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# How would i solve this equation x+4/x^3+8+x+2/x^2-2x+4=11/2x+4.

Topic: How would i solve this equation x+4/x^3+8+x+2/x^2-2x+4=11/2x+4.
July 18, 2019 / By Hyacinth
Question: x+4/x^3+8+x+2/x^2-2x+4=11/2x+4 is a problem I have and i was wondering if someone could help me to understand the steps needed to solve it. The more work the better, as this isnt for a test or anything, these are just a type of problem that has shown up a lot in my school work and i could use a hand understanding them. THank you

## Best Answers: How would i solve this equation x+4/x^3+8+x+2/x^2-2x+4=11/2x+4.

Edwyna | 4 days ago
Please use parentheses. I assume that you mean: ... (x+4) / (x^3+8) + (x+2) / (x^2-2x+4) = 11 / (2x+4) or (x+4) / [ (x+2)(x^2-2x+4) ] + (x+2) / (x^2-2x+4) = 11/(2x+4) or [ (x+4) + (x+2)(x+2) ] / [ (x+2)(x^2-2x+4) ] = 11/(2x+4) or [ x^2 + 5x + 8 ] / [ (x+2)(x^2-2x+4) ] = 11/(2x+4) or 2 ( x^2 + 5x + 8 ) (x+2) = 11 (x+2) (x^2 - 2x + 4) or 2 ( x^2 + 5x + 8 ) = 11 (x^2 - 2x + 4) ← x ≠ -2 or 9x^2 - 32x + 28 = 0 or (9x - 14) (x - 2) = 0 or x = { 14/9, 2 }
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Originally Answered: Graphing equation question? how would solve this linear equation?
2x + 3(11/3) = 12 3/1 * 11/3 = 33/3 When multiplying, just multiply across, numerator to numerator and denominator to denominator. (I know I put a 1 under the 3, all whole numbers have a 1 under them) (example : 3/1 = 3) or ( lets say 5 in fraction form is 5/1 ) (Another fraction example 2/5 * 3/4 mult across and you get 2 * 3 = 6 and 5 * 4 = 20 answer 6/20 then, simplify and get 3/10) Now simplify 33/3 = 11 2x + 11 = 12 Now subtract 11 from both sides. and you get 2x = 1 2x = 1 Now divide 2 on both sides and you get x = 1/2 ordered pair is (1/2 , 11/3)

Charlie
your question is completely ambiguous without parentheses to show what's the top and what's the bottom of the fractions. I think you might mean (x + 4)/(x³ + 8) + (x+2)/(x²-2x+4) = 11/(2x+4) in which case you need to know that x³ + 8 = (x + 2)(x² - 2x + 4) and so the lcd will be 2(x + 2)(x² - 2x + 4) we multiply all 3 fractions by that and cancel out all the denominators, leaving 2(x + 4) + 2(x + 2)(x + 2) = 11(x² - 2x + 4) 2x + 8 + 2x² + 8x + 8 = 11x² - 22x + 44 2x² + 10x + 16 = 11x² - 22x + 44 0 = 9x² - 32x + 28 now we look for factors of 9•28 = 252 that add up to -32, so 14 and 18 9x² - 18x - 14x + 28 = 0 9x(x - 2) - 14(x - 2) = 0 (9x - 14)(x - 2) = 0 so x = 14/9 or x = 2 make sure you check each solution; you can't use one that would make a denominator 0.
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Ariel
Factor first! And USE PARENTHESES!! (x + 4)/(x^3 + 8) + (x + 2)/(x^2 + 2x - 4) = 11/(2x + 4) (x + 4)/([x + 2)(x^2 - 2x + 4)] + (x + 2)/(x^2 - 2x + 4) = 11/[2(x + 2)] 2(x + 4) + 2(x + 2)(x + 2) = 11(x^2 - 2x + 4) 2x + 8 + 2x^2 + 8x + 8 = 11x^2 - 22x + 44 0 = 9x^2 - 32x + 28 0 = (9x - 14)(x - 2) 9x - 14 = 0 or x - 2 = 0 9x = 14..............x = 2 x = 14/9
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