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Inessa [10]
3 years ago
11

I can’t figure out the solution I think it is y < -10

Mathematics
1 answer:
Olenka [21]3 years ago
4 0

Hey there!

  • <em>With negatives (-) you have to switch your symbol to the other way</em>
  • \bold{\frac{y}{-2}\leq5}
  • Firstly, substitute the \bold{y\ value  } as an invisible 1
  • So, now we have to flip the equation around
  • \bold{\frac{-1}{2}y\leq5}
  • We have to \bold{multiply} by \bold{-2} on each of your sides
  • \bold{-2\times\frac{-1}{2}\leq-2\times5}
  • \bold{Cancel \ out:2\times\frac{-1}{2}y \ because \ it \ equal \ 1}
  • \bold{Keep: -2\times5 \ because \ it \ helps \ us \ find \ our \ answer}
  • \bold{-2\times5=-10}
  • \boxed{\boxed{\bold{Answer:y\geq-10}}}\checkmark

Good luck on your assignment  and enjoy your day!

~\frak{LoveYourselfFirst:)}

<em />

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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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