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| Question 188087This question is from textbook Introductory and Intermediate Algebra
 :  Hi i'm trying to work on my assignments again.I'm trying to find out if I did the right thing on first problem & please help me with the others.
 1) Add or subtract as indicated.SImplify the result if possible.
 [x/(x^2-10x+25)] - [(x-4) / (2x-10)]
 [x /(x-5)(x-5)] - [(x-4)/ 2(x-5)
 then I multiplied 1st equation by 2/2 & 2nd equation by x-5
 [x /(x-5)(x-5)x 2/2] - [(x-4)/ 2(x-5) x (x-5)/(x-5)]
 so
 [2.x /2(x-5)(x-5)] - [(x-4)(x-5)/ 2(x-5)(x-5)]
 since denominators are same now
 2x-x^2+5x+4x-20 / 2(x-5)(x-5)
 -x^2+11x-20 / 2(x-5)(x-5)
 I guess i cannot simplify it more.Did I do the right thing?
 
 2) SOlve each rational equation. If an equation has no solution, so state.
 [(x-3) /(x-2)] + [(x+1) /(x+3)] = [(2x^2 - 15) / (x^2 +x -6)]
 
 3) Use a rational equations to solve. Each problem involving motion.
 * A pool can be filled by one pipe in 3hours and by a second pipe in 6hours.How long will it take using both pipes to fill the pool?
 
 
 This question is from textbook Introductory and Intermediate Algebra
 
 Answer by stanbon(75887)
      (Show Source): 
You can put this solution on YOUR website! 1) Add or subtract as indicated.SImplify the result if possible. [x/(x^2-10x+25)] - [(x-4) / (2x-10)]
 [x /(x-5)(x-5)] - [(x-4)/ 2(x-5)
 lcd = 2(x-5)^2
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 Rewrite each fraction with the lcd as its denominator:
 [2x/lcd] - [(x-4)(x-5)/lcd]
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 Combine the numerators over the lcd:
 [2x- (x^2-9x+20)]/lcd
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 Simplify:
 [-x^2 + 11x - 20]/[2(x-5)^2]
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 Your answer is correct.
 ============================================
 2) SOlve each rational equation. If an equation has no solution, so state.
 [(x-3) /(x-2)] + [(x+1) /(x+3)] = [(2x^2 - 15) / (x^2 +x -6)]
 Multiply both sides by (x-2)(x+3) to get:
 [(x^2-9) + (x-2)(x+1)] = 2x^2-15
 Simplify:
 [x^2-9 + x^2-x-2] = 2x^2-15
 -11 - x = -15
 x = 4
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 3) Use a rational equations to solve. Each problem involving motion.
 A pool can be filled by one pipe in 3 hours and by a second pipe in 6hours.How long will it take using both pipes to fill the pool?
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 1st pipe DATA:
 Time = 3 hrs/job ; rate = 1/3 job/hr
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 2nd pipe DATA:
 Time = 6 hrs/job ; rate = 1/6 job/hr
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 together DATA:
 Time = x hrs/job ; rate = 1/x job/hr
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 Equation:
 rate + rate = together rate
 1/3 + 1/6 = 1/x
 Multiply thru by 6x to get:
 2x + x = 6
 3x = 6
 x = 2 hrs (time for the pipes to do the job together)
 ==========================================================
 Cheers,
 Stan H.
 
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