To solve the system of equations:
- 3x + 5y = 1 (Equation 1)
- 7x + 4y = -13 (Equation 2)
We can use the method of substitution or elimination. Here, I will use substitution.
First, let's solve Equation 1 for \(y\):
\[ 5y = 1 - 3x \] \[ y = \frac{1 - 3x}{5} \]
Now we substitute this expression for \(y\) into Equation 2:
\[ 7x + 4\left(\frac{1 - 3x}{5}\right) = -13 \]
Multiply through by 5 to eliminate the fraction:
\[ 35x + 4(1 - 3x) = -65 \] \[ 35x + 4 - 12x = -65 \] \[ 23x + 4 = -65 \] \[ 23x = -65 - 4 \] \[ 23x = -69 \] \[ x = -3 \]
Now we'll substitute \(x = -3\) back into the equation we derived for \(y\):
\[ y = \frac{1 - 3(-3)}{5} \] \[ y = \frac{1 + 9}{5} \] \[ y = \frac{10}{5} = 2 \]
Thus, the y-value of the solution to the system of equations is:
\[ \boxed{2} \]
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