We are given a system of two linear equations with two variables, $x$ and $y$. We need to solve for $x$ and $y$. The system of equations is: $\frac{1}{2}x + \frac{3}{4}y = 5$ $\frac{5}{6}x - \frac{2}{3}y = 3$

1. Problem Description

We are given a system of two linear equations with two variables, $x$ and $y$. We need to solve for $x$ and $y$. The system of equations is:

$\frac{1}{2}x + \frac{3}{4}y = 5$

$\frac{5}{6}x - \frac{2}{3}y = 3$

2. Solution Steps

First, let's multiply the first equation by 4 to eliminate the fractions:

$4(\frac{1}{2}x + \frac{3}{4}y) = 4(5)$

$2x + 3y = 20$ (Equation 1)

Next, let's multiply the second equation by 6 to eliminate the fractions:

$6(\frac{5}{6}x - \frac{2}{3}y) = 6(3)$

$5x - 4y = 18$ (Equation 2)

Now we have a system of two linear equations without fractions:

$2x + 3y = 20$

$5x - 4y = 18$

We can solve this system using either substitution or elimination. Let's use elimination. Multiply Equation 1 by 5 and Equation 2 by 2:

$5(2x + 3y) = 5(20)$

$10x + 15y = 100$ (Equation 3)

$2(5x - 4y) = 2(18)$

$10x - 8y = 36$ (Equation 4)

Subtract Equation 4 from Equation 3 to eliminate $x$:

$(10x + 15y) - (10x - 8y) = 100 - 36$

$10x + 15y - 10x + 8y = 64$

$23y = 64$

$y = \frac{64}{23}$

Now, substitute the value of $y$ back into Equation 1 to solve for $x$:

$2x + 3(\frac{64}{23}) = 20$

$2x + \frac{192}{23} = 20$

$2x = 20 - \frac{192}{23}$

$2x = \frac{460 - 192}{23}$

$2x = \frac{268}{23}$

$x = \frac{268}{2 \cdot 23}$

$x = \frac{134}{23}$

3. Final Answer

$x = \frac{134}{23}$ and $y = \frac{64}{23}$