150_soln - Prerequisite Solutions Math 150 1 Solve for x...

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Prerequisite Solutions Math 150 1. Solve for x: ( ) ( ) 5 2 x 3 4 2 x 3 = + First, multiply by the least common denominator, which is 20: ( ) ( ) 5 2 x 3 20 4 2 x 3 20 = + () () 2 x 3 4 2 x 3 5 = + ( ) ( ) 2 x 12 2 x 15 = + Then, distribute and combine like terms: 24 x 12 30 x 15 = + 30 24 x 12 x 15 = 54 x 3 = 18 3 54 x = =
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2. Solve for r: nt ) n r 1 ( P B + = () () () () r n 1 P B n n r n 1 P B n r 1 P B 1 n r 1 1 P B n r 1 n r 1 P B n r 1 P B n r 1 P B P ) n r 1 ( P P B nt 1 nt 1 nt 1 nt 1 nt nt nt 1 nt 1 nt nt 1 nt nt = = = + = + = + = + = + = + =
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3. Solve for x: 162 x 36 x 2 2 = + First set the equation equal to zero by adding 162 to both sides; then we will be able to factor this to find the answers: 0 162 x 36 x 2 162 162 162 x 36 x 2 2 2 = + + + = + + Now we can divide both sides by 2. We want to do this because it is easier to work with smaller numbers, and we chose 2 because 2 is the greatest common factor to all of the terms: 0 81 x 18 x 2 0 2 162 x 36 x 2 2 2 = + + = + + Now we can factor the trinomial: ( )( ) 0 9 x 9 x = + + x + 9 = 0 or x + 9 = 0 So: x = -9 4. Solve for x and y: 0 y x 3 0 y 2 x = = First, we want to eliminate one of the terms by subtracting one equation from the
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150_soln - Prerequisite Solutions Math 150 1 Solve for x...

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