What is the solution to the system of equations below?


x + 3 y = 15 and 4 x + 2 y = 30
(6, 3)
(3, 6)
(7, –6)
(–6, 7)

1 answer

To solve the system of equations given:

  1. \( x + 3y = 15 \) (Equation 1)
  2. \( 4x + 2y = 30 \) (Equation 2)

Let's start with Equation 1 and express \( x \) in terms of \( y \):

\[ x = 15 - 3y \]

Next, we will substitute this expression for \( x \) into Equation 2.

Substituting \( x \):

\[ 4(15 - 3y) + 2y = 30 \]

Now distribute \( 4 \):

\[ 60 - 12y + 2y = 30 \]

Combine like terms:

\[ 60 - 10y = 30 \]

Subtract \( 60 \) from both sides:

\[ -10y = 30 - 60 \] \[ -10y = -30 \]

Divide by \( -10 \):

\[ y = 3 \]

Now that we have \( y \), substitute it back into the expression for \( x \):

\[ x = 15 - 3(3) \] \[ x = 15 - 9 \] \[ x = 6 \]

Thus, the solution to the system of equations is:

\[ (x, y) = (6, 3) \]

The correct answer is (6, 3).