Sure, let's solve the given system of linear equations step by step using the method of elimination or substitution.
We are given the system:
[tex]\[
\begin{cases}
3x + 2y = -1 \\
2x - 5y = 31
\end{cases}
\][/tex]
### Step 1: Eliminate one of the variables
We can eliminate [tex]\( y \)[/tex] by finding common coefficients for [tex]\( y \)[/tex] in both equations, and then subtracting or adding the equations.
Let's multiply the first equation by 5 and the second equation by 2 so that the coefficients of [tex]\( y \)[/tex] in both equations are opposites:
[tex]\[
\begin{cases}
5(3x + 2y) = 5(-1) \\
2(2x - 5y) = 2(31)
\end{cases}
\][/tex]
This simplifies to:
[tex]\[
\begin{cases}
15x + 10y = -5 \\
4x - 10y = 62
\end{cases}
\][/tex]
### Step 2: Add the equations to eliminate [tex]\( y \)[/tex]
Now we add the two new equations together:
[tex]\[
(15x + 10y) + (4x - 10y) = -5 + 62
\][/tex]
Simplifying this:
[tex]\[
15x + 4x + 10y - 10y = 57
\][/tex]
This reduces to:
[tex]\[
19x = 57
\][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Divide both sides by 19 to solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{57}{19}
\][/tex]
[tex]\[
x = 3
\][/tex]
### Step 4: Substitute [tex]\( x \)[/tex] back into one of the original equations
Now that we have [tex]\( x = 3 \)[/tex], substitute it back into the first original equation to solve for [tex]\( y \)[/tex]:
[tex]\[
3(3) + 2y = -1
\][/tex]
[tex]\[
9 + 2y = -1
\][/tex]
Subtract 9 from both sides:
[tex]\[
2y = -1 - 9
\][/tex]
[tex]\[
2y = -10
\][/tex]
Divide both sides by 2:
[tex]\[
y = \frac{-10}{2}
\][/tex]
[tex]\[
y = -5
\][/tex]
### Solution
Thus, the solution to the system of equations is:
[tex]\[
x = 3,\, y = -5
\][/tex]
So, the solution in coordinate form is:
[tex]\[
(3, -5)
\][/tex]
