Asked by Tim
Perform the following operations and prove closure. Show your work.
x over x+3 + x+2 over x+5
x+4 over x^2+5x+6 • x+3 over x^2−16
2 over x^2−9 − 3x over x^2−5x+6
x+4 over x^2−5x+6 / x^2−16 over x+3
Compare and contrast division of integers to division of rational expressions.
x over x+3 + x+2 over x+5
x+4 over x^2+5x+6 • x+3 over x^2−16
2 over x^2−9 − 3x over x^2−5x+6
x+4 over x^2−5x+6 / x^2−16 over x+3
Compare and contrast division of integers to division of rational expressions.
Answers
Answered by
mathhelper
Next time please type your problems in a more normal way
I will do the 3rd:
2/(x^2−9) − 3x/(x^2−5x+6)
step1: try to factor
= 2/((x+3)(x-3)) - 3x/((x-2)(x-3))
so the LCD is (x+3)(x-3)(x-2)
= (2(x-2) - 3x(x+3) )/( (x+3)(x-3)(x-2))
= (2x - 4 - 3x^2 - 3x) / ( (x+3)(x-3)(x-2))
= -(3x^2 + x + 4) / ( (x+3)(x-3)(x-2))
there should have been a restriction stated throughout, namely
x ≠ ± 3, 2
I will do the 3rd:
2/(x^2−9) − 3x/(x^2−5x+6)
step1: try to factor
= 2/((x+3)(x-3)) - 3x/((x-2)(x-3))
so the LCD is (x+3)(x-3)(x-2)
= (2(x-2) - 3x(x+3) )/( (x+3)(x-3)(x-2))
= (2x - 4 - 3x^2 - 3x) / ( (x+3)(x-3)(x-2))
= -(3x^2 + x + 4) / ( (x+3)(x-3)(x-2))
there should have been a restriction stated throughout, namely
x ≠ ± 3, 2
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