# Video: Pack 5 β’ Paper 3 β’ Question 14

Pack 5 β’ Paper 3 β’ Question 14

05:46

### Video Transcript

The following equation can be written as a single fraction in the form ππ₯ plus π over π₯ squared minus 25, where π and π are integers. Calculate the value of π and π.

In this question, we should keep an eye on the form we need. However, we shouldnβt get too hung up on it until the very end. Instead, itβs useful to think what we do know about this equation.

We have two fractions which weβre looking to add together and an integer which weβre going to subtract. So itβs sensible to start by finding a common denominator. We can find the common denominator by multiplying the denominators of the two equations that we have. That gives us π₯ plus five multiplied by π₯ minus five.

What weβll do next is multiply the numerator and the denominator of the first fraction by π₯ minus five. Remember, weβre allowed to do this because π₯ minus five over π₯ minus five is just one. Weβre finding an equivalent fraction.

To create that common denominator for our second fraction, weβll multiply both the numerator and the denominator by π₯ plus five. Once again, this is just the same as multiplying by one and creating an equivalent fraction.

Itβs slightly different with the integer. Itβs useful to start by writing the integer as a fraction. Three is the same as three over one. Then we can multiply both the numerator and the denominator by π₯ plus five multiplied by π₯ minus five. Once again, this is just the same as multiplying by one.

At this stage, itβs really useful to rewrite your expressions slightly by putting them into brackets. This will help you remember that, next, you need to expand the brackets.

Letβs start by expanding the denominator. This is actually the difference of two squares, which is a process you can learn off by heart. Multiplying the first expression in each bracket gives us π₯ squared. Next, we multiply the outer part of each expression. π₯ multiplied by negative five is negative five π₯. Multiplying the inner part of each expression gives us five multiplied by π₯, which is five π₯. And multiplying the last part of each expression is five multiplied by negative five, which is negative 25.

Notice how negative five π₯ plus five π₯ is zero. And this part of the expression cancels out. Weβre left with π₯ squared minus 25. This process is called the difference of two squares because we end up with two square numbers, which we are subtracting one from the other. In maths, the difference means subtraction. I can now change the denominator of each expression to π₯ squared minus 25. Notice the final fraction is being multiplied by one, which is simply π₯ squared minus 25.

Now letβs expand the top part of each expression. Two π₯ multiplied by π₯ is two π₯ squared. Two π₯ multiplied by negative five is negative 10π₯. Seven multiplied by π₯ is seven π₯. And seven multiplied by negative five is negative 35. We can simplify this a little further by combining negative 10π₯ plus seven π₯. And our first fraction becomes two π₯ squared minus three π₯ minus 35 all over π₯ squared minus 25.

Repeating this process for the second fraction, π₯ multiplied by π₯ is π₯ squared. π₯ multiplied by five is five π₯. Negative four multiplied by π₯ is negative four π₯. And negative four multiplied by five is negative 20. Once again, we can collect together like terms by adding five π₯ minus four π₯, which is just π₯.

Remember, we already worked out that π₯ plus five multiplied by π₯ minus five was π₯ squared minus 25. So the numerator of this final fraction becomes three lots of π₯ squared minus 25. We can simplify this further by multiplying both parts of the bracket by three, giving us three π₯ squared minus 75.

Now letβs substitute each fraction back into the original expression. Once we have some fractions which have a common denominator, we can apply the addition or subtraction operation to the numerators. Notice how weβre subtracting the final fraction. So weβll need to be really careful when it comes to our signs. Letβs combine this into one fraction.

Since weβre subtracting the final fraction, we end up with negative three π₯ squared. But we also end up doing minus negative 75, which means our fraction becomes plus 75. Two π₯ squared plus π₯ squared minus three π₯ squared is zero. So the π₯ squareds disappear. Negative three π₯ plus π₯ is negative two π₯. And negative 35 minus 20 plus 75 is positive 20.

Remember, the question was asking us to calculate the value of π and π. So itβs not enough just to leave our answer in its fraction form. Referring back to the question, π is the coefficient of π₯, the number in front of π₯, which here is negative two. π is the constant, which is 20.