### Video Transcript

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

So we have a pair of simultaneous
equations or a system of linear equations in two variables π and π. And weβre told that we must use the
elimination method in order to solve this system of equations. Weβll 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 π. They both have positive two π. Now your first thought may be that
we can, therefore, eliminate the π-variable by adding the two equations
together. But letβs see what that looks
like.

On the left-hand side, three π
plus four π gives seven π. We then have positive two π plus
another positive two π, which gives positive four π. And on the right-hand side, we have
14 plus 16, which is equal to 30. So we have the equation seven π
plus four π equals 30. This equation still involves both
variables, so we havenβt achieved our aim of eliminating one, which means that
adding the two equations together wasnβt the correct step to take.

Instead, letβs try subtracting one
equation from the other. And as the coefficient of the other
valuable, π, is greater in equation 2 than it is in equation 1, Iβm going to try
subtracting equation 1 from equation 2. On the left-hand side, four π
minus three π gives π. We then have two π minus two
π. So that cancels out to zero. And on the right-hand side, 16
minus 14 is two. So we have π equals two. Weβve eliminated the
π-variable. And in fact, we found the solution
for π 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 πβs, 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 itβs the πβs that we were
interested in here. As the signs of the πβs were the
same, we eliminated them by subtracting one equation from the other.

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

So we have our solution to the
simultaneous equations: π is equal to two and π 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. Thatβs equation 2. Substituting π equals two and π
equals four into the left-hand side of equation 2 gives four times two plus two
times four. Thatβs 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 weβve checked is π equals two and π equals four.