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3 years ago

Which of these strategies would eliminate a variable in the system of equations?

Mathematics

1 answer:

BartSMP [9]3 years ago

3 0

Answer:

what are the strategies

Step-by-step explanation:

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What is the difference? StartFraction 2 x + 5 Over x squared minus 3 x EndFraction minus StartFraction 3 x + 5 Over x cubed minu

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Answer:

$\frac{(x+5)(x+2)}{x(x-3)(x+3)}$

Step-by-step explanation:

The given expression is

$\frac{2x+5}{x^{2} -3x} -\frac{3x+5}{x^{3} -9x} -\frac{x+1}{x^{2}-9}$

First, we need to factor each denominator

$\frac{2x+5}{x(x-3)} -\frac{3x+5}{x(x+3)(x-3)} -\frac{x+1}{(x-3)(x+3)}$

So, the least common factor (LCF) is $x(x-3)(x+3)$ , because they are the factors that repeats.

Now, we diviide the LCF by each denominator, to then multiply it by each numerator.

$\frac{(x+3)(2x+5)-3x-5-x(x+1)}{x(x-3)(x+3)} =\frac{2x^{2}+5x+6x+15-3x-5-x^{2}-x }{x(x-3)(x+3)}\\\frac{x^{2}+7x+10}{x(x-3)(x+3)}$

Then, we factor the numerator, to do so, we need to find two numbers which product is 10 and which sum is 7.

$\frac{(x+5)(x+2)}{x(x-3)(x+3)}$

Therefore, the expression is equivalent to

$\frac{(x+5)(x+2)}{x(x-3)(x+3)}$

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