.
$-2x + 3y - 4z = 8$
$5x - 3y + 5z = -8$
$7x - 3y + 3z = 8$

Explanation:

Step1: Eliminate y with Eq1+Eq2

Add equation 1 and equation 2 to cancel out $y$:
$$(-2x + 3y - 4z) + (5x - 3y + 5z) = 8 + (-8)$$
Simplify to get:
$$3x + z = 0 \quad \text{(Equation 4)}$$

Step2: Eliminate y with Eq3-Eq2

Subtract equation 2 from equation 3 to cancel out $y$:
$$(7x - 3y + 3z) - (5x - 3y + 5z) = 8 - (-8)$$
Simplify to get:
$$2x - 2z = 16$$
Divide by 2:
$$x - z = 8 \quad \text{(Equation 5)}$$

Step3: Solve for x with Eq4+Eq5

Add equation 4 and equation 5:
$$(3x + z) + (x - z) = 0 + 8$$
Simplify to get:
$$4x = 8$$
Solve for $x$:
$$x = \frac{8}{4} = 2$$

Step4: Solve for z using x=2

Substitute $x=2$ into Equation 5:
$$2 - z = 8$$
Solve for $z$:
$$-z = 8 - 2 = 6 \implies z = -6$$

Step5: Solve for y using x,z

Substitute $x=2$ and $z=-6$ into Equation 1:
$$-2(2) + 3y - 4(-6) = 8$$
Calculate and simplify:
$$-4 + 3y + 24 = 8$$
$$3y + 20 = 8$$
$$3y = 8 - 20 = -12$$
$$y = \frac{-12}{3} = -4$$

Answer:

$x=2$, $y=-4$, $z=-6$