X2 x 2 0 o solve this quadratic equation x2 x 2

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Unformatted text preview: bstitute x + 2 into the quadratic function. x2 − (x + 2) = 0 o Solve this quadratic equation. x2 − x − 2 = 0 (x + 1) (x − 2) = 0 x = −1 or x =2 o Substitute each of these x-values into the y = −1 + 2 or y=1 or linear function. Answer: x = −1 and y = 1 or x = 2 and y = 4 y = 2+2 y =4 check! Could this example be solved with the substitution method using different decisions? Unit 3: Day 2 notes - Solving Systems of Equations Algebraically (Part 1) Page 2 of 2 exercise: Solve using The Substitution Method. Find the exact values. a) y = x2 − 4x + 2 b) 3x + 2y − 11 = 0 x = −1 and y = 7 c) or y = x2 + 2 y = (x + 1)2 − 4 y = −2x2 + 7 7 x = 2 and y = 1 4 x= d) -1 ± 31 -1 ± 4 31 and y = 3 9 y = x2 − 4x + 2 2x − y + 1 = 0 x − 2y − 8 = 0 x = 1 and y = 3 no solution PRE-CALCULUS 11 Unit 3 – Day 3: SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (Part 2) SOLVING SYSTEMS ALGEBRAICALLY The solution of a system of equations can be solved algebraically: • with the substitution method, or • with the elimination method. SOLVING SYSTEMS OF EQUATIONS WITH THE ELIMINATION METHOD To solve a system of equations algebraically using The Elimination Method: • Write both equation so that like-terms line up. • Eliminate one of the variables by adding (subtracting) the equations; all the terms of one of the variables must be eliminated. o The coefficients of the eliminated terms must have opposite (equal) values. o It may be necessary to multiply one or both equations to get suitable coefficients. • Solve this equation; find the roots - the values of this first variable. • Substitute each of these roots into an equation with both variables. • Solve these equations; find the value of the second variable. example: Solve this system of equations graphically. x2 − y = 5 2x2 o Multiply both sides of the quadratic function by 2. Multiply both sides of the linear function by −1. and x − 2y = 4 − 2y = 10 −x + 2y = −4 o Add the equations. 2x 2 − x o Solve this quadratic equation. 2 x2 − x − 6 = 0 =6 (2x + 3) (x − 2) = 0 x = −3 or 2 o Substitute each of these x-values into one of the functions. 3 ( − 2 ) − 2y = 4 or (2) − 2y = 4 −3 − 4y = 8 or 11 or y=− 4 Answer: x = − 3 and y = − 11 2 4 or x = 2 and y = −1 x =2 − 2y = 2 y = −1 check! Unit 3: Day 3 notes - Solving Systems of Equations Algebraically (Part 2) Page 2 of 2 exercise: Solve using The Elimination Method. Find the exact values. a) x2 − 4x − y + 2 = 0 b) 3x + 2y − 11 = 0 x = −1 and y = 7 c) or no solution 2x2 + 4x − y + 3 = 0 7 x = 2 and y = 1 4 x2 − 2y − 6 = 0 2x2 − 4y + 3 = 0 x2 + 2x − 2y + 6 = 0 x = 0 and y = 3 d) or x = −2 and y = 3 −x2 + 2y + 6 = 0 2x2 − 4y − 12 = 0 infinitely many solutions PRE-CALCULUS 11 Unit 3 – Day 4: SOLVING PROBLEMS WITH SYSTEMS OF EQUATIONS Solving problems using algebra: 1) Define variables introduc...
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This document was uploaded on 02/16/2014 for the course MATH Pre-Calcul at Holy Cross Regional High School.

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