Let's solve the system of equations using the substitution method. The system of equations is:
[tex]\[
\begin{aligned}
5x + 4y &= 7 \quad \text{(Equation 1)} \\
y &= 2x + 5 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
Step 1: Substitute Equation 2 into Equation 1
From Equation 2, we have:
[tex]\[
y = 2x + 5
\][/tex]
We can substitute this expression for \( y \) into Equation 1:
[tex]\[
5x + 4(2x + 5) = 7
\][/tex]
Step 2: Simplify and solve for \( x \)
Distribute the \( 4 \) inside the parentheses:
[tex]\[
5x + 8x + 20 = 7
\][/tex]
Combine the \( x \) terms:
[tex]\[
13x + 20 = 7
\][/tex]
Subtract 20 from both sides to isolate the \( x \) term:
[tex]\[
13x = 7 - 20
\][/tex]
Simplify the right side:
[tex]\[
13x = -13
\][/tex]
Divide both sides by 13:
[tex]\[
x = \frac{-13}{13}
\][/tex]
So,
[tex]\[
x = -1
\][/tex]
Step 3: Substitute \( x = -1 \) back into Equation 2 to solve for \( y \)
We use Equation 2 to find the corresponding \( y \) value:
[tex]\[
y = 2(-1) + 5
\][/tex]
Simplify:
[tex]\[
y = -2 + 5
\][/tex]
Therefore,
[tex]\[
y = 3
\][/tex]
Solution
The solution to the system of equations is:
[tex]\[
\begin{array}{l}
x = -1 \\
y = 3
\end{array}
\][/tex]
So, the point [tex]\((x, y) = (-1, 3)\)[/tex] satisfies both equations in the given system.
