### Video Transcript

Using elimination, solve the simultaneous equations five π₯ plus four π¦ equals 27, three π₯ plus 12π¦ equals 45.

To solve this pair of simultaneous equations means we need to find the values of the two variables, in this case π₯ and π¦, that satisfy both equations simultaneously. Weβre told that we need to use the elimination method, which means that somehow we need to eliminate or remove one of the variables. We need to create a new equation in one variable only.

In order to do this, we need to make the coefficients of one of the variables, either π₯ or π¦, the same in both equations, or at least the same magnitude. The coefficients of π₯ in the two equations are five and three, and the coefficients of π¦ are positive four and positive 12. As four is a factor of 12, the most straightforward way to do this is to multiply the entire first equation by three, as this will give 12π¦. We must remember to multiply every term in equation by three so that the equation we create is equivalent to the one we started with. Five π₯ multiplied by three is 15π₯, four π¦ multiplied by three is 12π¦, and 27 multiplied by three is 81. So weβve created a new equation, which weβll label as equation three: 15π₯ plus 12π¦ equals 81, which is equivalent to our first equation.

If we now write equation two below this, we can see that we have exactly the same number of π¦βs in the two equation. The coefficients of π¦ are both positive 12. So we can eliminate the π¦-variable if we subtract one equation from the other. The coefficient of π₯, the other variable, is larger in equation three. So letβs subtract equation two from equation three. When we do this, we have 15π₯ minus three π₯, which is 12π₯; 12π¦ minus 12π¦, which is zero; and 81 minus 45, which is 36. This equation simplifies to 12π₯ equals 36. And so we have eliminated the π¦-variable and given an equation in one variable only.

To solve this equation, we divide both sides by 12. And we find that π₯ is equal to three. So we found the value of one of the two variables. To find the value of the other variable π¦, we take this value of π₯ and we substitute it into any of our equations, either equation one or equation two, or our new equation, equation three. It doesnβt really matter which we choose. But equation one might be the most straightforward because the coefficients and the numbers involved are smallest. Substituting π₯ equals three then gives five multiplied by three plus four π¦ equals 27. Thatβs 15 plus four π¦ equals 27. And then subtracting 15 from each side, we find that four π¦ is equal to 12. We can then divide both sides of the equation by four to find that π¦ is equal to three.

We found our solution then. Both π₯ and π¦ are equal to three. We should check our answer though. As we found the value of π¦ by substituting π₯ into equation one, we should check our values of both π₯ and π¦ in equation two. Substituting the value three for both π₯ and π¦ gives three multiplied by three plus 12 multiplied by three. Thatβs nine plus 36, which is equal to 45. As this is the same value as the value on the right-hand side of equation two, this confirms that our solution is correct.

We could also have recalled the acronym SSS, which stands for same signs subtract, to help us in this problem. This reminds us that if the coefficients of a variable are the same size and have the same sign in both equations, then we can eliminate this variable by subtracting one equation from the other, as we did here.

Using elimination then, we found that the solution to this pair of simultaneous equations is π₯ equals three and π¦ equals three.