Solving Quadratic/Linear Systems Algebraically Integrated A2/trig.

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Presentation transcript:

Solving Quadratic/Linear Systems Algebraically Integrated A2/trig

Come on! Solve it! Solve for “x”  x 2 = 3x – 18 Solve for “x” and “y” using substitution method x + 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 “x 2 ” term and the other just has an “x”.  If there are NO “x 2 ” 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 = x 2 + 2x + 1 -x + y = 3 y = x 2 + 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 = x 2 + 2x + 1 -x + y = 3 y = x 2 + 2x + 1 (OK) -x + y = 3 +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 “x 2 ” term is positive. y = x 2 + 2x + 1 y = x + 3 x 2 + 2x + 1 = x + 3 -x -x x 2 + x + 1 = x 2 + x – 2 = 0

How to solve a Quadratic/Linear System Step 3:  Factor the quadratic equation and solve for both values of “x” Ex: x 2 + x – 2 = 0 ( )( ) = 0 (x )(x ) = 0 (x + 2)(x – 1) = 0 x + 2 = 0x – 1 = 0 x = -2 x = 1

Solve a Quadratic/Linear System  Since 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 -(-2) + y = 3-(1) + y = y = y = y = 1y = 4 (-2, 1)(1, 4)

YOU CAN DO IT!!!! 1) y = x 2 – 4x + 3 y = x – 1 2) x 2 – y = 5 y = 3x - 1

solutions 1) (4,3), (1,0) y = x 2 – 4x + 3 y = x – 1 2)(4,11), (-1,-4) x 2 – y = 5 y = 3x - 1

A circle and a line Try this one!

A circle and a line Solution: (-2, 7) and (2,3)