Solve the system using any method. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, express [tex]$x$[/tex] and [tex]$y$[/tex] in terms of the real number [tex][tex]$a$[/tex][/tex].)

[tex]\[
\begin{array}{c}
\left\{
\begin{array}{l}
-x + 3y = 32 \\
7x + 3y = -8
\end{array}
\right. \\
(x, y) = (\square)
\end{array}
\][/tex]



Answer :

To solve the given system of equations:
[tex]\[ \begin{cases} -x + 3y = 32 \\ 7x + 3y = -8 \end{cases} \][/tex]

we can use the method of elimination to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex].

1. Eliminate variable [tex]\( y \)[/tex]:
To eliminate [tex]\( y \)[/tex], we subtract the first equation from the second equation. This gives us a system of equations with only [tex]\( x \)[/tex] as the variable. Let's write down both equations:

[tex]\[ \begin{cases} -x + 3y = 32 \quad \text{(Equation 1)} \\ 7x + 3y = -8 \quad \text{(Equation 2)} \end{cases} \][/tex]

Subtract Equation 1 from Equation 2:

[tex]\[ (7x + 3y) - (-x + 3y) = -8 - 32 \][/tex]

Simplify the resulting equation:

[tex]\[ 7x + 3y + x - 3y = -8 - 32 \][/tex]

[tex]\[ 8x = -40 \][/tex]

Solve for [tex]\( x \)[/tex]:

[tex]\[ x = -\frac{40}{8} = -5 \][/tex]

So, [tex]\( x = -5 \)[/tex].

2. Solve for [tex]\( y \)[/tex]:
Now that we have [tex]\( x = -5 \)[/tex], substitute this value into one of the original equations to solve for [tex]\( y \)[/tex]. Let's use Equation 1:

[tex]\[ -x + 3y = 32 \][/tex]

Substitute [tex]\( x = -5 \)[/tex]:

[tex]\[ -(-5) + 3y = 32 \][/tex]

Simplify:

[tex]\[ 5 + 3y = 32 \][/tex]

Isolate [tex]\( y \)[/tex]:

[tex]\[ 3y = 32 - 5 \][/tex]

[tex]\[ 3y = 27 \][/tex]

[tex]\[ y = \frac{27}{3} = 9 \][/tex]

So, [tex]\( y = 9 \)[/tex].

Therefore, the solution to the system of equations is:
[tex]\[ (x, y) = (-5, 9) \][/tex]