## Algebra 1 Tutorial

*Quadratic-Linear Equations*

#### Intro

When we do quadratic-linear equations, it is the same as systems of linear equations. We have to apply the same principle of substituting y or x, depending on the linear equation itself. We can do without substituting and rather treat the equations as equations to be done algebraically. We have to find the intersecting coordinates or points intersecting on the graph to get the answer.

#### Sample Problem

x^2+y=9

y=x+9

Find the solutions of x and its intersecting points.

#### Solution

There are two ways to do this.

First way:

x^2+y=9

y=x+9

1. Subtract x^2 from both sides.

x^2+y=9

-x^2 -x^2

y1=-x^2+9

2. Set y1 and y2 equal and add x^2 and subtract 9 from both sides.

y1=y2

-x^2+9=x+9

+x^2-9 +x^2-9

0= x^2+x

3. Factor out x^2+x to get the solutions for x.

0= x(x+1)

x= 0

x+1= 0

x=-1

4. Substitute x for these two solutions to get the intersection points or in other words, where they meet.

y= x+9

x= 0 x=-1

y= 0+9 y=-1+9

y=9 (0,9) y=-8 (-1,8)

Second way:

x^2+y=9

y=x+9

1. Substitute y for x+9.

x^2+x+9=9

2. Subtract 9 from both sides.

x^2+x+9-9=9-9

x^2+x=0

Repeat step 3 and 4 from the first way because you will get the same results.

# About The Author

Math, Biology, And History Instructor |

Hello. I am Kisung Yoon, and I am a senior in high school. I am currently in the ARISTA Honor Society, and I have tutored my fellow students from my school in Algebra I, Algebra 2, Global History, US History, and Biology. Not only that, I also tutored my younger sisters on the standardized tests mai... |

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