Asked by Lesley
Subtracting
,division and multiplication of fractions
A. 16p^2-4q^2/(6p^2-7pq+2q^2) *6p^2-13pq+6q^2/(4p^2-4pq-3q^2)
B. a^2+a-2/(6a-12) / a+2/(a^2-3a+2)
C.3/(x-3) -x^2/(6+x-x^2) -2/(x+2)
D.x+1/(x^3-1) - 1/(x^2+x+1)
Key
,division and multiplication of fractions
A. 16p^2-4q^2/(6p^2-7pq+2q^2) *6p^2-13pq+6q^2/(4p^2-4pq-3q^2)
B. a^2+a-2/(6a-12) / a+2/(a^2-3a+2)
C.3/(x-3) -x^2/(6+x-x^2) -2/(x+2)
D.x+1/(x^3-1) - 1/(x^2+x+1)
Key
Answers
Answered by
Reiny
You will need brackets in the numerator as well
in A
B appears to have a typo
C) notice the middle term denominator factors and it becomes the LCD
3/(x-3) - x^2/((3-x)(2+x)) - 2/(x+2)
Do the others in the same way
hint for D
after you insert the brackets in the top you get
(x+1)/((x-1)(x^2 + x + 1) - 1/(x^2 + x + 1)
so x^3-1 actually becomes the common denominator
= -3/(3-x) - x^2/((3-x)(2+x)) - 2/(x+2)
= (-3(2+x) - x^2 - 2(3-x) )/((3-x)(2+x))
= ( -6 - 3x - x^2 - 6 + 2x)/((3-x)(2+x))
= (-x^2 - x - 12)/((3-x)(2+x))
= (x^2 + x + 12)/((x-3)(2+x))
in A
B appears to have a typo
C) notice the middle term denominator factors and it becomes the LCD
3/(x-3) - x^2/((3-x)(2+x)) - 2/(x+2)
Do the others in the same way
hint for D
after you insert the brackets in the top you get
(x+1)/((x-1)(x^2 + x + 1) - 1/(x^2 + x + 1)
so x^3-1 actually becomes the common denominator
= -3/(3-x) - x^2/((3-x)(2+x)) - 2/(x+2)
= (-3(2+x) - x^2 - 2(3-x) )/((3-x)(2+x))
= ( -6 - 3x - x^2 - 6 + 2x)/((3-x)(2+x))
= (-x^2 - x - 12)/((3-x)(2+x))
= (x^2 + x + 12)/((x-3)(2+x))
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