We are asked to find the solution to the system of linear equations using the substitution method. The given system is: $\frac{1}{3}x - \frac{1}{9}y = 4$ (1) $\frac{3}{4}x + \frac{4}{5}y = 9$ (2)

1. Problem Description

We are asked to find the solution to the system of linear equations using the substitution method. The given system is:

$\frac{1}{3}x - \frac{1}{9}y = 4$ (1)

$\frac{3}{4}x + \frac{4}{5}y = 9$ (2)

2. Solution Steps

First, we solve equation (1) for $x$.

$\frac{1}{3}x - \frac{1}{9}y = 4$

Multiply both sides by 9 to eliminate fractions:

$9(\frac{1}{3}x - \frac{1}{9}y) = 9(4)$

$3x - y = 36$

$3x = y + 36$

$x = \frac{1}{3}y + 12$ (3)

Next, substitute equation (3) into equation (2):

$\frac{3}{4}x + \frac{4}{5}y = 9$

$\frac{3}{4}(\frac{1}{3}y + 12) + \frac{4}{5}y = 9$

$\frac{1}{4}y + 9 + \frac{4}{5}y = 9$

$\frac{1}{4}y + \frac{4}{5}y = 0$

Multiply both sides by 20 to eliminate fractions:

$20(\frac{1}{4}y + \frac{4}{5}y) = 20(0)$

$5y + 16y = 0$

$21y = 0$

$y = 0$

Now, substitute $y=0$ into equation (3) to find $x$:

$x = \frac{1}{3}(0) + 12$

$x = 0 + 12$

$x = 12$

Therefore, the solution to the system is $x = 12$ and $y = 0$.

3. Final Answer

(12, 0)