Video: Pack 1 β€’ Paper 2 β€’ Question 1

Pack 1 β€’ Paper 2 β€’ Question 1

04:48

Video Transcript

Solve the simultaneous equations for π‘₯ and 𝑦: seven π‘₯ plus two 𝑦 equals negative three and π‘₯ minus three 𝑦 equals 16.

We can solve a pair of linear simultaneous equations by elimination or by substitution. We are firstly gonna look at the method of elimination. Our first step is to make the coefficients of π‘₯ or the coefficients of 𝑦 the same. This means that we want the same number in front of both π‘₯ terms or both 𝑦 terms. We could multiply equation two by seven, as both coefficients of π‘₯ would then be seven.

However, in this case, we’re going to multiply the top equation by three and the bottom equation by two. Multiplying the top equation by three gives us 21π‘₯ plus six 𝑦 equals negative nine. And multiplying the second equation by two gives us two π‘₯ minus six 𝑦 equals 32.

As the signs in front of the six 𝑦s are different, we’re going to add the two equations. Six 𝑦 plus minus six 𝑦 is equal to zero. Adding the π‘₯ terms gives us 23π‘₯, as 21π‘₯ plus two π‘₯ is 23π‘₯. Negative nine plus 32 is equal to 23. Dividing both sides of this equation by 23 gives us a value of π‘₯ equal to one.

We now need to substitute π‘₯ equals one back into one of the original equations, either equation one or equation two. In this case, we’re going to substitute π‘₯ equals one into equation one. This gives us seven multiplied by one plus two 𝑦 equals negative three.

Simplifying this leaves us with seven plus two 𝑦 equals negative three. Subtracting seven from both sides of this equation gives us two 𝑦 equals negative 10. And finally, dividing both sides by two gives us a value of 𝑦 equal to negative five. This means that the solution to the simultaneous equations by elimination is π‘₯ equals one and 𝑦 equals negative five.

We’re now going to look at solving the simultaneous equations by substitution. Our first step here is to make either π‘₯ or 𝑦 the subject of one of the equations. In this case, we’re going to make π‘₯ the subject of equation two.

Adding three 𝑦 to both sides of equation two gives us π‘₯ is equal to 16 plus three 𝑦. We now need to substitute this into equation one. This gives us seven multiplied by 16 plus three 𝑦 plus two 𝑦 equals negative three.

Expanding or multiplying out the brackets gives us 112 plus 21𝑦, as seven multiplied by 16 is 112 and seven multiplied by three 𝑦 is 21𝑦. Grouping our like terms gives us 112 plus 23𝑦 equals negative three. Subtracting 112 from both sides of this equation gives us 23𝑦 equals negative 115. And finally, dividing both sides of this equation by 23 gives us an answer of 𝑦 equals negative five.

We now need to substitute 𝑦 equals negative five back into equation three to calculate π‘₯. π‘₯ is equal to 16 plus three multiplied by negative five. Three multiplied by negative five is negative 15. So we’re left with π‘₯ equals 16 minus 15. Therefore, π‘₯ is equal to one.

Solving the simultaneous equations by substitution gave us the same answers: π‘₯ equals one and 𝑦 equals negative five. We can check these answers by substituting the values into equation one, equation two, or equation three.

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