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Solve the following system of equations by elimination. Verify the solution.

\(\left\{ \begin{array}{l}
 2y = x + 3 \\
 4y = x^2 + 6x - 15 \end{array} \right.\)


Multiply \(2y = x + 3\) by \(2\), so there is the same coefficient on the \(y\)-terms in each equation.

\(\begin{align}
 2(2y &= x + 3) \\
 4y &= 2x + 6 \\
 \end{align}\)

Now subtract the two equations.

\(\begin{align}
  4y &= x^2 + 6x - 15 \\
  - (4y &= \hspace{19pt} 2x + \hspace{4pt} 6) \\
\hline {} \\
  0 &= x^2 + 4x - 21 \\
 \end{align}\)

\(0 = (x + 7)(x - 3)\)
\(\begin{align}
 x + 7 &= 0 \\
 x &= -7 \end{align}\)

\(\begin{align}
 2y = x + 3 \\
 2y = (-7) + 3 \\
 2y = -4 \\
 y = -2 \end{align}\)
\(\begin{align}
 x - 3 &= 0 \\
 x &= 3 \end{align}\)

\(\begin{align}
 2y = x + 3 \\
 2y = (3) + 3 \\
 2y = 6 \\
 y = 3 \end{align}\)

The solutions to the system are \((-7, -2)\) and \((3, 3)\).

Verify for \((-7, -2)\).

\(2y = x + 3\)

Left Side Right Side
\[\begin{array}{r}
 2y \\
 2( - 2) \\
  - 4
 \end{array}\]
\[\begin{array}{l}
 x + 3 \\
 ( - 7) + 3 \\
  - 4
 \end{array}\]
             LS = RS
\(4y = x^2 + 6x - 15\)

Left Side Right Side
\[\begin{array}{r}
 4y \\
 4( - 2) \\
  - 8  \end{array}\]
\[\begin{array}{l}
 x^2 + 6x - 15 \\
 ( - 7)^2 + 6( - 7) - 15 \\
 49 - 42 - 15 \\
  - 8
 \end{array}\]
             LS = RS

Verify for \((3, 3)\).

\(2y = x + 3\)

Left Side Right Side
\[\begin{array}{r}
 2y \\
 2( 3) \\
  6
 \end{array}\] 		\[\begin{array}{l}
 x + 3 \\
 ( 3) + 3 \\
  6
 \end{array}\]
             LS = RS
\(4y = x^2 + 6x - 15\)

Left Side Right Side
\[\begin{array}{r}
 4y \\
 4( 3) \\
 12  \end{array}\] 		\[\begin{array}{l}
 x^2 + 6x - 15 \\
 ( 3)^2 + 6( 3) - 15 \\
 9 + 18 - 15 \\
  12
 \end{array}\]
             LS = RS