### Video Transcript

Using elimination, solve the simultaneous equations negative five π₯ plus six π¦ equals 13, five π₯ plus two π¦ equals 11.

Weβve been given a pair of linear simultaneous equations in two variables, π₯ and π¦. To solve these simultaneous equations, which we can label as equations one and two, means we need to find the values of π₯ and π¦ that satisfy both of these equations.

The question specifies that we need to use the elimination method. This means that we need to eliminate or remove one of the variables. We need to create a new equation which is in one variable only. We do this by making the coefficients of one of the variables, 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. So one can be positive and the other negative.

Looking carefully at the two equations weβve been given, we see that the coefficient of π₯ in the first equation is negative five, whereas the coefficient of that same variable π₯ in the second equation is positive five. This means that we will be able to eliminate the variable π₯ from this pair of equations. But how do we do this?

Well, as the coefficients of π₯ are the same magnitude but have different signs, if we add the two equations together, these terms will cancel each other out. Thatβs because negative five π₯ plus five π₯ gives zero π₯, or simply zero. So the equation weβre left with will have no π₯-term. Adding the rest of the terms together, we have six π¦ plus two π¦, which is eight π¦, and on the right-hand side 13 plus 11, which is 24. So the sum of these two equations is eight π¦ equals 24, an equation in π¦ only.

Weβve eliminated the π₯-variable. And we can now solve this equation to determine the value of π¦. By dividing both sides of the equation by eight, we find that π¦ is equal to three. So we found the value of one of the variables. To find the value of the second variable π₯, we substitute this value of π¦ that we found into either of the two equations.

Letβs choose to substitute it into the second equation because all of the coefficients here are positive, which might make things a little simpler. Substituting π¦ equals three into equation two then gives five π₯ plus two multiplied by three is equal to 11. That gives five π₯ plus six equals 11. And then subtracting six from each side gives the equation five π₯ is equal to five. Finally, we can divide both sides of this equation by five. And we find that π₯ is equal to one.

Weβve found the solution to this pair of simultaneous equations then, but we should check our answer. As we substituted π¦ equals three into equation two, we can now check our answer for both π₯ and π¦ by substituting both values into equation one. Substituting π₯ equals one and π¦ equals three into the left-hand side of this equation gives negative five multiplied by one plus six multiplied by three. Thatβs negative five plus 18, or 18 minus five, which is equal to 13. And as this is equal to the value on the right-hand side of this equation, this confirms that our solution is correct.

So, using the elimination method, which we achieved by adding the two equations together, weβve solved the simultaneous equations and found that π₯ is equal to one and π¦ is equal to three.