### Video Transcript

Using elimination, solve the simultaneous equations negative five π₯ plus seven π¦ equals nine, 15π₯ plus 12π¦ equals 39.

Weβre asked to solve this pair of simultaneous equations using the elimination method. This means that we need to make the coefficients of one of our variables, in this case either π₯ or π¦, the same. In fact, they donβt have to be exactly the same; they can instead be the same size or magnitude but different signs.

In the equations weβve been given, the coefficients of π₯ are negative five and 15 and the coefficients of π¦ are positive seven and positive 12. So neither the coefficients of π₯ nor the coefficients of π¦ are the same or at least the same size in our equations at the moment. This means that weβll need to create at least one equivalent equation by multiplying one or perhaps both of our equations by some constant.

As five is a factor of 15, the simplest way to do this is going to be to multiply equation one by three because then the coefficient of π₯ will be negative 15. We need to multiply every term in equation one by three. And when we do, we get negative 15π₯ plus 21π¦ equals 27. This is an equivalent equation to equation one. And we can call it equation three. Weβll write equation two below it.

And when we do, we should notice that we now have negative 15π₯ in our first equation and positive 15π₯ in the second. This means that if we add the two equations together, we will eliminate the π₯-variable because negative 15π₯ plus 15π₯ is zero π₯ or simply zero. Adding the equations term by term then gives zero plus 33π¦ equals 66, or simply 33π¦ equals 66. And so we have eliminated the π₯-variables and created an equation in π¦ only.

We can solve this equation for π¦ by dividing both sides by 33. And it gives π¦ is equal to two. We have therefore found the value of one of the two variables. To find the value of the other variable, we can take this value of π¦ and substitute it into any of our three equations. Letβs substitute into equation two because all of the coefficients are positive here, so it might be a little bit more straightforward.

This gives 15π₯ plus 12 multiplied by two is equal to 39. That gives 15π₯ plus 24 equals 39. And subtracting 24 from each side, we find that 15π₯ is equal to 15. To solve for π₯, we divide both sides of the equation by 15, and we find that π₯ is equal to one.

So weβve solved this pair of simultaneous equations: π₯ is equal to one and π¦ is equal to two. But we should check our answer. As we substituted π¦ equals two into the second equation to find the value of π₯, we should now substitute both values into the first equation and check it does indeed work.

Substituting π₯ equals one and π¦ equals two into the expression on the left-hand side of this equation gives negative five multiplied by one plus seven multiplied by two. That gives negative five plus 14, or 14 minus five, which is equal to nine. As this is the same as the value on the right-hand side of equation one, this confirms that our solution is correct. Using the elimination method then, weβve found that the solution to this pair of simultaneous equations is π₯ equals one and π¦ equals two.