Question

Pls help! Will mark Branliest!! 50 points for whoever answers!!!!!!

Answer 1

Answer:

(1, - 1, 3 )

Step-by-step explanation:

x - 2y + 2z = 9 → (1)

- x + 3y = - 4 → (2)

2x - 5y + 3z = 16 → (3)

Add (1) and (2) term by term to eliminate x

y + 2z = 5 → (4)

Multiply (2) by 2

- 2x + 6y = - 8 → (5)

Add (3) and (5) term by term to eliminate x

y + 3z = 8 → (6)

Subtract (6) from (4) term by term to eliminate y

- z = - 3 ( multiply both sides by - 1 )

z = 3

Substitute z = 3 into (4)

y + 2(3) = 5

y + 6 = 5 ( subtract 6 from both sides )

y = - 1

Substitute y = - 1, z = 3 into (1) and solve for x

x - 2(- 1) + 2(3) = 9

x + 2 + 6 = 9

x + 8 = 9 ( subtract 8 from both sides )

x = 1

solution is (1, - 1, 3 )

Answer 2

Answer: See below

Step-by-step explanation:

Isolate x for x-2y+2z=9

3x=9+2y

$x=\frac{9+2y}{3}$....(1)

Substitute (1) into the 2nd equation

$\begin{bmatrix}-\frac{9+2y}{3}+3y=-4\\ 2\cdot \frac{9+2y}{3}-5y+3z=16\end{bmatrix}$

$\frac{3\left(-9+7y\right)}{3}=3\left(-4\right)$

7y=-3

y=-3/7

Substitute y=-3/7

$\begin{bmatrix}3z+\frac{18-11\left(-\frac{3}{7}\right)}{3}=16\end{bmatrix}$

$\begin{bmatrix}3z+\frac{53}{7}=16\end{bmatrix}$

Isolate z by substituting it

$3z+\frac{53}{7}=16$

$\frac{3z}{3}=\frac{\frac{59}{7}}{3}$

$z=\frac{59}{21}$

For x =9+2y/3 substitute z=59/21 and y=-3/7

$x=\frac{9+2\left(-\frac{3}{7}\right)}{3}$

$x=\frac{19}{7}$

$x=\frac{19}{7},\:z=\frac{59}{21},\:y=-\frac{3}{7}$