# Video: AQA GCSE Mathematics Higher Tier Pack 1 • Paper 2 • Question 25

AQA GCSE Mathematics Higher Tier Pack 1 • Paper 2 • Question 25

06:41

### Video Transcript

Solve the following equation: 𝑥 over three plus 𝑥 minus 14 over two 𝑥 minus five is equal to two. Give your solutions to two decimal places. You must show your working.

Let’s remind ourselves how we add fractions. Usually, we find the lowest common denominator of the two fractions. We don’t know the lowest common denominator for these two fractions since they’re algebraic. But we can find a common denominator.

To find a common denominator, we find the product of the two denominators. In this case, a common denominator is three multiplied by two 𝑥 minus five.

To expand these brackets, we need to multiply each term on the inside by the three on the outside. Three multiplied by two is six. So three multiplied by two 𝑥 is six 𝑥. Three multiplied by five is 15. So three multiplied by negative five is negative 15. And our common denominator is six 𝑥 minus 15.

Remember though when we create the common denominator, we need to ensure that we’re also multiplying the numerator to form equivalent fractions. We multiply the denominator of the first fraction by two 𝑥 minus five. So we need to do that to the numerator. That gives us 𝑥 multiplied by two 𝑥 minus five all over six 𝑥 minus 15.

Once again, we can expand by multiplying each term inside this bracket by the 𝑥 on the outside. 𝑥 multiplied by two 𝑥 is two 𝑥 squared and 𝑥 multiplied by negative five is negative five 𝑥.

Similarly, we multiply the denominator of the second fraction by three. So we also need to multiply the numerator by three. That’s three multiplied by 𝑥 minus 14 all over six 𝑥 minus 15.

Once again, we expand the brackets on the numerator as before. And we get three 𝑥 minus 42 all over six 𝑥 minus 15.

Now that we have two fractions with the same denominator, we can add their numerators. We have two 𝑥 squared minus five 𝑥 over six 𝑥 minus 15 plus three 𝑥 minus 42 over six 𝑥 minus 15. That’s two 𝑥 squared minus five 𝑥 plus three 𝑥 minus 42 all over six 𝑥 minus 15.

And of course, we can simplify the numerator by collecting like terms. Negative five 𝑥 plus three 𝑥 is negative two 𝑥. And the expression on the left-hand side of our original equation becomes two 𝑥 squared minus two 𝑥 minus 42 all over six 𝑥 minus 15.

Let’s put this back into the original equation. We’re going to need to clear some space. We have two 𝑥 squared minus two 𝑥 minus 42 over six 𝑥 minus 15 is equal to two.

Now, this doesn’t look particularly nice. But we can get rid of the nasty looking denominator on the left-hand side by multiplying through by six 𝑥 minus 15.

Remember if we do that, we cancel out the denominator on the left-hand side. We don’t then need to multiply the numerator by six 𝑥 minus 15. So we’re just left with two 𝑥 squared minus two 𝑥 minus 42. And that’s equal to two multiplied by six 𝑥 minus 15. Expanding those brackets like we have been, we get 12𝑥 minus 30 on the right-hand side of this equation.

Now, you might have noticed that we have a quadratic expression on the left-hand side. We can’t solve this until we make sure that we have a quadratic equation that’s equal to zero.

So let’s begin by subtracting 12𝑥 from both sides. That gives us two 𝑥 squared minus 14𝑥 minus 42 is equal to negative 30. We’re then gonna add 30 to both sides of the equation. And we get two 𝑥 squared minus 14𝑥 minus 12 is equal to zero.

Notice how each term in this equation is a multiple of two. So we can divide everything by two. And if we do that, we get 𝑥 squared minus seven 𝑥 minus six is equal to zero.

Now that our quadratic equation is in the right form, we can solve it. We usually try to factorise. But there is a hint that this might not be suitable in the question. We’re told to give our solutions to two decimal places.

So that tells us that the expression 𝑥 squared minus seven 𝑥 minus six will not be factorable. Instead, we’ll need to use the quadratic formula and this is something you need to know by heart.

For a quadratic equation of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero, 𝑥 has two solutions. It’s negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎.

Let’s compare our equation to the general form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. 𝑎 is the coefficient of 𝑥 squared. It’s the number of 𝑥 squareds we have, which in this case is one. 𝑏 is the coefficient of 𝑥. It’s the number of 𝑥s we have. In this case, that’s negative seven. And 𝑐 is the constant. Here, that’s negative six.

Let’s substitute these into the formula for the quadratic equation. Doing so and we get negative negative seven plus or minus the square root of negative seven squared minus four multiplied by one multiplied by negative six all over two multiplied by one.

You’ll notice that all the negative numbers are in brackets here. That’s particularly important for the negative seven squared. It ensures your calculator knows it’s squaring a negative number.

We can split it up and type in the addition and subtraction part as shown. But it is sensible to evaluate each part first. Negative negative seven is seven, negative seven squared is 49, and negative four multiplied by one multiplied by negative six is 24. So we have seven plus or minus the square root of 49 plus 24 over two.

Let’s calculate the addition part first. If we type this into our calculator, we get 7.7720. And then, the second solution for 𝑥 is the subtraction part. And that gives us negative 0.7720.

Remember we need to give our answers to two decimal places. In each number, the second digit after the decimal point is a seven. We look to the digit immediately to the right of this. This is called the deciding digit.

Since in both cases, the deciding digit is two, it’s less than five, we round the number down. And we’ve solved the equation. Correct to two decimal places, 𝑥 is equal to 0.77 or negative 0.77.