### 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.