Systems of Linear and Quadratic Equations | Mathematics for Grade 12 PDF Download

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 x²)

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

Example: Solve these two equations:

• y = x² - 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

• x² - 5x + 7 = 2x + 1

Simplify into "= 0" format (like a standard Quadratic Equation)

• Subtract 2x from both sides: x² - 7x + 7 = 1
• Subtract 1 from both sides: x² - 7x + 6 = 0

Solve the Quadratic Equation!

• Start with: x² - 7x + 6 = 0
• Rewrite -7x as -x-6x: x² - x - 6x + 6 = 0
• Then: x(x-1) - 6(x-1) = 0
• Then: (x-1)(x-6) = 0
  

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

Hence, the two points are (1,3) and (6,13)

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)
  

Example: Solve these two equations:

• y - x² = 7 - 5x
• 4y - 8x = -21

Make both equations into "y=" format:

• First equation is: y - x² = 7 - 5x
  • Add x² to both sides: y = x² + 7 - 5x
• Second equation is: 4y - 8x = -21
  • Add 8x to both sides: 4y = 8x - 21
  • Divide all by 4: y = 2x - 5.25

Set them equal to each other

• x² - 5x + 7 = 2x - 5.25

Simplify into "= 0" format (like a standard Quadratic Equation)

• Subtract 2x from both sides: x² - 7x + 7 = -5.25
• Add 5.25 to both sides: x² - 7x + 12.25 = 0

Solve the Quadratic Equation!

Using the Quadratic Formula from Quadratic Equations:

• x = [ -b ± √(b²-4ac) ] / 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 = 2x-5.25 = 1.75

Hence, (3.5,1.75)

Real World Example

The cannon ball flies through the air, following a parabola:
y = 2 + 0.12x - 0.002x²

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:

• 0.15x = 2 + 0.12x - 0.002x²

Simplify into "= 0" format:

• Bring all terms to left: 0.002x² + 0.15x - 0.12x - 2 = 0
• Simplify: 0.002x² + 0.03x - 2 = 0
• Multiply by 500: x² + 15x - 1000 = 0

Solve the Quadratic Equation:

• Split 15x into -25x + 40x: x² -25x + 40x - 1000 = 0
• Then: x(x-25) + 40(x-25) = 0
• Then: (x+40)(x-25) = 0
• x = -40 or 25

The negative answer can be ignored, so x = 25
Use the linear equation to calculate matching "y" value:

• y = 0.15 x 25 = 3.75

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 x² + y² = 25
And the straight line 3y - 2x = 6

First put the line in "y=" format:

• Move 2x to right hand side: 3y = 2x + 6
• Divide by 3: y = 2x/3 + 2

NOW, Instead of making the circle into "y=" format, we can use substitution (replace "y" in the quadratic with the linear expression):

• Put y = 2x/3 + 2 into circle equation: x² + (2x/3 + 2)2 = 25
• Expand: x² + 4x²/9 + 2(2x/3)(2) + 2² = 25
• Multiply all by 9: 9x² + 4x² + 2(2x)(2)(3) + (9)(22) = (9)(25)
• Simplify: 13x² + 24x + 36 = 225
• Subtract 225 from both sides: 13x² + 24x - 189 = 0

Now it is in standard Quadratic form, let's solve it:

• 13x²+ 24x - 189 = 0
• Split 24x into 63x - 39x: 13x² + 63x - 39x - 189 = 0
• Then: x(13x + 63) - 3(13x + 63) = 0
• Then: (x - 3)(13x + 63) = 0
• So: x = 3 or -63/13

Now work out y-values:
Substitute x = 3 into linear equation:

• 3y - 6 = 6
• 3y = 12
• y = 4
• So one point is (3, 4)

Substitute x = -63/13 into linear equation:

• 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)