# Video: GCSE Mathematics Foundation Tier Pack 5 • Paper 2 • Question 13

GCSE Mathematics Foundation Tier Pack 5 • Paper 2 • Question 13

06:21

### Video Transcript

Part a) Finish the table of values for 𝑦 equals three over two 𝑥 plus two. Part b) Draw the graph of 𝑦 equals three over two 𝑥 plus two for values of 𝑥 between negative two and two. Part c) Use your graph to find the value of 𝑥 when 𝑦 equals 2.3.

Our equation 𝑦 equals three over two 𝑥 plus two will be a straight line as it is in the form 𝑦 equals 𝑚𝑥 plus 𝑐. 𝑚 is the gradient, in this case three over two, and 𝑐 is the 𝑦-intercept, in this case two.

The gradient tells us how far in the 𝑦 direction we move for every one unit in the 𝑥 direction. In this case, for every one unit in the 𝑥 direction we move three over two or 1.5 units in the 𝑦 direction. The intercept tells us where our straight line crosses the 𝑦-axis, in this case plus two. The values of 𝑥 in the table are increasing by one.

As our gradient is three over two or 1.5, this means that our 𝑦-values must increase by 1.5. Every time we add one to the 𝑥-values, we need to add 1.5 to the 𝑦-values. Negative one plus 1.5 is equal to 0.5, 0.5 plus 1.5 is equal to two, and two plus 1.5 equals 3.5. It is also worth checking that 3.5 plus 1.5 equals five, which it does.

An alternative method to calculate the missing 𝑦 numbers in the table would be to substitute the corresponding 𝑥-values into the equation.

Substituting 𝑥 equals negative one into the equation 𝑦 equals three over two 𝑥 plus two gives us three over two multiplied by negative one plus two. Three over two multiplied by negative one is negative three over two or negative 1.5. Adding two to this gives us 0.5. This means that our first value is correct.

Next, we can substitute zero into the equation. This gives us three over two multiplied by zero plus two. Anything multiplied by zero equals zero. Adding two to this gives us an answer of two.

Finally, we can substitute 𝑥 equals one into the equation three over two multiplied by one plus two. This is equal to 3.5. Therefore, the missing 𝑦-values in the table was 0.5, two, and 3.5.

The second part of our question asks us to draw the graph of 𝑦 equals three over two 𝑥 plus two for 𝑥-values between negative two and two. On the 𝑦-axis, 10 squares represents one unit. Therefore, one square would equal 0.1 as one divided by 10 is equal to 0.1. Our scale on the 𝑥-axis is exactly the same. 10 squares equals one unit. So one little square will equal 0.1.

We now need to plot the five coordinates from the table: negative two, negative one; negative one, 0.5; zero, two; one, 3.5; and two, five. To plot the first point, we go along the corridor or the 𝑥-axis to negative two and down the stairs or down the 𝑦-axis to negative one. For the second point, we go along the 𝑥-axis to negative one and up the 𝑦-axis to 0.5. The third point is at zero, two — along the 𝑥-axis to zero and up the 𝑦-axis to two. We plot the two points one, 3.5 and two, five in the same way.

Joining these five points with a straight line shows the graph of 𝑦 equals three over two 𝑥 plus two between negative two and two on the 𝑥-axis. Notice at this point that our 𝑦-intercept is two. The graph crosses the 𝑦-axis at two. The gradient of the line is 1.5 or three over two. For every one unit we move on the 𝑥-axis, we go up 1.5 or three over two units on the 𝑦-axis.

The final part of our question asked us to use the graph to find the value of 𝑥 when 𝑦 is equal to 2.3. We find 2.3 on the 𝑦-axis. This is three small squares above two, go across until we hit our line, and then go vertically downwards. This gives us an answer of 0.2 as it is two little squares to the right of zero. When 𝑦 is equal to 2.3, 𝑥 is equal to 0.2.

We can check this answer by substituting 𝑥 equals 0.2 into our equation. If our answer is correct, we’ll get a 𝑦-value of 2.3. Three over two or 1.5 multiplied by 0.2 is 0.3. Adding two to this gives us an answer of 2.3. This means that our answer of 𝑥 equals 0.2 is correct.