Solve the system of equations.

3x+6y=21
–5x+2y=13 (1 point)

1 answer

To solve the system of equations

  1. \( 3x + 6y = 21 \)
  2. \( -5x + 2y = 13 \)

we can use the substitution or elimination method. Here, we'll use the substitution method by first solving the first equation for \( y \).

Starting with the first equation:

  1. \( 3x + 6y = 21 \)

We can simplify this equation by dividing all terms by 3:

\[ x + 2y = 7 \]

Now, we can solve for \( y \):

\[ 2y = 7 - x \] \[ y = \frac{7 - x}{2} \]

Next, we will substitute this expression for \( y \) into the second equation:

  1. \( -5x + 2y = 13 \)

Substituting \( y \):

\[ -5x + 2\left(\frac{7 - x}{2}\right) = 13 \]

Now simplify:

\[ -5x + (7 - x) = 13 \] \[ -5x + 7 - x = 13 \] \[ -6x + 7 = 13 \]

Now, subtract 7 from both sides:

\[ -6x = 6 \]

Dividing by -6 gives:

\[ x = -1 \]

Now, we will substitute \( x = -1 \) back into the equation for \( y \):

\[ y = \frac{7 - (-1)}{2} \] \[ y = \frac{7 + 1}{2} = \frac{8}{2} = 4 \]

Thus, the solution to the system of equations is:

\[ \boxed{(-1, 4)} \]

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

For the first equation: \[ 3(-1) + 6(4) = -3 + 24 = 21 \quad \text{(True)} \]

For the second equation: \[ -5(-1) + 2(4) = 5 + 8 = 13 \quad \text{(True)} \]

Both equations are satisfied with \( x = -1 \) and \( y = 4 \). Thus, the solution \( \boxed{(-1, 4)} \) is correct.