# Question Video: Dividing Algebraic Expressions Mathematics

Simplify 2π₯/(π₯ + 3) Γ· 2π₯/(5π₯ + 15).

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### Video Transcript

Simplify two π₯ over π₯ plus three divided by two π₯ over five π₯ plus 15.

So to solve this problem, the first thing we need to do is remind ourselves of how we divide fractions. So, if we have π over π divided by π over π, this is equal to π over π multiplied by π over π, so the reciprocal of the second fraction, which is equal to ππ over ππ. You might also remember the skill using the phrase βkeep it, change it, flip it.β Keep the first fraction, change the sign to multiply, and flip the second fraction. So, in our question, what this means is that our expression is gonna become two π₯ over π₯ plus three multiplied by five π₯ plus 15 over two π₯. And thatβs because thatβs reciprocal of two π₯ over five π₯ plus 15.

So, the next thing we always look to do in this type of problem is take a look at numerators and denominators and see if any of them could be factored. And in fact, yes, in this problem we can see that the right-hand side numerator can be factored. And if we factor five π₯ plus 15, well, what we can do is take out five as a factor. So, when we do that, weβll have five multiplied by π₯ plus three. So now our expression is two π₯ over π₯ plus three multiplied by five multiplied by π₯ plus three over two π₯. Now, the purpose of this factoring is to see if we can uncover any common factors in the numerator and denominator. And here we have because we have π₯ plus three on the left-hand side is the denominator. And we have π₯ plus three on the right-hand side in our numerator. So, what this means is that we can divide through by that common factor, as we have done here.

But also, if we take a look at this question, we can also see another common factor. And that common factor is two π₯ because on the numerator on the left-hand side, we have two π₯ and on the denominator on the right-hand side we have two π₯. So, we divide through by this as well. So now, what weβre left with is one over one multiplied by five multiplied by one over one which neatly leaves us with a result of five.