Module 6: Notes and Solutions to Select Practice Problems. The examples in the Course Guide are very much like the problems in the homework so you should look at those as well as the solutions here. These are solutions for Practice Problems-Unit 1. (3) -a - 3 = -10 You should perform opposite/inverse operations until you get the variable alone on one side of the equal sign. The opposite of -3 is +3 so: -a - 3 +3 = -10 +3 -a = -7 The variable is not alone yet. There is a minus sign next to it so we are not finished. You can get rid of the minus sign by multiplying both sides of the equation by -1. (-1)(-a) =( -7)(-1) Thus, a = 7 Check: -a - 3 = -10 so -7 - 3 = -10 must be true.√ (14) 3e 9 5 There are two ways to solve this equation. 1) There are two operations here. The most 2) There is a fraction in front of the obvious is probably the division by 5. The variable. Multiply a fraction by it's opposite of division is multiplication so reciprocal (Check out Module 5 in the multiply everything by 5. Course Guide if you want more info on this topic) and you get 1. Multiplying by 1 gives you the variable by itself. Since this 5 3e 95 is an equation, you must multiply 5 everything by the reciprocal. -3e = -45 5 3e 5 9 The only operation left is multiplication by 3 5 3 (-3). The opposite of multiplication is division so divide everything by (-3) As you can see, the variable is alone now and all that's left is the simplification of the right-hand side (rhs) of the equation. 3e 45 3 3 9 5 45 e 15 Thus e = 15 3 3 Check: 3e 3(15) 45 9 so 9 5 5 5 (39) -14 = 5p + 18 Don't be fooled by the appearance of the problem. This one could just as easily been written as 5p + 18 = -14. Anyway, the solution of the original problem follows. -14 - 18 = 5p + 18 -18 - 32 5p so p = -32/5 5 5 32 14 5 18 Check: -14 = 5p + 18 so must be true. √ 5 14 32 18 (47) 7 3x 2 4 Here's another one that looks backwards. 3x 2 4 3x 72 22 4 3x 9 4 4 3x 4 9 3 4 3 x 12 7 Check: 7 3x 312 2 so 7 2 must be true. 4 4 36 7 2 √ 4 7 92 (60) 2b - 9 = 5b This one has one variable in two different places. You have to get them together to solve the equation. Adding the opposite of either term to both sides of the equation will do it. 2b - 2b - 9 = 5b - 2b -9 = 3b 9 3b so b = -3 3 3 Check: 2b - 9 = 5b so 2(-3) - 9 = 5(-3) must be true. -6 -9 = -15 √ (68) y - (2 + 2y) = 6 You need to combine like terms so you have to remove the parenthesis. Remember to distribute the negative sign. y - (2 + 2y) = 6 y - 2 - 2y = 6 -2 - y = 6 -2 + 2 - y = 6 + 2 -y = 8 so y = -8 Check: y - (2 + 2y) = 6 so -8 - (2 + 2[-8]) = 6 must be true. -8 - (2 + [-16]) = 6 -8 - (-14) = 6 -8 + 14 = 6 √ (74) 3( x 4) 9 2 I don't really like working with fractions so I usually get rid of them when I can so multiply everything by 2 first. 2 3( x 4) 92 2 We now have the equation: -3(x - 4) = -18 -3x + 12 = -18 -3x + 12 - 12 = -18 - 12 -3x = -30 x = 10 Check: 3( x 4) 3(10 4) 9 so 9 must be true. 2 2 3(10 4) 9 2 30 12 9 √ 2 18 9 2 Notes for Unit 2. There's always a lot of buzz about the Quadratic Formula. b b 2 4ac where ax2 + bx + c = 0 and a 0 . x 2a It is just another formula that you have used in the calculator exercise from Module 2. Put the numbers in for a, b and c and follow the order of operations. The Quadratic Formula can be used to find the solutions to all quadratic equations (in standard form) so each solution below will use both methods (when applicable) and check the answers. 4. x2 + 5x + 6 = 0 This quadratic equation is factorable. You can also use the Quadratic Formula: x2 + 5x + 6 = 0 a = 1, b = 5, c = 6 (x + 2)(x + 3) = 0 x In order for this product to be equal to zero: 5 5 2 4(1)(6) 5 25 24 2(1) 2 5 1 5 1 2 2 5 1 4 x 2 2 2 5 1 6 x 3 2 2 Check: x2 + 5x + 6 = 0 so (-2)2 + 5(-2) + 6 = 0 and (-3)2 + 5(-3) + 6 = 0. x (x + 2) must equal zero so x = -2 (x + 3) must equal zero so x = -3 4 + (-10) + 6 = 0 √ 9 + (-15) + 6 = 0 √ 11. 16t2 - 25 = 0 This quadratic equation is factorable. 16t2 - 25 = 0 (4t - 5)(4t + 5) = 0 You can also use the Quadratic Formula: In order for this product to be equal to zero: 0 0 2 4(16)( 25) 1600 x 2(16) 32 40 5 x 32 4 40 5 x 32 4 a = 16, b = 0, c = -25 (4t - 5) = 0 4t - 5 + 5 = 0 + 5 4t = 5 so t = 5/4 OR (4t + 5) = 0 4t + 5 - 5 = 0 - 5 4t = -5 so t = -5/4 Check: 16t2 - 25 = 0 so 16(5/4)2 - 25 = 0 (Check is the same for x = -5/4 ) 16(25/16) - 25 = 0 25 - 25 = 0 √ 2x2 - x = 0 18. This quadratic equation is factorable. You can also use the Quadratic Formula: 2x2 - x = 0 so x(2x - 1) = 0 a = 2, b = -1 , c = 0 In order for this product to be equal to zero: (1) (1) 2 4(2)(0) 1 1 2 ( 2 ) 4 x=0 11 OR x 0 2x - 1 = 0 4 2x - 1 + 1 = 0 + 1 11 1 x 2x = 1 so x = 1/2 4 2 2 2 2 Check: 2x - x = 0 so 2(0) - 0 = 0 √ and 2(1/2) - (1/2) = 0 x 1 1 2 0 4 2 √ 1 1 0 2 2 26. 8x2 + 3x - 12 = 0 Here's one that cannot be factored (try it and you'll see) so you must use the Quadratic Formula: a = 8, b = 3, c = -12 x 3 32 4(8)( 12) 3 9 384 3 393 2(8) 16 16 3 393 3 393 OR 16 16 Check: I'm only going to do one of these since the math follows the same steps for both. 8x2 + 3x - 12 = 0 so Thus, x 2 3 393 3 393 3 12 0 8 16 16 3 393 3 393 3 393 3 12 0 2 16 16 9 3 393 3 393 393 9 3 393 12 32 032 32 16 402 6 393 2 9 3 393 12(32) 0 402 6 393 (18) 6 393 384 0 402 402 0 You are not required to do each problem using both methods but you should be able to use each method to solve a quadratic equation.

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# Module 6: