Video: Pack 3 β€’ Paper 3 β€’ Question 11

Pack 3 β€’ Paper 3 β€’ Question 11

03:32

Video Transcript

Solve the simultaneous equations four π‘₯ plus three 𝑦 is equal to 10; negative three π‘₯ plus five 𝑦 is equal to negative 22.

We’ll begin just by labelling the equations as one and two so that we can refer to them more easily. We’re going to use the elimination method to answer this question, which means we want to first eliminate either the π‘₯ or the 𝑦 terms. We’ll eliminate the π‘₯ terms, which means we need to make the coefficients of π‘₯ the same in both equations.

As the coefficients are currently four and negative three, we need to find the lowest common multiple of four and three, which is 12. To get from four to 12, we have to multiply by three. So we multiply the whole of equation one by three, giving 12π‘₯ plus nine 𝑦 is equal to 30. To get from three to 12, we have to multiply by four. So I multiply the whole of equation two by four, giving negative 12π‘₯ plus 20𝑦 is equal to negative 88. Now, we notice that the coefficients of π‘₯ are almost the same, but one is positive and one is negative.

An equally valid approach would have been to eliminate the 𝑦 terms. So we could have multiplied equation one by five and equation two by three. Both equations will then have 15𝑦. Assuming we perform all of the next stages correctly, this method would lead us to the same answer.

Returning to our equations, we know that because the π‘₯ terms have different signs, we eliminate the π‘₯s by adding the two equations together. 12π‘₯ plus negative 12π‘₯ gives zero π‘₯, nine 𝑦 plus 20𝑦 gives 29𝑦, and 30 plus negative 88 gives negative 58. So we have 29𝑦 is equal to negative 58 and we’ve eliminated the π‘₯ terms. To find the value of 𝑦, we need to divide both sides of the equation by 29, giving 𝑦 is equal to negative two.

So we’ve solved the equations for 𝑦. And now, we need to find the value of π‘₯. To do so, we need to substitute the value of 𝑦 that we’ve just found into any of the four equations. It doesn’t matter which one we choose. Let’s substitute into equation one, which was four π‘₯ plus three 𝑦 is equal to 10. This gives four π‘₯ plus three multiplied by negative two is equal to 10.

Three multiplied by negative two is negative six. So we have four π‘₯ minus six is equal to 10. And we can solve this equation to find the value of π‘₯. Adding six to both sides gives four π‘₯ is equal to 16. And dividing both sides by four gives π‘₯ is equal to four. The solution to the simultaneous equations is π‘₯ equals four, 𝑦 equals negative two.

A sensible step to do at the end of this question would be to check our answer by substituting the values we found for π‘₯ and 𝑦 into the second equation. Negative three π‘₯ would become negative three multiplied by four and five 𝑦 would become five multiplied by negative two. This gives negative 12 minus 10 which simplifies to negative 22. This is what we’re expecting negative three π‘₯ plus five 𝑦 to be equal to. And therefore, we can be confident in our answer.

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