Video: AQA GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 2 β€’ Question 11

a) Complete the table of values for 𝑦 = 4 βˆ’ 2π‘₯. b) Draw the graph of 𝑦 = 4 βˆ’ 2π‘₯ on the grid for the values of π‘₯ from βˆ’ 2 to 4. c) Solve π‘₯ = 4 βˆ’ 2π‘₯.

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Video Transcript

This question has three parts. Part a) Complete the table of values for 𝑦 equals four minus two π‘₯. Part b) Draw the graph of 𝑦 equals four minus two π‘₯ on the grid for the values of π‘₯ from negative two to four. And c) Solve π‘₯ equals four minus two π‘₯.

Starting with part a), we know that our 𝑦-values equal four minus two times our π‘₯-values. If we want to find 𝑦 when π‘₯ equals negative two, we substitute π‘₯ in our equation with the value negative two. By order of operations, we should multiply negative two times negative two first, which is positive four. In this case, 𝑦 equals four plus four. 𝑦 equals eight. Next, we want to consider what 𝑦 is if π‘₯ is zero. So we plug in zero for our π‘₯-value. Two times zero equals zero. And that means when π‘₯ equals zero, 𝑦 equals four. And we plugged that into our table. We could continue using the substitution method.

However, at this point, we also notice something else. Every time π‘₯ is increased by one, 𝑦 is decreased by two. And that means we can find our next value on the table by taking two and subtracting two. When π‘₯ equals two, 𝑦 equals zero. We can confirm this with substitution. Two times two equals four. And four minus four equals zero. Again, we can just subtract two. Zero minus two is negative two. And negative two minus two equals negative four. If you’re going to use this method, you need to check and make sure that your π‘₯-values are increasing by the same amount each time.

Now that we’ve completed our table, part b) wants us to draw the graph of 𝑦 equals four minus two π‘₯. We can use some of the points for our table to help us. We can use a few of the coordinates from our table to help us sketch this graph. When π‘₯ equals negative two, 𝑦 equals eight. When π‘₯ equals zero, 𝑦 equals four. So we have a point at zero, four. When π‘₯ equals two, 𝑦 equals zero. And we have a point at two, zero. And finally, when π‘₯ equals four, 𝑦 equals negative four. We’ll sketch a line to connect these four points. This is the graph of 𝑦 equals four minus two π‘₯, from negative two to four.

Part c) is asking us to solve π‘₯ equals four minus two π‘₯. We wanna solve for π‘₯. But we have an π‘₯-variable on both sides of the equation. And we need to try and get them on the same side. To do that, we’ll add two π‘₯ to both sides of our equation. π‘₯ plus two π‘₯ equals three π‘₯. And negative two π‘₯ plus two π‘₯ cancels out. So we’re left with four on the right side. We can now see that three π‘₯ equals four. We then divide by three on both sides of the equation.

Four divided by three we can write as four over three, four-thirds.

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