Section_1.5-Equations

Section_1.5-Equations - Section 1.5 Equations Linear...

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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
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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
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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
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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 . (b) We have 3 x 2 - 12 x + 7 = 0 3 x 2 - 12 x = - 7 3( x 2 - 4 x ) = - 7 x 2 - 4 x = - 7 3 x 2 - 4 x + 4 = - 7 3 + 4 ( x - 2) 2 = - 7 3 + 4 = - 7 3 + 12 3 = - 7 + 12 3 = 5 3 x - 2 = ± 5 3 x = 2 ± 5 3 The solutions are x = 2 - 5 3 and x = 2 + 5 3 .
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