R E V I E W P R O B L E M S F O R C H A P T E R 2 49Consider finally some equations involving fractions. Recall that the fraction a/b is notdefined if b = 0. If b 0, however, then a/b = 0 is equivalent to a = 0.E X A M P L E 3 Solve the following equations:(a)1 – K2√1 + K2 = 0 (b)45 + 6r – 3r2(r4 + 2)3/2 = 0 (c)x2 – 5x√x2 – 25 = 0Solution:(a) The denominator is never 0, so the fraction is 0 when 1-K2 = 0, that is when K = ±1.(b) Again the denominator is never 0. The fraction is 0 when 45 + 6r – 3r2 = 0, that is3r2 – 6r – 45 = 0. Solving this quadratic equation, we find that r = -3 or r = 5.(c) The numerator is equal to x(x – 5), which is 0 if x = 0 or x = 5. At x = 0 thedenominator is √-25, which is not defined, and at x = 5 the denominator is 0. Weconclude that the equation has no solutions.P R O B L E M S F O R S E C T I O N 2 . 5Solve the equations in Problems 1–2: 1. (a) x(x + 3) = 0(d) √2x + 5 = 0(b) x3(1 + x2)(1 – 2x) = 0(e) x2 + 1x(x + 1)= 0(c) x(x – 3) = x – 3(f)x(x + 1)x2 + 1 = 0 ⊂SM ⊃2. (a) 5 + x2(x – 1)(x + 2)= 0 (b) 1 +2xx2 + 1 = 0(c)(x + 1)1/3 – 1 3x(x + 1)-2/3(x + 1)2/3 = 0 (d)xx – 1+ 2x = 0⊂SM ⊃3. Examine what conclusions can be drawn about the variables if: (a) z2(z – a) = z3(a + b), a 0(b) (1 + λ)µx = (1 + λ)yµ(d) ab – 2b – λb(2 – a) = 0(c)-λ1 – µ2 λ1 + µ=R E V I E W P R O B L E M S F O R C H A P T E R 2In Problems 1–2, solve each of the equations.1. (a) 3x – 20 = 16 (b) -5x + 8 + 2x = -(4-x) (c) -6(x – 5) = 6(2 – 3x)(d)4 – 2×3= -5-x (e)52x – 1=12 – x(f) √x – 3 = 650 C H A P T E R 2 / I N T R O D U C T O R Y T O P I C S I I : E Q U A T I O N S⊂SM ⊃2. (a) x – 3x – 4=x + 3x + 4(b)3(x + 3)x – 3– 2 = 9xx2 – 9 +27(x + 3)(x – 3)(c)2×3=2x – 33+5 x(d)x – 5x + 5– 1 =1 x–11x + 20×2 – 5×3. Solve the following equations for the variables specified:(a) x = 23(y – 3) + y for y (b) ax – b = cx + d for x(c) AK√L = Y0 for L (d) px + qy = m for y(e)11 + r– a11 + r+ b= c for r (f) P x(P x + Q)-1/3 + (P x + Q)2/3 = 0 for x4. Consider the macro model(i) Y = C + I¯ + G, (ii) C = b(Y – T ), (iii) T = tYwhere the parameters b and t lie in the interval (0, 1), Y is the gross domestic product (GDP),C is consumption, I¯ is total investment, T denotes taxes, and G is government expenditure.(a) Express Y and C in terms of I¯, G, and the parameters.(b) What happens to Y and C as t increases?⊂SM ⊃5. Solve the following equations for the variables indicated: (a) 3K-1/2L1/3 = 1/5 for K(c) p – abxb-(b) (1 + r/100)t = 2 for rρ + λb-ρ*-1/ρ = c for b 01 = 0 for x0 (d) )(1 – λ)a-6. Solve the following quadratic equations: (a) z2 = 8z(d) 12p2 – 7p + 1 = 07. Solve the following equations:(b) x2 + 2x – 35 = 0(e) y2 – 15 = 8y p2 + 5p – 14 = 0(f) 42 = x2 + x(a) (x2 – 4)√5 – x = 0 (b) (x4 + 1)(4 + x) = 0 (c) (1 – λ)x = (1 – λ)y8. Johnson invested $1500, part of it at 15% interest and the remainder at 20%. His total yearlyincome from the two investments was $275. How much did he invest at each rate?9. If 53x = 25y+2 and x – 2y = 8, then what is x – y?HARDER PROBLEM10. Solve the following systems of equations:(a)2 x+3 y= 43 x–2 y= 19(b)3√x + 2√y = 22√x – 3√y = 14(c)x2 + y2 = 134×2 – 3y2 = 24

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