Question

algebra 2: review: three-variable linear systems #3 solve the systems by using either the substitution method or the elimination method. 1) \\(\

$$\begin{cases}12x + 6y + 7z = -35 \\\\ 7x - 5y - 6z = 200 \\\\ x + y = -10\\end{cases}$$

\\) 2) \\(\

$$\begin{cases}-2x + y + z = -2 \\\\ 5x + 3y + 3z = 71 \\\\ 4x - 2y - 3z = 1\\end{cases}$$

\\)

Explanation:

System 1:

Step1: Isolate $y$ from third equation

$y = -10 - x$

Step2: Substitute $y$ into first equation

$12x + 6(-10 - x) + 7z = -35$
Simplify: $12x - 60 - 6x + 7z = -35 \implies 6x + 7z = 25$

Step3: Substitute $y$ into second equation

$7x - 5(-10 - x) - 6z = 200$
Simplify: $7x + 50 + 5x - 6z = 200 \implies 12x - 6z = 150 \implies 2x - z = 25$

Step4: Solve $2x - z = 25$ for $z$

$z = 2x - 25$

Step5: Substitute $z$ into $6x + 7z = 25$

$6x + 7(2x - 25) = 25$
Simplify: $6x + 14x - 175 = 25 \implies 20x = 200 \implies x = 10$

Step6: Find $y$ with $x=10$

$y = -10 - 10 = -20$

Step7: Find $z$ with $x=10$

$z = 2(10) - 25 = -5$

System 2:

Step1: Eliminate $z$ using eq2 and eq3

Add $5x + 3y + 3z = 71$ and $4x - 2y - 3z = 1$:
$9x + y = 72$

Step2: Solve for $y$

$y = 72 - 9x$

Step3: Eliminate $z$ using eq1 and eq2

Multiply $-2x + y + z = -2$ by 3: $-6x + 3y + 3z = -6$
Subtract from $5x + 3y + 3z = 71$:
$11x = 77 \implies x = 7$

Step4: Find $y$ with $x=7$

$y = 72 - 9(7) = 72 - 63 = 9$

Step5: Find $z$ using eq1

$-2(7) + 9 + z = -2 \implies -14 + 9 + z = -2 \implies z = 3$

Answer:

1. $x=10$, $y=-20$, $z=-5$
2. $x=7$, $y=9$, $z=3$