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solve the following system. \\(\\begin{cases}-3y=-9 \\\\ 2x + 5y-2z=-3 …

Question

solve the following system.
\\(\

$$\begin{cases}-3y=-9 \\\\ 2x + 5y-2z=-3 \\\\ 3x + 5y=-3\\end{cases}$$

\\)
x =

y =

z =

Explanation:

Step1: Solve for y from first - equation

Divide both sides of $-3y=-9$ by $-3$.
$y = \frac{-9}{-3}=3$

Step2: Substitute y into third - equation

Substitute $y = 3$ into $3x + 5y=-3$. We get $3x+5\times3=-3$, which simplifies to $3x+15=-3$.
Subtract 15 from both sides: $3x=-3 - 15=-18$.
Then divide both sides by 3: $x=\frac{-18}{3}=-6$

Step3: Substitute x and y into second - equation

Substitute $x=-6$ and $y = 3$ into $2x + 5y-2z=-3$.
We have $2\times(-6)+5\times3-2z=-3$.
Simplify the left - hand side: $-12 + 15-2z=-3$, which is $3-2z=-3$.
Subtract 3 from both sides: $-2z=-3 - 3=-6$.
Divide both sides by $-2$: $z=\frac{-6}{-2}=3$

Answer:

$x=-6$
$y = 3$
$z = 3$