`2x+5y=2y-6`
`5x+2y=7`
In the system of equations above, what is the value of the product xy?

Answer

To solve the system of equations given by:

1. \( 2x + 5y = 2y - 6 \)
2. \( 5x + 2y = 7 \)

we will follow these steps:

### Step 1: Rearrange the first equation
We start with the first equation:

\[ 2x + 5y = 2y - 6 \]

Subtract \(2y\) from both sides:

\[ 2x + 5y - 2y = -6 \]

This simplifies to:

\[ 2x + 3y = -6 \]

Let's label this as Equation 1.

### Step 2: Keep the second equation as it is
The second equation remains:

\[ 5x + 2y = 7 \]

Let's label this as Equation 2.

### Step 3: Eliminate one variable
To eliminate one variable, we can multiply Equation 1 and Equation 2 by suitable coefficients so that the coefficients of \(y\) are the same.

Multiply Equation 1 by \(2\):

\[ 2(2x + 3y) = 2(-6) \]
\[ 4x + 6y = -12 \quad \text{(Equation 3)} \]

Multiply Equation 2 by \(3\):

\[ 3(5x + 2y) = 3(7) \]
\[ 15x + 6y = 21 \quad \text{(Equation 4)} \]

### Step 4: Subtract Equation 3 from Equation 4
Now we subtract Equation 3 from Equation 4:

\[ (15x + 6y) - (4x + 6y) = 21 - (-12) \]

This simplifies to:

\[ 15x - 4x + 6y - 6y = 21 + 12 \]
\[ 11x = 33 \]

### Step 5: Solve for \(x\)
Now, divide both sides by \(11\):

\[ x = \frac{33}{11} = 3 \]

### Step 6: Substitute \(x\) back to find \(y\)
Now that we have \(x\), we can substitute it back into either Equation 1 or Equation 2. We will use Equation 2:

\[ 5x + 2y = 7 \]

Substituting \(x = 3\):

\[ 5(3) + 2y = 7 \]
\[ 15 + 2y = 7 \]

### Step 7: Solve for \(y\)
Subtract \(15\) from both sides:

\[ 2y = 7 - 15 \]
\[ 2y = -8 \]

Now divide by \(2\):

\[ y = \frac{-8}{2} = -4 \]

### Step 8: Find the product \(xy\)
Now we have \(x = 3\) and \(y = -4\). The product \(xy\) is:

\[ xy = 3 \times (-4) = -12 \]

Thus, the value of the product \(xy\) is \(-12\).

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