The numerator must always be 1 degree less than the denominator The numerator must always be 1 degree less than the denominator. A, B, C,..., and Z are constants.

The numerator must always be 1 degree less than the denominator The numerator must always be 1 degree less than the denominator. That is why there are 2 terms on the top.

Write the partial fraction decomposition of Factor the denominator. 1 Write a fraction for each factor. The factors are all LINEAR! Every fraction needs the LCD Multiply top and bottom by what is missing. Multiply out the tops and remove the bottoms, they are all the same. 0x2 + Substitute the values in for A, B, and C. Collect all like terms. x2 terms: 0 = A + B x terms: 1 = -2A – B + C Constants: 2 = A Back substitute. 1 = -2(2) – (-2) + C 1 = -2 + C 3 = C 2 X 0 = 2 + B -2 = B

Modified Zero Product Rule, another way to find A, B, and C. Let x = 0 Let x = 1 Let x = 2

Write the partial fraction decomposition of The denominator is factored. Linear Quadratic Write a fraction for each factor. Every fraction needs the LCD Multiply top and bottom by what is missing. Multiply out the tops and remove the bottoms, they are all the same. Collect all like terms. x3 terms: 0 = A + C x2 terms: 0 = B + D x terms: 1 = 4A; A = ¼ Constants: 4 = 4B; B = 1 Modified Zero Product Rule Back substitute. 0 = ¼ + C - ¼ = C 0 = 1 + D -1 = D Let x = 0 Doesn’t work well with Quad. Factors!

+ Solve by Elimination. Back Substitute to find y. ( )(-1) ( )(-1) + Calculator Check. Set = to y.

+ ( )(-1) Solve the following systems of equations: Back Substitute to find x. Solve by Elimination. Solve y. ( )(-1) + Calculator Check. Set = to y. Graph, 2nd Trace, Intersection Zoom 6 Standard Window Zoom 5 Square Window 1st Curve? Y2 2nd Curve? Y3 1st Curve? Y1 2nd Curve? Y3 1st Curve? Y2 2nd Curve? Y3 1st Curve? Y1 2nd Curve? Y3

Solve the following systems of equations: Solve by Substitution. Multiply all terms by x2 Set = 0 and write in descending order. Divide by2. Factor like a quadratic. Calculator check set up for Y =

Solve the following systems of equations: Identify restrictions on answers for x and y. x is a base… x > 0 and x can’t = 1. y is in the log … y > 0 . Convert both equations to exponential. Solve by Substitution.

Section 8.7 Systems of Inequalities. There will either be _________ solutions or ______________ solutions. The solution is the OVERLAP NO INFINITE Solve the following systems of equations: 2 ways to graph. Solve for y TRUE Find x and y intercepts. ( _____ , 0 ) ( 0 , _____) Graph like y = mx + b When solved for y, the > symbol tells us to shade above the line. If we have a < symbol, then we shade below the line. Remember…if there is no equal to line, then the line is a dashed line. 4 Plot the intercepts and draw the line. Test the origin to see if the inequality is true. If it is true, we shade in that direction. If not true, we shade away from the origin. 2

Solve the following systems of equations: Identify the graphs. Circle The solution is the OVERLAP Parabola r = 5, Center @ ( 0, 0 ) The < symbol means we want all the points inside the circle. The > symbol means we want to shade all the points outside the circle. Vertex at ( 0, -5 ) with our 1, 3, 5, ... pattern. y < means we shade below the curve.

Ax + By = C; Standard Form Note. Solve the following systems of equations: Find the x and y intercepts. Ax + By = C; Standard Form Note. The slope is m = -A / B

Write the system of inequalities. Take care of the easy lines first, horizontal and vertical lines. The slope is m = -2/1 Y-intercept (0, 4), b = 4 y = -2x + 4

( 2, 1 ) gives the maximum value of 8. Find the Maximum value of z = 3x + 2y 5 Subject to the following constraints. 4 z = 3x + 2y 3 ( 0, 2 ): z = 3(0) + 2(2) = 4 ( 0, 2 ) 2 ( 2, 1 ): z = 3(2) + 2(1) = 8 ( 1, 0 ): z = 3(1) + 2(0) = 3 ( 2, 1 ) 1 ( 1, 0 ) 1 2 3 4 5 ( 2, 1 ) gives the maximum value of 8.

Max: Points farthest from the origin. Find the maximum and minimum value of z = 5x + 7y ( 1, 5 ): z = 5(1) + 7(5) = 40 ( 6, 3 ): z = 5(6) + 7(3) = 51 ( 6, 3 ) is the point for max. Subject to the following constraints 12 Quad 1 only. 10 Find intercepts. 8 Find point & slope. 6 ( 1, 5 ) 4 ( 6, 3 ) ( 0, 2 ) Intercepts are fractions…To find a point, use multiples of 2 and 5 to get a sum that is 27. 2 + 25 = 27. Therefore x = 1 and y = 5 for this to work. Find the slope and graph the line. 2 ( 3, 0 ) 2 4 6 8 10 12 Min: Points closest to the origin. ( 0, 2 ): z = 5(0) + 7(2) = 14 ( 3, 0 ): z = 5(3) + 7(0) = 15 ( 0, 2 ) is the point for min.