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All courses Math II · A-REI.7 7 of 73
Solve simple linear-quadratic systems algebraically and graphically.

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

Problem
Solve the linear-quadratic system by substitution: y=x², y=x+2.
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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

Set the two expressions for y equal

\[x^2~=~x+2\]

Both equations are equal to \(y\), so they must also be equal to each other.

02

Solve the quadratic equation

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

Move all terms to one side and factor to find the x-values where the line and parabola intersect.

03

Find the x-values

\[x~=~2~\text{or}~x~=~-1\]

Set each factor equal to zero to get the two possible x-coordinates.

04

Substitute to get the ordered pairs

\[x~=~2~\Rightarrow~y~=~4~\text{and}~x~=~-1~\Rightarrow~y~=~1,~\text{so}~(-1,1),~(2,4)\]

Use either equation to find the matching y-value for each x-value, then write the intersection points as ordered pairs.

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Another way

  1. Check both ordered pairs in the original equations: for \((-1,1)\), \(1=(-1)^2\) and \(1=-1+2\); for \((2,4)\), \(4=2^2\) and \(4=2+2\).

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Common mistake

Solving \(x^2=x+2\) correctly but stopping at \(x=-1\) and \(x=2\). The problem asks for solution points, so you must also find the y-values.