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

1. Problem Description

We are given a system of two linear equations with two variables, $x$ and $y$.

The system is:

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

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

We need to solve for $x$ and $y$.

2. Solution Steps

Let the first equation be (1) and the second equation be (2).

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

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

Multiply equation (1) by 4 to eliminate the fractions.

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

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

Multiply equation (2) by 6 to eliminate the fractions.

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

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

Now we have the simplified system:

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

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

Multiply equation (3) by 5 and equation (4) by 2 to eliminate $x$.

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

$10x + 15y = 100$ (5)

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

$10x - 8y = 36$ (6)

Subtract equation (6) from equation (5):

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

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

$23y = 64$

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

Substitute $y = \frac{64}{23}$ into equation (3):

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

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

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

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

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

$x = \frac{268}{23} \times \frac{1}{2}$

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

3. Final Answer

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

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