# 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 π¦: seven π₯ plus two π¦ equals negative three and π₯ minus three π¦ 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 π¦ the same. This means that we want the same number in front of both π₯ terms or both π¦ terms. We could multiply equation two by seven, as both coefficients of π₯ would then be seven.

However, in this case, weβ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 π¦ equals negative nine. And multiplying the second equation by two gives us two π₯ minus six π¦ equals 32.

As the signs in front of the six π¦s are different, weβre going to add the two equations. Six π¦ plus minus six π¦ 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, weβre going to substitute π₯ equals one into equation one. This gives us seven multiplied by one plus two π¦ equals negative three.

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

Weβre now going to look at solving the simultaneous equations by substitution. Our first step here is to make either π₯ or π¦ the subject of one of the equations. In this case, weβre going to make π₯ the subject of equation two.

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

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

We now need to substitute π¦ 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 weβ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 π¦ equals negative five. We can check these answers by substituting the values into equation one, equation two, or equation three.