Question

solve the following system of equations for all three variables.
$x + 3y + 2z = 1$
$-x + 3y + 6z = 1$
$-x - 4y - 10z = -8$

Explanation:

Step1: Add Eq1 and Eq2

Add equation 1 ($x + 3y + 2z = 1$) and equation 2 ($-x + 3y + 6z = 1$) to eliminate $x$:
$$(x - x) + (3y + 3y) + (2z + 6z) = 1 + 1$$
$$6y + 8z = 2$$
Simplify by dividing by 2:
$$3y + 4z = 1 \quad \text{(Eq4)}$$

Step2: Subtract Eq2 from Eq3

Subtract equation 2 from equation 3 ($-x - 4y - 10z = -8$) to eliminate $x$:
$$(-x + x) + (-4y - 3y) + (-10z - 6z) = -8 - 1$$
$$-7y - 16z = -9 \quad \text{(Eq5)}$$

Step3: Solve Eq4 and Eq5 for $y,z$

Multiply Eq4 by 4:
$$12y + 16z = 4 \quad \text{(Eq6)}$$
Add Eq5 and Eq6:
$$(-7y + 12y) + (-16z + 16z) = -9 + 4$$
$$5y = -5$$
$$y = -1$$
Substitute $y=-1$ into Eq4:
$$3(-1) + 4z = 1$$
$$-3 + 4z = 1$$
$$4z = 4$$
$$z = 1$$

Step4: Substitute $y,z$ into Eq1

Substitute $y=-1, z=1$ into $x + 3y + 2z = 1$:
$$x + 3(-1) + 2(1) = 1$$
$$x - 3 + 2 = 1$$
$$x - 1 = 1$$
$$x = 2$$

Answer:

$x=2$, $y=-1$, $z=1$