Question 201759: X^2-3x-4/X^2-16 / 3x+3/X^2+9x+20 please help thank you inadvance
Found 2 solutions by RAY100, jsmallt9:
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! ( x^2 -3x -4) / (x^2-16) / (3x+3) / (x^2 +9x+20)
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{ (x^2 -3x-4) / (x^2 -16)} * { (x^2 +9x +20)/ (3x+3) } ,,,invert and multiply
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factor
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{ (x+1)(x-4) / (x+4)(x-4) } * { (x+5)(x+4) / 3 (x+1) }
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simply by canceling like terms
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(x+5) / 3
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check, let x=0,, original and answer = 5/3,,,ok
Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website! Please put parentheses around each numerator and each denominator. Without them there are way too many ways to interpret:
X^2-3x-4/X^2-16 / 3x+3/X^2+9x+20
I'll give you some general tips:- If division of fractions is indicated, change the division to multiplying by the reciprocal of the second fraction
- Factor the numerators and denominators.
- See if the fractions will reduce by canceling common factors (remembering to "cross cancel" if there fractions being multiplied).
- If fractions need to be added or subtracted
- You will need common denominators of course.
- The Lowest Common Denominator (LCD) will be the product of all the different factors in the various denominators.
- To turn each denominator into the LCD, multiply the numerator and denominator of each fraction by whatever factors of the LCD are "missing" from its denominator, if any.
- Multiply out the numerators but leave the denominators factored (so we can reduce the answer later more easily).
- Now that the denominators are the same, add or subtract the numerators.
- Factor the new numerator, if possible
- Reduce your final fraction by canceling common factors, if any
- After reducing, multiply out any numerators or denominators are still factored
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