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What is the simplifie form of : [2x/3(x+3)]/[7x/(x+3)(x-5)]. I know that inorder to divide radicals you flip the second denominator and numerator so the new equation would be : [2x/3(x+3)]*[(x+3)(x-5)/7x] However I do not know how to simplify the rest . Help ?
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cancel out the (x+3)s [2x/3]*[(x-5)/(7x)] cancel out the xs [2/3][(x-5)/7] multiply them together: 2*(x-5)/21
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\[\frac{2x}{3(x+3)}\div\frac{7x}{(x+3)(x-5)}=\frac{2x}{3(x+3)}\times\frac{(x+3)(x-5)}{7x}\]
Now cancel x and x+3 and get: \[\frac{2(x-5)}{21}\]
[2x/3(x+3)]/[7x/(x+3)(x-5)]. = [2x/3(x+3)] * (x+3)(x-5)/ 7x] = 2 ( x- 5) / 21
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