Question

I'm studying for my algebra test and I'm having trouble solving a system of linear equations. I was given the equations 6x - 5y = 27 and 3x + 5y = 36, but I'm not sure how to find the values of x and y. Can you find a solution to this system of equations and provide the step-by-step process used to solve it?

Answer 1

Sure! Let's solve the system of linear equations by using the method of elimination. The equations you have are:

1. \( 6x - 5y = 27 \) (Equation 1)
2. \( 3x + 5y = 36 \) (Equation 2)

### Step 1: Align the equations
We want to eliminate one of the variables. Looking at the coefficients of \(y\), we can see that if we add the two equations, the \(y\) terms will cancel out, because one is \( -5y \) and the other is \( +5y \).

### Step 2: Multiply Equation 2
To make it easier to eliminate \(y\), we can multiply Equation 2 by 1 (which doesn't change it), but we notice that \(y\) coefficients are already opposites:

\[
3x + 5y = 36 \quad \text{(remains the same)}
\]

### Step 3: Add the equations
Now we can add Equation 1 and Equation 2:

\[
(6x - 5y) + (3x + 5y) = 27 + 36
\]

This simplifies to:

\[
6x + 3x - 5y + 5y = 63
\]

\[
9x = 63
\]

### Step 4: Solve for \(x\)
Now, to find \(x\), divide both sides by 9:

\[
x = \frac{63}{9} = 7
\]

### Step 5: Substitute \(x\) back into one of the equations
Next, we can substitute \(x = 7\) back into either of the original equations to solve for \(y\). We'll use Equation 2:

\[
3(7) + 5y = 36
\]

This simplifies to:

\[
21 + 5y = 36
\]

### Step 6: Solve for \(y\)
Now, subtract 21 from both sides:

\[
5y = 36 - 21
\]

\[
5y = 15
\]

Now divide by 5:

\[
y = \frac{15}{5} = 3
\]

### Final Solution
The solution to the system of equations is:
\[
x = 7, \quad y = 3
\]

### Verification
To verify, we can substitute \(x = 7\) and \(y = 3\) back into both original equations:

**For Equation 1:**
\[
6(7) - 5(3) = 42 - 15 = 27 \quad \text{(True)}
\]

**For Equation 2:**
\[
3(7) + 5(3) = 21 + 15 = 36 \quad \text{(True)}
\]

Both equations hold true, so the solution is verified. Therefore, the final answer is:

\[
\boxed{(7, 3)}
\]