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All courses Algebra I · A-REI.7 53 of 76
Solve simple linear-quadratic systems algebraically and graphically.

Solve a linear-quadratic system by substituting the line into the quadratic

Problem
Solve the system \(y=x^{2}\) and \(y=x+2\) by substitution. Equate the two expressions for \(y\), solve every resulting \(x\)-value, and substitute back to form the complete ordered-pair set.
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Hint

Since both expressions equal \(y\), set them equal to each other: \(x^2=x+2\).

After you solve for the x-values, substitute each one back into either equation to get the matching y-values.

Solution walkthrough

01

Equate the two outputs

\[y=x^2~\text{and}~y=x+2~->~x^2=x+2\]

At a system solution, both equations have the same x and the same y. Since each right side equals y, set them equal.

02

Solve the resulting quadratic

\[\begin{aligned} x^2-x-2=0 \\ (x-2)(x+1)=0 \\ x=2~\text{or}~x=-1 \end{aligned}\]

Move all terms to one side, factor using numbers -2 and 1, and apply the zero-product property.

03

Find the matching outputs

\[\begin{aligned} x=2~->~y=2^2=4~\text{and}~y=2+2=4 \\ x=-1~->~y=(-1)^2=1~\text{and}~y=-1+2=1 \end{aligned}\]

Substitute each x into both original equations. Each pair of outputs agrees, confirming the intersection coordinates.

04

State the complete ordered-pair set

\[(-1,1),(2,4)\]

The system solutions must be ordered pairs, not just x-values. The complete set is (-1,1) and (2,4).

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Another way

  1. Substitute y=x+2 into y=x², solve the same quadratic, then use the line to calculate both y-values.

!

Common mistake

Do not stop after finding x=-1 and x=2. A system solution includes the matching y-coordinate for each x-value.

Solution walkthrough video