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Unformatted text preview: Section 1.5 Equations Linear Equations EXAMPLES: 1. Solve the equation 7 x 4 = 3 x + 8 Solution: We have 7 x 4 = 3 x + 8 (7 x 4) + 4 = (3 x + 8) + 4 7 x = 3 x + 12 7 x 3 x = (3 x + 12) 3 x 4 x = 12 1 4 4 x = 1 4 12 x = 3 2. Solve for M the equation F = G mM r 2 . Solution: We have F = ( Gm r 2 ) M = ( r 2 Gm ) F = ( r 2 Gm )( Gm r 2 ) M = r 2 F Gm = M The solution is M = r 2 F Gm . 3. The surface area A of the closed rectangular box can be calculated from the length l, the width w, and the height h according to the formula A = 2 lw + 2 wh + 2 lh Solve for w in terms of the other variables in this equation. 1 3. The surface area A of the closed rectangular box can be calculated from the length l, the width w, and the height h according to the formula A = 2 lw + 2 wh + 2 lh Solve for w in terms of the other variables in this equation. Solution: We have A = (2 lw + 2 wh ) + 2 lh A 2 lh = 2 lw + 2 wh A 2 lh = (2 l + 2 h ) w A 2 lh 2 l + 2 h = w The solution is w = A 2 lh 2 l + 2 h . Quadratic Equations EXAMPLES: 1. Solve the equation x 2 + 5 x = 24 . Solution: We have x 2 + 5 x = 24 x 2 + 5 x 24 = 0 ( x 3)( x + 8) = 0 x 3 = 0 or x + 8 = 0 x = 3 or x = 8 The solutions are x = 3 and x = 8 . 2. Solve each equation: (a) x 2 = 5 (b) ( x 4) 2 = 5 2 2. Solve each equation: (a) x 2 = 5 (b) ( x 4) 2 = 5 Solution: (a) We have x = 5 . (b) We have ( x 4) 2 = 5 x 4 = 5 x = 4 5 The solutions are x = 4 5 and x = 4 + 5 . 3. Solve each equation: (a) x 2 8 x + 13 = 0 (b) 3 x 2 12 x + 7 = 0 3 3. Solve each equation: (a) x 2 8 x + 13 = 0 (b) 3 x 2 12 x + 7 = 0 Solution: (a) We have x 2 8 x + 13 = 0 x 2 8 x = 13 x 2 8 x + 16 = 13 + 16 ( x 4) 2 = 3 x 4 = 3 x = 4 3 The solutions are x = 4 3 and x = 4 + 3 ....
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This note was uploaded on 01/23/2012 for the course MATH 8650 taught by Professor Kiryltsishchanka during the Spring '12 term at NYU.
 Spring '12
 KIRYLTSISHCHANKA
 Calculus, Algebra, Linear Equations, Equations

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