Video: Dividing Algebraic Expressions

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

02:12

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‘Ž over 𝑏 divided by 𝑐 over 𝑑, this is equal to π‘Ž over 𝑏 multiplied by 𝑑 over 𝑐, so the reciprocal of the second fraction, which is equal to π‘Žπ‘‘ over 𝑏𝑐. You might also remember the skill using the phrase β€œkeep it, change it, flip it.” 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 that’s because that’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, we’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 we’re left with is one over one multiplied by five multiplied by one over one which neatly leaves us with a result of five.

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