To solve the given system of linear equations using substitution, follow these steps:
Given the system:
[tex]\[
\begin{aligned}
x & = 3y + 10 \quad \text{(Equation 1)} \\
-3x + 5y & = -18 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
1. From Equation 1:
[tex]\[
x = 3y + 10
\][/tex]
We already have [tex]\( x \)[/tex] expressed in terms of [tex]\( y \)[/tex].
2. Substitute [tex]\( x \)[/tex] from Equation 1 into Equation 2:
[tex]\[
-3(3y + 10) + 5y = -18
\][/tex]
3. Simplify the equation:
[tex]\[
-9y - 30 + 5y = -18
\][/tex]
Combine the terms involving [tex]\( y \)[/tex]:
[tex]\[
-4y - 30 = -18
\][/tex]
4. Isolate [tex]\( y \)[/tex]:
[tex]\[
-4y = -18 + 30
\][/tex]
[tex]\[
-4y = 12
\][/tex]
Divide both sides by -4:
[tex]\[
y = -3
\][/tex]
5. Substitute [tex]\( y = -3 \)[/tex] back into Equation 1 to find [tex]\( x \)[/tex]:
[tex]\[
x = 3(-3) + 10
\][/tex]
[tex]\[
x = -9 + 10
\][/tex]
[tex]\[
x = 1
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
x = 1
\][/tex]
[tex]\[
y = -3
\][/tex]
