### Video Transcript

Solve the simultaneous equations two π₯ minus four π¦ equals 14, three π₯ plus π¦ equals seven.

To solve this pair of simultaneous equations means to find the values of π₯ and π¦ that work in both equations at the same time. We need to use a formal algebraic method to solve these equations, not just trial and error. There are couple of different methods that we could use we will look at one in detail, and then mention another method that we could have used. Iβll begin by labelling the two equations as equation one and equation two just so that we have some shorthand with which to refer to them. The method that weβre going to look at in detail is called the elimination method. And this is where we try to eliminate either π₯ or π¦ from the equations so that we then have an equation in only π₯ or only π¦ which we can solve.

To do this, we need to make the coefficients, so thatβs the number in front of either π₯ or π¦, the same. Now as the coefficient of π₯ in the first equation is two, but itβs three in the second equation, weβd have to multiply both equations by some number in order to eliminate π₯. However, as thereβs only one π¦ in the second equation, if we multiply equation two by four, weβll have four π¦, which will match up with the four π¦ in the first equation. We have to multiply every single term in equation two by four. So three π₯ multiplied by four gives 12π₯, π¦ multiplied by four gives four π¦, and seven multiplied by four gives 28. Iβm just going to write equation one which we havenβt changed out again underneath, and what youβll notice is that we now have positive four π¦ in the first equation and negative four π¦ in the second.

If we now add equation one and equation three together, the positive four π¦ in equation three will cancel out with the negative four π¦ in equation one, so weβll just be left with an equation in terms of π₯, which means weβve eliminated the π¦ variable.
12π₯ plus two π₯ gives 14π₯, positive four π¦ plus negative four π¦ cancels out, and on the right of the equation 28 plus 14 gives 42. Now you may be familiar with the acronym SSS, which stands for same sign subtract. We use this to help us work out whether we should be adding or subtracting the two equations to eliminate one variable. If the signs were the same, we would have subtracted one equation from the other. However, in this case, our signs of positive four π¦ and negative four π¦ were different, which is why we added rather than subtracted.

We now have an equation in terms of π₯ only. And to solve this equation, weβll need to divide both sides by 14, which gives π₯ is equal to 42 over 14. Now 42 over 14 can actually be simplified to 21 over seven as both the numerator and denominator of this fraction are even numbers. So to work out what 42 divided by 14 is, we actually just need to work out what 21 divided by seven is. One of our times tables is that three multiplied by seven is equal to 21. So this means that 21 divided by seven is equal to three. And therefore 42 divided by 14 is also equal to three. So we found the value of π₯: π₯ is equal to three. Weβre only halfway there though because we also need to find the value of π¦.

To do so, weβll need to substitute the value that weβve just calculated for π₯ into either of the two equations. Now Iβm going to choose to substitute into equation two as we have a positive rather than a negative number of π¦ in this equation, which will make it a little easier. The equation is three π₯ plus π¦ equals seven. So substituting π₯ equals three gives three multiplied by three plus π¦ equals seven. Three multiplied by three is nine. So the equation becomes nine plus π¦ equals seven. To solve this equation, weβll need to subtract nine from each side. So nine plus π¦ minus nine is just π¦. And on the right, seven minus nine is negative two because seven minus seven gets us to zero and then we have to subtract a further two. So now we found the value of: π¦ π¦ is equal to negative two.

Now itβs always a good idea to check our solution if we can. So what weβll do is weβll substitute the values that we found for π₯ and π¦ into equation one. The left-hand side of equation one is two π₯ minus four π¦. So this becomes two multiplied by three, because π₯ was three minus, four multiplied by negative two, because π¦ is negative two. Two multiplied by three is six, and four multiplied by negative two is negative eight. So now we have six minus negative eight. Those two negative signs next to one another together form a positive. So we have six plus eight, which is equal to 14. This is the value on the right of equation one. So this tells us that our solution for π₯ and π¦ is correct.

Now I said that there was more than one method we could use, and weβll just briefly look at a method called the substitution method. This might be useful because equation two has just one π¦ in it. We can rearrange equation two to give an expression for π¦ in terms of π₯. We need to subtract three π₯ from each side, and it gives π¦ is equal to seven minus three π₯. What we do next is we substitute this expression for equation π¦ into equation one. So equation one was two π₯ minus four π¦ equals 14. And as π¦ is equal to seven minus three π₯, equation one becomes two π₯ minus four lots of seven minus three π₯ is equal to 14. Weβve now eliminated the π¦ variable from equation one in a different way and given ourselves an equation in terms of π₯ only, which we can solve.

I wonβt go through all of the steps for this method, but you would need to expand the bracket first of all, being careful with the negative sign, and then solve the equation in a similar way to like we did on the left of the screen to find the value of π₯. It would still give π₯ is equal to three. Once we knew that π₯ is equal to three, we could substitute this into the equation π¦ equals seven minus three π₯ and solve to find the value of π¦, which again would be negative two. Whichever method we use to solve this simultaneous equations, as long as we apply all the algebraic steps correctly, will give the same solution of π₯ equals three π¦ equals negative two.