### Video Transcript

Use the elimination method to solve the given simultaneous equations: π₯ minus two π¦ equals negative two and two π₯ plus three π¦ equals 10.

Our first step is to ensure the coefficients of π¦, the numbers in front of the π¦, are the same. On the top equation, multiplying two π¦ by three gives us six π¦. And on the bottom equation, multiplying three π¦ by two also gives us six π¦. It is important that we multiply every term in the equation by three in the top and by two on the bottom. This ensures that the equation is balanced.

For example, on the bottom equation, multiplying two π₯ by two gives us four π₯. Multiplying three π¦ by two gives us six π¦. And multiplying 10 by two gives us 20. As the question asked us to solve this by elimination, we now need to eliminate the π¦-values or π¦ terms. In order to do this, weβre gonna add the two equations as negative six π¦ plus positive six π¦ gives us zero. Adding the two equations gives us seven π₯ equals 14. This is because three π₯ plus four π₯ is seven π₯. And negative six plus 20 equals 14. Dividing both sides of this equation by seven gives us an answer of π₯ equals two.

To solve these simultaneous equations, we also need a π¦-value. Weβve worked out π₯. Now we have to work out what π¦ is. In order to do this, we are going to substitute π₯ equals two into one of the equations. We can pick any of the four equations that you see. The two that we started with or the two that have got coefficient six π¦. In this case, Iβm going to sub the π₯ equals two into the equation two π₯ plus three π¦ equals 10. Substituting the two for π₯ then gives us two multiplied by two plus three π¦ equals 10. Two multiplied by two is four. And then subtracting four from both sides of our equation leaves us with three π¦ equals six. Dividing both sides of this equation by three leaves us a final answer of π¦ equals two.

Our solution can also be written as an ordered pair or coordinate. In this case, the ordered pair two, two would be the solution of the two simultaneous equations: π₯ minus two π¦ equals negative two and two π₯ plus three π¦ equals 10. This could be demonstrated on a graph as well as algebraically.