Sure, I'll provide a step-by-step solution using substitution to solve the given system of equations. We have the following system:
[tex]\[
\begin{aligned}
3x + 5y &= 1 \quad \text{(Equation 1)} \\
y &= 2x - 5 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
### Step 1: Substitute Equation 2 into Equation 1
First, we substitute the expression for [tex]\( y \)[/tex] from Equation 2 into Equation 1.
Given:
[tex]\[ y = 2x - 5 \][/tex]
Substitute [tex]\( y = 2x - 5 \)[/tex] into [tex]\( 3x + 5y = 1 \)[/tex]:
[tex]\[
3x + 5(2x - 5) = 1
\][/tex]
### Step 2: Simplify the Equation
Now, solve for [tex]\( x \)[/tex]:
[tex]\[
3x + 10x - 25 = 1
\][/tex]
Combine like terms:
[tex]\[
13x - 25 = 1
\][/tex]
Add 25 to both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[
13x - 25 + 25 = 1 + 25 \implies 13x = 26
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{26}{13} \implies x = 2
\][/tex]
### Step 3: Substitute [tex]\( x \)[/tex] back into Equation 2
Now that we have [tex]\( x = 2 \)[/tex], substitute this value back into Equation 2 to find [tex]\( y \)[/tex]:
[tex]\[
y = 2(2) - 5
\][/tex]
Calculate [tex]\( y \)[/tex]:
[tex]\[
y = 4 - 5 \implies y = -1
\][/tex]
### Conclusion
The solution to the system of equations is:
[tex]\[
\boxed{x = 2, \; y = -1}
\][/tex]
