Example 1
Completion requirements
| Example 1 |
Multimedia |
Solve the given system of equations by elimination. Verify the solution.
\(\left\{ \begin{array}{l}
y = x^2 - 6x - 6.5 \\
2y = 4x^2 + 5 \end{array} \right.\)
\(\left\{ \begin{array}{l}
y = x^2 - 6x - 6.5 \\
2y = 4x^2 + 5 \end{array} \right.\)
Start by expressing the two equations in the same format. The second equation does not have an \(x\)-term, so include \(0x\).
\(2y = 4x^2 + 0x + 5\)
Now that both equations are in the same format, multiply one or both equations by a constant, such that the coefficient on one variable is the same in both equations. It will usually be easiest to do this for a variable that is not squared, \(y\) in this case.
\(\begin{align}
y &= x^2 - 6x - 6.5 \\
2\left( y \right) &= 2\left( {x^2 - 6x - 6.5} \right) \\
2y &= 2x^2 - 12x - 13
\end{align}\)
Now that both equations contain \(2y\), the \(y\)’s will be eliminated if the two equations are subtracted. This will leave an equation with only \(x\)’s, which can be solved.
\(\begin{align}
2y &= 4x^2 + \thinspace \thinspace 0x + \thinspace \thinspace 5 \\
- (2y &= 2x^2 - 12x - 13) \\
\hline {} \\
0 &= 2x^2 + 12x + 18 \\
\end{align}\)
\(\begin{align}
0 &= 2\left( {x^2 + 6x + 9} \right) \\
0 &= 2\left( {x + 3} \right)^2 \\
\end{align}\)
\(\begin{align}
x + 3 &= 0 \\
x &= -3 \\
\end{align}\)
Now that the \(x\)-value is known, substitute it into an equation that contains both \(x\) and \(y\) to determine the \(y\)-value.
\(\begin{align}
y &= x^2 - 6x - 6.5 \\
y &= \left( { - 3} \right)^2 - 6\left( { - 3} \right) - 6.5 \\
y &= 20.5 \end{align}\)
The solution is \((-3, 20.5)\).
Verification:
\(2y = 4x^2 + 0x + 5\)
Now that both equations are in the same format, multiply one or both equations by a constant, such that the coefficient on one variable is the same in both equations. It will usually be easiest to do this for a variable that is not squared, \(y\) in this case.
\(\begin{align}
y &= x^2 - 6x - 6.5 \\
2\left( y \right) &= 2\left( {x^2 - 6x - 6.5} \right) \\
2y &= 2x^2 - 12x - 13
\end{align}\)
Now that both equations contain \(2y\), the \(y\)’s will be eliminated if the two equations are subtracted. This will leave an equation with only \(x\)’s, which can be solved.
\(\begin{align}
2y &= 4x^2 + \thinspace \thinspace 0x + \thinspace \thinspace 5 \\
- (2y &= 2x^2 - 12x - 13) \\
\hline {} \\
0 &= 2x^2 + 12x + 18 \\
\end{align}\)
\(\begin{align}
0 &= 2\left( {x^2 + 6x + 9} \right) \\
0 &= 2\left( {x + 3} \right)^2 \\
\end{align}\)
\(\begin{align}
x + 3 &= 0 \\
x &= -3 \\
\end{align}\)
Now that the \(x\)-value is known, substitute it into an equation that contains both \(x\) and \(y\) to determine the \(y\)-value.
\(\begin{align}
y &= x^2 - 6x - 6.5 \\
y &= \left( { - 3} \right)^2 - 6\left( { - 3} \right) - 6.5 \\
y &= 20.5 \end{align}\)
The solution is \((-3, 20.5)\).
Verification:
\(y = x^2 - 6x - 6.5\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r} y \\ 20.5 \end{array}\] |
\[\begin{array}{l} x^2 - 6x - 6.5 \\ \left( { - 3} \right)^2 - 6\left( { - 3} \right) - 6.5 \\ 20.5 \end{array}\] |
| LS = RS | |
\(2y = 4x^2 + 5\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r} 2y \\ 2\left( {20.5} \right) \\ 41 \end{array}\] |
\[\begin{array}{l} 4x^2 + 5 \\ 4\left( { - 3} \right)^2 + 5 \\ 41 \end{array}\] |
| LS = RS | |