Solve the system of equations. 3x - 4y = 40

5x - 5y = 55

1 answer

To solve the system of equations:

  1. \( 3x - 4y = 40 \) (Equation 1)
  2. \( 5x - 5y = 55 \) (Equation 2)

First, we can simplify Equation 2. Dividing every term by 5 gives us:

\[ x - y = 11 \tag{Equation 2 simplified} \]

Next, we can express \( x \) in terms of \( y \) from Equation 2:

\[ x = y + 11 \tag{Equation 3} \]

Now, we can substitute Equation 3 into Equation 1:

\[ 3(y + 11) - 4y = 40 \]

Expanding this gives:

\[ 3y + 33 - 4y = 40 \]

Combining like terms:

\[ -y + 33 = 40 \]

To isolate \( y \), we can subtract 33 from both sides:

\[ -y = 40 - 33 \]

\[ -y = 7 \]

Multiplying by -1 gives:

\[ y = -7 \]

Now, we can substitute \( y = -7 \) back into Equation 3 to find \( x \):

\[ x = -7 + 11 \]

\[ x = 4 \]

Now we have the solution:

\[ x = 4, \quad y = -7 \]

To verify, we can substitute these values into the original equations:

  1. For Equation 1: \[ 3(4) - 4(-7) = 12 + 28 = 40 \quad \text{(True)} \]

  2. For Equation 2: \[ 5(4) - 5(-7) = 20 + 35 = 55 \quad \text{(True)} \]

Thus, the solution to the system of equations is:

\[ \boxed{(4, -7)} \]