Question

left parenthesis, minus, 6, x(−6x plus, 2, y+2y equals= minus, 14, right parenthesis−14)
left parenthesis, minus, 5, x(−5x plus, 5, y+5y equals= 5, right parenthesis5)
minus, 6, x−6x plus, 2, y+2y equals= minus, 14−14
minus, 5, x−5x plus, 5, y+5y equals= 55
-4
-4xx plus+
8
8yy equals=
24
24
Answer
Attempt 1 out of 5

Answer 1

It appears you are working with a system of equations, but the formatting is somewhat unclear. Let's break down what you have shared into a clearer format.

You seem to have two equations:

1. \( -6x + 2y = -14 \)
2. \( -5x + 5y = 5 \)

Let's simplify and solve these equations step by step.

**Equation 1:**
\(-6x + 2y = -14\)

To isolate \(y\), we can add \(6x\) to both sides:

\[
2y = 6x - 14
\]

Now divide by 2:

\[
y = 3x - 7
\]

**Equation 2:**
\(-5x + 5y = 5\)

We can isolate \(y\) similarly by adding \(5x\) to both sides:

\[
5y = 5x + 5
\]

Now divide by 5:

\[
y = x + 1
\]

**Now we have two expressions for \(y\):**

1. \(y = 3x - 7\)
2. \(y = x + 1\)

**Setting them equal to find \(x\):**

\[
3x - 7 = x + 1
\]

Subtract \(x\) from both sides:

\[
3x - x - 7 = 1
\]

Simplifying gives:

\[
2x - 7 = 1
\]

Adding 7 to both sides:

\[
2x = 8
\]

Dividing both sides by 2:

\[
x = 4
\]

**Now, substituting \(x = 4\) back into either equation to find \(y\). We'll use \(y = x + 1\):**

\[
y = 4 + 1 = 5
\]

Thus, the solution to the system of equations is:

\[
x = 4, \quad y = 5
\]

If you have any further questions or additional context regarding your equations or problems, feel free to provide that!