Solving Quadratic/Linear Systems

AlgebraicallyIntegrated A2/trig

Come on! Solve it!

• Solve for “x” x2

= 3x – 18

• Solve for “x” and “y” using substitution methodx + y = 5

2x = 4y – 8

Solving a Quadratic/Linear System

• A quadratic/linear system is when there are two equation but one equation has an “x2” term and the other just has an “x”. If there are NO “x2” terms then the system is

called a linear system.

• To solve a quadratic/linear system you need to use the “substitution method” and factoring method of a quadratic equation.

How to solve a Quadratic/Linear System

Step 1:

Start by taking both equations and set the both equal to “y”.

Use algebraic skills to manipulate the equation and move terms from one side of the equal sign to the other

Ex:

y = x2 + 2x + 1

-x + y = 3

y = x2 + 2x + 1 (OK)

How to solve a Quadratic/Linear System

Step 1:

Start by taking both equations and set the both equal to “y”.

Use algebraic skills to manipulate the equation and move terms from one side of the equal sign to the other

Ex:

y = x2 + 2x + 1

-x + y = 3

y = x2 + 2x + 1 (OK)

-x + y = 3

+x +x

y = x + 3 (OK)

How to solve a Quadratic/Linear System

Step 2:

Now that both equations are equal to “y”, set them equal to each other eliminating the “y” and having one equation containing only “x’s”

Use algebraic skills to manipulate the equation so that one side is equal to “0”. Make sure the “x2” term is positive.

y = x2 + 2x + 1

y = x + 3

x2 + 2x + 1 = x + 3

-x -x

x2 + x + 1 = 3

-3 -3

x2 + x – 2 = 0

How to solve a Quadratic/Linear System

Step 3:

Factor the quadratic equation and solve for both values of “x”

Ex:

x2 + x – 2 = 0

( )( ) = 0

(x )(x ) = 0

(x + 2)(x – 1) = 0

x + 2 = 0 x – 1 = 0

x = -2 x = 1

Solve a Quadratic/Linear SystemSince there are 2

values of “x” you must solve for 2 values of “y” – one for each “x”.

It is probably best to use the linear equation substitute in to.

After you solve for “y” there are two solutions and express your answer as a coordinate.

x = -2 x = 1

-x + y = 3 -x + y = 3

-(-2) + y = 3 -(1) + y = 3

2 + y = 3 -1 + y = 3

-2 -2 +1 +1

y = 1 y = 4

(-2, 1) (1, 4)

YOU CAN DO IT!!!!

1)

y = x2 – 4x + 3

y = x – 1

2)

x2 – y = 5

y = 3x - 1

solutions

1) (4,3), (1,0)

y = x2 – 4x + 3

y = x – 1

2)(4,11), (-1,-4)

x2 – y = 5

y = 3x - 1

A circle and a line

• Try this one!

(x +2)2 +(y−3)2 =16x+ y=5

A circle and a line

• Solution:

(-2, 7) and (2,3)

(x +2)2 +(y−3)2 =16x+ y=5