Video: Solving Simultaneous Equations by Elimination

Use the elimination method to solve the simultaneous equations 3π‘Ž + 2𝑏 = 14, 4π‘Ž + 2𝑏 = 16.

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Video Transcript

Use the elimination method to solve the simultaneous equations three π‘Ž plus two 𝑏 equals 14, four π‘Ž plus two 𝑏 equals 16.

So we have a pair of simultaneous equations or a system of linear equations in two variables π‘Ž and 𝑏. And we’re told that we must use the elimination method in order to solve this system of equations. We’ll label our two equations as equation 1 and equation 2 for ease of referencing them. And looking at the two equations, we notice, first of all, that they have exactly the same coefficient of 𝑏. They both have positive two 𝑏. Now your first thought may be that we can, therefore, eliminate the 𝑏-variable by adding the two equations together. But let’s see what that looks like.

On the left-hand side, three π‘Ž plus four π‘Ž gives seven π‘Ž. We then have positive two 𝑏 plus another positive two 𝑏, which gives positive four 𝑏. And on the right-hand side, we have 14 plus 16, which is equal to 30. So we have the equation seven π‘Ž plus four 𝑏 equals 30. This equation still involves both variables, so we haven’t achieved our aim of eliminating one, which means that adding the two equations together wasn’t the correct step to take.

Instead, let’s try subtracting one equation from the other. And as the coefficient of the other valuable, π‘Ž, is greater in equation 2 than it is in equation 1, I’m going to try subtracting equation 1 from equation 2. On the left-hand side, four π‘Ž minus three π‘Ž gives π‘Ž. We then have two 𝑏 minus two 𝑏. So that cancels out to zero. And on the right-hand side, 16 minus 14 is two. So we have π‘Ž equals two. We’ve eliminated the 𝑏-variable. And in fact, we found the solution for π‘Ž at the same time. The correct way to eliminate one variable then was to subtract one equation from the other. And the reason for this is that the coefficients of the variable we were trying to eliminate, that is the 𝑏’s, are identical in both equations and they have the same sign.

There is a helpful acronym that we can use to help us remember this, SSS. It stands for if we have the same signs, then we subtract. We must remember that it is the signs of the variable we are looking to eliminate that is important. So it’s the 𝑏’s that we were interested in here. As the signs of the 𝑏’s were the same, we eliminated them by subtracting one equation from the other.

Now that we found the value of π‘Ž, we need to find the value of 𝑏, which we can do by substituting our value of π‘Ž into either of the two equations. Let’s choose equation one. We have three multiplied by two plus two 𝑏 is equal to 14. That’s six plus two 𝑏 equals 14. And subtracting six from each side gives two 𝑏 is equal to eight. We then solve for 𝑏 by dividing each side of the equation by two, giving 𝑏 equals four.

So we have our solution to the simultaneous equations: π‘Ž is equal to two and 𝑏 is equal to four. But we should check our answer, which we can do by substituting the pair of values we found into the other equation. That’s equation 2. Substituting π‘Ž equals two and 𝑏 equals four into the left-hand side of equation 2 gives four times two plus two times four. That’s eight plus eight, which is equal to 16, the value on the right-hand side of equation 2. So this confirms that our solution is correct.

We need to remember then that helpful acronym SSS, which stands for if the signs of the variable we want to eliminate are the same, then we subtract. Of course, the reverse is also true. If the signs of the variable we want to eliminate are different, then we add. Our solution to this set of simultaneous equations which we’ve checked is π‘Ž equals two and 𝑏 equals four.

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