Video: Pack 3 β€’ Paper 2 β€’ Question 19

Pack 3 β€’ Paper 2 β€’ Question 19

02:09

Video Transcript

Show that 𝑒 over 𝑣 minus one squared plus 𝑒 over 𝑣 minus one can be simplified to 𝑒𝑣 over 𝑣 minus one squared, with 𝑣 is not equal to one.

This may look nasty. But, really, it’s asking us to add two fractions together. It just so happens that these fractions contain algebra. Let’s think about how we would normally add fractions.

We’d first find a common denominator and then we create equivalent fractions before finally adding the numerators. In fact, we need to follow roughly the same process here. Notice how both of the dominators contain 𝑣 minus one. We can multiply the denominator of the second fraction by 𝑣 minus one to get 𝑣 minus one squared. That’s our common denominator.

If we do that, we must also multiply the numerator of the second fraction by 𝑣 minus one. That gives us 𝑒 over 𝑣 minus one squared β€” that first fraction hasn’t changed β€” plus 𝑒 multiplied by 𝑣 minus one all over 𝑣 minus one squared.

Notice how the denominator in our answer now looks like the denominator required of us in the question. A sensible next step is to expand these brackets. Remember we can expand a single bracket by multiplying the term on the outside by everything on the inside: 𝑒 multiplied by 𝑣 is 𝑒𝑣 and 𝑒 multiplied by negative one is negative 𝑒. The second fraction, therefore, becomes 𝑒𝑣 minus 𝑒 all over 𝑣 minus one squared.

Now that the denominators are the same, we can add the numerators. That gives us 𝑒 plus 𝑒𝑣 minus 𝑒 all over 𝑣 minus one squared. We should then try to simplify by collecting like terms. 𝑒 minus 𝑒 is simply zero.

Our expression simplifies to 𝑒𝑣 over 𝑣 minus one all squared as required.

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