To solve the system of equations:
[tex]\[
\begin{array}{l}
4x + 5y = 12 \\
3x - 2y = 32
\end{array}
\][/tex]
Follow these steps:
1. Write down the two equations:
[tex]\[
4x + 5y = 12 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
3x - 2y = 32 \quad \text{(Equation 2)}
\][/tex]
2. Isolate one of the variables: You can choose to isolate either [tex]\( x \)[/tex] or [tex]\( y \)[/tex]. For this solution, let's isolate [tex]\( y \)[/tex] from Equation 1.
[tex]\[
4x + 5y = 12 \implies 5y = 12 - 4x \implies y = \frac{12 - 4x}{5}
\][/tex]
3. Substitute the expression for [tex]\( y \)[/tex] into Equation 2:
[tex]\[
3x - 2\left(\frac{12 - 4x}{5}\right) = 32
\][/tex]
4. Simplify the equation:
[tex]\[
3x - \frac{24 - 8x}{5} = 32
\][/tex]
[tex]\[
3x - \frac{24}{5} + \frac{8x}{5} = 32
\][/tex]
5. Multiply every term by 5 to clear the denominator:
[tex]\[
5 \cdot 3x - 5 \cdot \frac{24}{5} + 5 \cdot \frac{8x}{5} = 5 \cdot 32
\][/tex]
[tex]\[
15x - 24 + 8x = 160
\][/tex]
6. Combine like terms:
[tex]\[
15x + 8x - 24 = 160
\][/tex]
[tex]\[
23x - 24 = 160
\][/tex]
7. Solve for [tex]\( x \)[/tex]:
[tex]\[
23x = 160 + 24
\][/tex]
[tex]\[
23x = 184
\][/tex]
[tex]\[
x = \frac{184}{23}
\][/tex]
[tex]\[
x = 8
\][/tex]
8. Substitute [tex]\( x = 8 \)[/tex] back into the expression for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{12 - 4x}{5} = \frac{12 - 4 \cdot 8}{5} = \frac{12 - 32}{5} = \frac{-20}{5} = -4
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
x = 8, \quad y = -4
\][/tex]
