We are given a system of two linear equations with two variables, $v$ and $w$: $2v + 6w = -36$ $5v + 2w = 1$ We need to find the values of $v$ and $w$ that satisfy both equations.

1. Problem Description

We are given a system of two linear equations with two variables, $v$ and $w$:

$2v + 6w = -36$

$5v + 2w = 1$

We need to find the values of $v$ and $w$ that satisfy both equations.

2. Solution Steps

We can use the elimination method to solve this system of equations. First, we multiply the first equation by 1 and the second equation by -3 to eliminate the $w$ variable:

Equation 1: $2v + 6w = -36$

Equation 2: $5v + 2w = 1$

Multiply Equation 1 by 1:

$1(2v + 6w) = 1(-36)$

$2v + 6w = -36$

Multiply Equation 2 by -3:

$-3(5v + 2w) = -3(1)$

$-15v - 6w = -3$

Now add the modified equations:

$(2v + 6w) + (-15v - 6w) = -36 + (-3)$

$2v - 15v + 6w - 6w = -39$

$-13v = -39$

Divide by -13 to solve for $v$:

$v = \frac{-39}{-13}$

$v = 3$

Now that we have the value of $v$, we can substitute it back into either of the original equations to solve for $w$. Let's use the second equation:

$5v + 2w = 1$

$5(3) + 2w = 1$

$15 + 2w = 1$

$2w = 1 - 15$

$2w = -14$

$w = \frac{-14}{2}$

$w = -7$

Thus, $v = 3$ and $w = -7$.

3. Final Answer

The solution is $v = 3$ and $w = -7$.