Question

solve the following system.\\\

$$\begin{cases}-y - 2z = 1\\\\2x - 4y + 5z = -10\\\\y = 3\\end{cases}$$

\\\\(x = \\) \\(\boxed{}\\)\\(y = \\) \\(\boxed{}\\)\\(z = \\) \\(\boxed{}\\)

Explanation:

Step1: Substitute \( y = 3 \) into the first equation

We have the first equation \(-y - 2z = 1\). Substitute \( y = 3 \) into it:
\(-3 - 2z = 1\)

Step2: Solve for \( z \) from the first equation

Add 3 to both sides of \(-3 - 2z = 1\):
\(-2z = 1 + 3\)
\(-2z = 4\)
Divide both sides by -2:
\(z = \frac{4}{-2} = -2\)

Step3: Substitute \( y = 3 \) and \( z = -2 \) into the second equation

The second equation is \(2x - 4y + 5z = -10\). Substitute \( y = 3 \) and \( z = -2 \):
\(2x - 4\times3 + 5\times(-2) = -10\)

Step4: Solve for \( x \) from the second equation

Simplify the left - hand side:
\(2x-12 - 10=-10\)
\(2x-22=-10\)
Add 22 to both sides:
\(2x=-10 + 22\)
\(2x = 12\)
Divide both sides by 2:
\(x=\frac{12}{2}=6\)

Answer:

\(x = 6\), \(y = 3\), \(z=-2\)