Systems of Linear and Quadratic Equations

A Organization of those two equations tin can be solved (notice 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
  • Ready them equal to each other
  • Simplify into "= 0" format (like a standard Quadratic Equation)
  • Solve the Quadratic Equation!
  • Employ the linear equation to summate matching "y" values, then nosotros go (10,y) points as answers

An example will help:

Example: Solve these two equations:

  • y = ten2 - 5x + seven
  • y = 2x + one

Make both equations into "y=" format:

They are both in "y=" format, so go straight to next step

Set them equal to each other

ten2 - 5x + vii = 2x + 1

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

Decrease 2x from both sides: x2 - 7x + 7 = 1

Subtract 1 from both sides: 102 - 7x + six = 0

Solve the Quadratic Equation!

(The hardest function for me)

You lot tin can read how to solve Quadratic Equations, only here we will factor the Quadratic Equation:

Start with: x2 - 7x + 6 = 0

Rewrite -7x every bit -x-6x: xtwo - x - 6x + 6 = 0

Then: x(x-1) - 6(x-1) = 0

Then: (x-one)(ten-6) = 0

linear and quadratic

Which gives us the solutions x=1 and 10=six

Apply the linear equation to calculate matching "y" values, so we get (x,y) points as answers

The matching y values are (likewise meet Graph):

  • for 10=one: y = 2x+1 = 3
  • for x=half dozen: y = 2x+1 = 13

Our solution: the two points are (1,three) and (half dozen,13)

I think of it equally three stages:

Combine into Quadratic Equation ⇒ Solve the Quadratic ⇒ Calculate the points

Solutions

There are 3 possible cases:

  • No real solution (happens when they never intersect)
  • Ane real solution (when the straight line just touches the quadratic)
  • Two real solutions (similar the example above)

linear and quadratic different intersections

Time for some other example!

Example: Solve these two equations:

  • y - 102 = 7 - 5x
  • 4y - 8x = -21

Brand both equations into "y=" format:

Kickoff equation is: y - xii = vii - 5x

Add x2 to both sides: y = x2 + 7 - 5x

2nd equation is: 4y - 8x = -21

Add together 8x to both sides: 4y = 8x - 21

Divide all by 4: y = 2x - five.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 = -five.25

Add five.25 to both sides: x2 - 7x + 12.25 = 0

Solve the Quadratic Equation!

Using the Quadratic Formula from Quadratic Equations:

linear and quadratic one intersection

  • x = [ -b ± √(bii-4ac) ] / 2a
  • x = [ vii ± √((-7)2-4×1×12.25) ] / 2×1
  • x = [ 7 ± √(49-49) ] / ii
  • x = [ 7 ± √0 ] / two
  • x = 3.v

Just 1 solution! (The "discriminant" is 0)

Use the linear equation to calculate matching "y" values, then nosotros become (x,y) points every bit answers

The matching y value is:

  • for x=iii.5: y = 2x-v.25 = 1.75

Our solution: (iii.5,ane.75)

Real World Case

Kaboom!

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

The state slopes upwards: y = 0.15x

Where does the cannon brawl land?

linear quadratic cannon shot

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: 0.002xii + 0.15x - 0.12x - two = 0

Simplify: 0.002x2 + 0.03x - 2 = 0

Multiply by 500: ten2 + 15x - 1000 = 0

Solve the Quadratic Equation:

Split 15x into -25x+40x: xii -25x + 40x - 1000 = 0

And then: x(x-25) + twoscore(10-25) = 0

And then: (x+40)(ten-25) = 0

x = -40 or 25

The negative answer can be ignored, and then x = 25

Employ the linear equation to calculate matching "y" value:

y = 0.15 x 25 = three.75

So the missive impacts the slope at (25, 3.75)

You lot can besides discover the answer graphically by using the Function Grapher:

linear quadratic graph.

Both Variables Squared

Sometimes BOTH terms of the quadratic can be squared:

Example: Notice the points of intersection of

The circle ten2 + y2 = 25

And the straight line 3y - 2x = half-dozen

line 3y-2x=6 vs circle x^2+y^2=25

First put the line in "y=" format:

Move 2x to right hand side: 3y = 2x + 6

Divide past three: y = 2x/3 + two

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

Put y = 2x/iii + 2 into circle equation: x2 + (2x/3 + 2)two = 25

Expand: x2 + 4x2/ix + two(2x/3)(2) + 22 = 25

Multiply all by 9: 9x2 + 4x2 + 2(2x)(two)(3) + (ix)(2two) = (9)(25)

Simplify: 13x2+ 24x + 36 = 225

Subtract 225 from both sides: 13xtwo+ 24x - 189 = 0

At present it is in standard Quadratic grade, let's solve it:

13x2+ 24x - 189 = 0

Split 24x into 63x-39x: 13x2+ 63x - 39x - 189 = 0

So: x(13x + 63) - 3(13x + 63) = 0

Then: (x - 3)(13x + 63) = 0

So: x = three or -63/13

Now work out y-values:

Substitute ten = 3 into linear equation:

  • 3y - half dozen = half dozen
  • 3y = 12
  • y = four
  • Then one point is (3, 4)

Substitute 10 = -63/13 into linear equation:

  • 3y + 126/13 = 6
  • y + 42/13 = 2
  • y = 2 - 42/13 = 26/13 - 42/13 = -sixteen/13
  • So the other point is (-63/xiii, -16/13)

line 3y-2x=6 vs circle x^2+y^2=25