Solve the System: Equations 6x + 4y = 18 and -2x + 3y = 20

To solve this problem, we'll use the elimination method:

• Step 1: Align the coefficients of $x$ to eliminate it.
• Step 2: Subtract one equation from the other to solve for the remaining variable.
• Step 3: Substitute the known variable back into one of the original equations to find the value of the second variable.

Now, let's work through each step:

Step 1: The given system of equations is:

$6x + 4y = 18$ (Equation 1)

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

To eliminate $x$, we want the coefficients of $x$ to be equal in magnitude. Multiply Equation 2 by 3 to match the $x$ coefficient in Equation 1:

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

This results in:

$-6x + 9y = 60$ (Equation 3)

Step 2: Add Equation 1 and Equation 3 to eliminate $x$:

$(6x + 4y) + (-6x + 9y) = 18 + 60$

This simplifies to:

$13y = 78$

Solving for $y$, we divide both sides by 13:

$y = \frac{78}{13} = 6$

Step 3: Substitute $y = 6$ back into Equation 1:

$6x + 4(6) = 18$

$6x + 24 = 18$

Subtract 24 from both sides:

$6x = 18 - 24$

$6x = -6$

Divide both sides by 6:

$x = \frac{-6}{6} = -1$

Therefore, the solution to the system of equations is $\mathbf{x = -1, y = 6}$.