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Systems of Equations

study tool for Algebra II
by

John Swallow

on 15 December 2011

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Transcript of Systems of Equations

1.4 Solving Absolute Value Equations Evaluate Expression Solve Equation No Solution One Solution |x| = -15 *absolute value cannot equal a negative number* Two Solutions |x| = 3x - 2 x = 3x - 2 x = -3x + 2 2 = 2x 4x = 2 1 = x x = 1/2 1 = 3(1) - 2 1/2 = 3(1/2) - 2 1 = 3 - 2 1/2 = 3/2 - 2 1 = 1 1/2 = -1/2 TRUE FALSE {1} Check note: x is on both sides! |x - 7| = 13 *split into positive and negative possibilities* x - 7 = 13 x - 7 = -13 x = 20 x = -6 {-6,20} 1.5 Solving Inequalities Solving Notes < or > - use an open circle -5x + 2 > -33 -5x > -35 x < 7 if you multiply or divide by a negative, switch the inequality sign solution set {-2, } {- ,2} < or > - use a closed circle Graphing 1.6 Solving Compound and Absolute Value Inequalities Absolute Value Notes AND - OR - shade at intersection shade both lines *absolute value inequalities, when separated into two become a compound inequality* |x-2| < 4 x-2 < 4 x-2 > -4 x < 6 x > -2 the negative inequality switches signs x > 7 7 0 -2 6 0 0 7 3.1 Solving Systems of Equations by Graphing consistent and independent consistent and dependent inconsistent (-1,2) graph both lines point of intersection is solution 3.2 Solving Systems of Equations Algebraically substitution elimination 3.3 Solving Systems of Inequalities by Graphing 3.4 Linear Programming 3.5 Solving Systems of Equations in Three Variables 4.8 Using Matrices to Solve Systems of Equations Compound Inequalities x - 9 < -5 x - 9 > -2 x < 4 4 4a - 9b = -20 5a - 4b = -30 -20 -30 -1 4 5 -9 -4 a b = * coefficient matrix variable matrix constant matrix to solve for the variable matrix, you have to multiply the constant matrix by the inverse of the coefficient matrix 4 -9 5 -4 -30 -20 * = b a using matrices to solve systems of equations is especially useful when the number of equations exceeds two *it's also useful because you can just type it into your calculator* 3.4 Linear Programming solve for variable in one equation substitute variable into other equation multiply one equation so that one of the variables cancels when the equations are added together *use substitution and elimination* use elimination to make a system of equations in two variables 6x + 2y + 4z = 2 3x + 4y - 8z = -3 -3x - 6y + 12z = 5 3x + 4y - 8z = -3 6x + 2y + 4z = 2 multiply by -2 -6x - 8y +16z = 6 6x + 2y + 4z = 2 -6y + 20z = 8 multiply by 2 -10y + 28z = 12 6x + 2y + 4z = 2 6x + 2y + 4z = 2 -3x - 6y + 12z = 5 -6x - 12y + 24z = 10 2|x| = 22 2|x| + 3 = 25 |x| = 11 -6y + 20z = 8 -10y + 28z = 12 solve the system of two equations -10y + 28z = 12 -6y + 20z = 8 multiply by 3 multiply by -5 30y - 100z = -40 -30y + 84z = 36 16z = -4 z = -1/4 30y - 100(-1/4) = -40 solve for the other variable 30y + 25 = -40 30y = -15 y = -1/2 y = -1/2 z = -1/4 substitute for the variables 6x + 2(-1/2) + 4(-1/4) = 2 6x - 2 = 2 6x = 4 x = 2/3 y = -1/2 z = -1/4 obviously this is a really long process it would be so much easier to just use a matrix 2r + s = 11 6r - 2s = -2 s = -2r +11 6r - 2(-2r +11) = -2 6r + 4r - 22 = -2 10r = 20 r = 2 s = -2(2) +11 s = -4 + 11 s = 7 substitute variable into other equation solve for variable in one equation plug in and solve for other variable plug in and solve for other variable (2,7) 6r - 2s = -2 2r + s = 11 multiply by 2 4r + 2s = 22 6r - 2s = -2 10r = 20 r = 2 2(2) + s = 11 4 + s = 11 s = 7 plug in and solve for other variable (2,7) < or > - solid line if you multiply or divide by a negative, switch the inequality sign < or > - dotted line < or < - shade below > or > - shade above the region where the shading overlaps is the solution Finding Vertices of the Region Rules for Coordinate Plane Inequalities determine vertices from graph find intersection on calculator solve system of equations for two lines OR OR
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