Sure! Let's solve the given system of equations step by step using the substitution method:
[tex]\[
\begin{cases}
2x + y = 20 \\
-5y = -6x + 12
\end{cases}
\][/tex]
First, we need to solve for [tex]\(y\)[/tex] in one of the equations. Let’s use the second equation:
[tex]\[
-5y = -6x + 12
\][/tex]
Divide both sides by [tex]\(-5\)[/tex] to isolate [tex]\(y\)[/tex]:
[tex]\[
y = \frac{6x - 12}{5}
\][/tex]
Now, substitute this expression for [tex]\(y\)[/tex] into the first equation:
[tex]\[
2x + y = 20
\][/tex]
[tex]\[
2x + \left(\frac{6x - 12}{5}\right) = 20
\][/tex]
To clear the fraction, multiply the entire equation by 5:
[tex]\[
5 \cdot 2x + 5 \cdot \frac{6x - 12}{5} = 20 \cdot 5
\][/tex]
[tex]\[
10x + 6x - 12 = 100
\][/tex]
Combine like terms:
[tex]\[
16x - 12 = 100
\][/tex]
Add 12 to both sides:
[tex]\[
16x = 112
\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{112}{16}
\][/tex]
[tex]\[
x = 7
\][/tex]
Next, substitute [tex]\(x = 7\)[/tex] back into the expression we found for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{6x - 12}{5}
\][/tex]
[tex]\[
y = \frac{6 \cdot 7 - 12}{5}
\][/tex]
[tex]\[
y = \frac{42 - 12}{5}
\][/tex]
[tex]\[
y = \frac{30}{5}
\][/tex]
[tex]\[
y = 6
\][/tex]
So, the solution to the system of equations is:
[tex]\[
x = 7, \quad y = 6
\][/tex]
