Systems of Linear and Quadratic Equations
A Linear Equation is an equation of a line.  
A Quadratic Equation is the equation of a parabola and has at least one variable squared (such as x^{2}) 

And together they form a System of a Linear and a Quadratic Equation 
A System of those two equations can be solved (find where they intersect), either:
 Graphically (by plotting them both on the Function Grapher and zooming in)
 or using Algebra
How to Solve using Algebra
 Make both equations into "y =" format
 Set them equal to each other
 Simplify into "= 0" format (like a standard Quadratic Equation)
 Solve the Quadratic Equation!
 Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers
An example will help:
Example: Solve these two equations:
 y = x^{2}  5x + 7
 y = 2x + 1
Make both equations into "y=" format:
They are both in "y=" format, so go straight to next step
Set them equal to each other
Simplify into "= 0" format (like a standard Quadratic Equation)
Solve the Quadratic Equation!
(The hardest part for me)
You can read how to solve Quadratic Equations, but here we will factor the Quadratic Equation:
Which gives us the solutions x=1 and x=6
Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers
The matching y values are (also see Graph):
 for x=1: y = 2x+1 = 3
 for x=6: y = 2x+1 = 13
Our solution: the two points are (1,3) and (6,13)
I think of it as three stages:
Combine into Quadratic Equation ⇒ Solve the Quadratic ⇒ Calculate the points
Solutions
There are three possible cases:
 No real solution (happens when they never intersect)
 One real solution (when the straight line just touches the quadratic)
 Two real solutions (like the example above)
Time for another example:
Example: Solve these two equations:
 y  x^{2} = 7  5x
 4y  8x = 21
Make both equations into "y=" format:
First equation is: y  x^{2} = 7  5x
Second equation is: 4y  8x = 21
Set them equal to each other
Simplify into "= 0" format (like a standard Quadratic Equation)
Solve the Quadratic Equation!
Using the Quadratic Formula from Quadratic Equations:
 x = [ b ± √(b^{2}4ac) ] / 2a
 x = [ 7 ± √((7)^{2}4×1×12.25) ] / 2×1
 x = [ 7 ± √(49^{}49) ] / 2
 x = [ 7 ± √0 ] / 2
 x = 3.5
Just one solution! (The "discriminant" is 0)
Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers
The matching y value is:
 for x=3.5: y = 2x5.25 = 1.75
Our solution: (3.5,1.75)
Real World Example
Kaboom!
The cannon ball flies through the air, following a parabola: y = 2 + 0.12x  0.002x^{2}
The land slopes upward: y = 0.15x
Where does the cannon ball land?
Both equations are already in the "y =" format, so set them equal to each other:
Simplify into "= 0" format:
Solve the Quadratic Equation:
The negative answer can be ignored, so x = 25
Use the linear equation to calculate matching "y" value:
So the cannonball impacts the slope at (25, 3.75)
You can also find the answer graphically by using the Function Grapher:
.
Both Variables Squared
Sometimes BOTH terms of the quadratic can be squared:
Example: Find the points of intersection of
The circle x^{2} + y^{2} = 25
And the straight line 3y  2x = 6
First put the line in "y=" format:
NOW, Instead of making the circle into "y=" format, we can use substitution (replace "y" in the quadratic with the linear expression):
Now it is in standard Quadratic form, let's solve it:
Now work out yvalues:
 3y  6 = 6
 3y = 12
 y = 4
 So one point is (3, 4)
 3y + 126/13 = 6
 y + 42/13 = 2
 y = 2  42/13 = 26/13  42/13 = 16/13
 So the other point is (63/13, 16/13)