We need to solve the following system of equations for $s$ and $t$ using the addition (elimination) method: $4s + 3t = 13$ $7s - 9t = 18$

1. Problem Description

We need to solve the following system of equations for $s$ and $t$ using the addition (elimination) method:

$4s + 3t = 13$

$7s - 9t = 18$

2. Solution Steps

We want to eliminate one of the variables, say $t$. To do this, we can multiply the first equation by 3 so that the coefficient of $t$ is $9t$, which will cancel with the $-9t$ in the second equation.

Multiply the first equation by 3:

$3(4s + 3t) = 3(13)$

$12s + 9t = 39$

Now we have the system:

$12s + 9t = 39$

$7s - 9t = 18$

Add the two equations:

$(12s + 9t) + (7s - 9t) = 39 + 18$

$12s + 7s + 9t - 9t = 57$

$19s = 57$

Divide by 19 to solve for $s$:

$s = \frac{57}{19}$

$s = 3$

Now substitute $s = 3$ into the first original equation to solve for $t$:

$4(3) + 3t = 13$

$12 + 3t = 13$

$3t = 13 - 12$

$3t = 1$

$t = \frac{1}{3}$

So the solution is $s = 3$ and $t = \frac{1}{3}$.

3. Final Answer

(3, 1/3)