The problem asks us to solve a system of two linear equations with two variables, $x$ and $y$, using the substitution method. The given system of equations is: $7x - \frac{2}{3}y = -37$ (1) $\frac{x}{4} + 5y = \frac{55}{4}$ (2)

1. Problem Description

The problem asks us to solve a system of two linear equations with two variables, $x$ and $y$, using the substitution method. The given system of equations is:

$7x - \frac{2}{3}y = -37$ (1)

$\frac{x}{4} + 5y = \frac{55}{4}$ (2)

2. Solution Steps

First, let's solve equation (2) for $x$:

$\frac{x}{4} + 5y = \frac{55}{4}$

Multiply both sides by 4:

$x + 20y = 55$

$x = 55 - 20y$ (3)

Now, substitute the expression for $x$ from equation (3) into equation (1):

$7(55 - 20y) - \frac{2}{3}y = -37$

$385 - 140y - \frac{2}{3}y = -37$

Multiply both sides by 3 to eliminate the fraction:

$3(385 - 140y - \frac{2}{3}y) = 3(-37)$

$1155 - 420y - 2y = -111$

$1155 - 422y = -111$

$-422y = -111 - 1155$

$-422y = -1266$

$y = \frac{-1266}{-422}$

$y = 3$

Now, substitute the value of $y$ back into equation (3) to find $x$:

$x = 55 - 20(3)$

$x = 55 - 60$

$x = -5$

Thus, the solution to the system is $x = -5$ and $y = 3$.

3. Final Answer

The solution is $(-5, 3)$.