Of course! Let's solve the system of equations using the substitution method step-by-step.
We have the system:
[tex]\[
\begin{cases}
\frac{1}{2}x + \frac{1}{2}y = 2 \cdot 5 \quad \text{(Equation 1)} \\
\frac{2}{5}x + y = 3 \cdot 5 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
First, simplify both equations by multiplying through by constants to clear the fractions.
Equation 1:
[tex]\[
\frac{1}{2}x + \frac{1}{2}y = 10 \Rightarrow x + y = 20 \quad \text{(1)}
\][/tex]
Equation 2:
[tex]\[
\frac{2}{5}x + y = 15 \Rightarrow 2x + 5y = 75 \quad \text{(2)}
\][/tex]
Now, solve Equation (1) for [tex]\(y\)[/tex]:
[tex]\[
x + y = 20 \Rightarrow y = 20 - x \quad \text{(3)}
\][/tex]
Substitute [tex]\(y = 20 - x\)[/tex] from Equation (3) into Equation (2):
[tex]\[
2x + 5(20 - x) = 75
\][/tex]
Distribute and simplify:
[tex]\[
2x + 100 - 5x = 75 \Rightarrow -3x + 100 = 75
\][/tex]
Isolate [tex]\(x\)[/tex]:
[tex]\[
-3x = 75 - 100 \Rightarrow -3x = -25 \Rightarrow x = \frac{-25}{-3} = \frac{25}{3} = 8.333\ldots \approx 8.333
\][/tex]
Now, substitute [tex]\(x = \frac{25}{3}\)[/tex] back into Equation (3) to find [tex]\(y\)[/tex]:
[tex]\[
y = 20 - \frac{25}{3}
\][/tex]
Convert 20 to a fraction with denominator 3:
[tex]\[
y = \frac{60}{3} - \frac{25}{3} = \frac{60 - 25}{3} = \frac{35}{3} = 11.666\ldots \approx 11.667
\][/tex]
So, the solution to the system of equations is:
[tex]\[
\left( \frac{25}{3}, \frac{35}{3} \right) \approx (8.333, 11.667)
\][/tex]
Thus, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are approximately:
[tex]\[
x \approx 8.333, \quad y \approx 11.667
\][/tex]
