Sure, let's solve the system of linear equations step by step:
Given system of equations:
1. [tex]\( 2x + 3y = -3 \)[/tex]
2. [tex]\( 3x + 5y = -9 \)[/tex]
Step 1: Start by multiplying each equation by a suitable number so that the coefficients of one of the variables are the same.
Let's eliminate [tex]\(y\)[/tex]. We can do this by multiplying the first equation by 5 and the second equation by 3 to make the coefficients of [tex]\(y\)[/tex] equal.
[tex]\[
\begin{align*}
5(2x + 3y) &= 5(-3) \\
3(3x + 5y) &= 3(-9)
\end{align*}
\][/tex]
Which simplifies to:
[tex]\[
\begin{align*}
10x + 15y &= -15 \quad \text{(1)} \\
9x + 15y &= -27 \quad \text{(2)}
\end{align*}
\][/tex]
Step 2: Subtract the second equation from the first to eliminate [tex]\(y\)[/tex].
[tex]\[
(10x + 15y) - (9x + 15y) = -15 - (-27)
\][/tex]
Simplifying this gives:
[tex]\[
10x + 15y - 9x - 15y = -15 + 27
\][/tex]
[tex]\[
x = 12
\][/tex]
So, [tex]\( x = 12 \)[/tex].
Step 3: Substitute [tex]\( x = 12 \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]. Let's use the first equation [tex]\( 2x + 3y = -3 \)[/tex]:
[tex]\[
2(12) + 3y = -3
\][/tex]
[tex]\[
24 + 3y = -3
\][/tex]
Subtract 24 from both sides:
[tex]\[
3y = -3 - 24
\][/tex]
[tex]\[
3y = -27
\][/tex]
Divide by 3:
[tex]\[
y = -9
\][/tex]
So, [tex]\( y = -9 \)[/tex].
Solution: The solution to the system of equations is [tex]\( x = 12 \)[/tex] and [tex]\( y = -9 \)[/tex].
Thus, the solution is:
[tex]\[
(x, y) = (12, -9)
\][/tex]
