We are given a system of two linear equations in two variables, $x$ and $y$, and we need to find the values of $x$ and $y$ that satisfy both equations. The system of equations is: $3x - 5y = 4$ $-2x + 6y = 18$

1. Problem Description

We are given a system of two linear equations in two variables, $x$ and $y$, and we need to find the values of $x$ and $y$ that satisfy both equations. The system of equations is:

$3x - 5y = 4$

$-2x + 6y = 18$

2. Solution Steps

We can solve this system of equations using either substitution or elimination. Here, we'll use elimination.

First, we multiply the first equation by 2 and the second equation by 3 to eliminate $x$.

Multiplying the first equation by 2, we get:

$2(3x - 5y) = 2(4)$

$6x - 10y = 8$

Multiplying the second equation by 3, we get:

$3(-2x + 6y) = 3(18)$

$-6x + 18y = 54$

Now, we add the two equations to eliminate $x$:

$(6x - 10y) + (-6x + 18y) = 8 + 54$

$8y = 62$

$y = \frac{62}{8} = \frac{31}{4}$

Now that we have the value of $y$, we can substitute it into one of the original equations to find the value of $x$. Let's use the first equation:

$3x - 5y = 4$

$3x - 5(\frac{31}{4}) = 4$

$3x - \frac{155}{4} = 4$

$3x = 4 + \frac{155}{4} = \frac{16}{4} + \frac{155}{4} = \frac{171}{4}$

$x = \frac{171}{4} \div 3 = \frac{171}{4} \cdot \frac{1}{3} = \frac{171}{12} = \frac{57}{4}$

So, $x = \frac{57}{4}$ and $y = \frac{31}{4}$.

3. Final Answer

$x = \frac{57}{4}, y = \frac{31}{4}$