### Video Transcript

Use the elimination method to solve
the simultaneous equations two-thirds π₯ plus three-quarters π¦ equals 17 over 24,
four-thirds π₯ plus five-thirds π¦ equals three over two.

Solving this pair of simultaneous
equations means we need to find the values of the two variables π₯ and π¦ that
satisfy both equations. Weβre told that we need to use the
elimination method, which means we need to find some way of eliminating or removing
one of the variables from these equations so that we create a new equation in one
variable only. The way we do this is by making the
coefficients of either π₯ or π¦ the same in both equations. In fact, they donβt have to be
exactly the same. They can be the same size or
magnitude, but one can be positive and one can be negative. Weβll see why this helps in a
moment.

For now, we observe that the
coefficients of π₯ in the two equations are two-thirds and four-thirds and the
coefficients of π¦ are three-quarters and five-thirds. If we were to multiply the entirety
of equation one by the constant two, then the coefficient of π₯ would be two-thirds
multiplied by two, which is four-thirds. And thatβs the same as the
coefficient of π₯ in the second equation. We have to multiply every term in
the equation by two though, which has the effect of increasing the numerator of each
fraction by a factor of two. So we have four-thirds of π₯ plus
six over four or six-quarters of π¦ is equal to 34 over 24. We can label this as equation three
and then rewrite equation two below it.

What we now see is that the
coefficient of π₯ is exactly the same in the two equations. This means that if we subtract one
equation from the other, then the π₯-terms will be eliminated because if we have any
number of π₯ and then we subtract that same number of π₯, weβre left with no
π₯βs. So, letβs subtract equation two
from equation three. When we do that, the π₯-terms
cancel as weβve already said. We have six over four π¦ minus five
over three π¦, which we could write as six over four minus five over three all
multiplied by π¦. And on the right-hand side, we have
34 over 24 minus three over two. We can simplify some of the
fractions. Six over four is equivalent to
three over two, and 34 over 24 is equivalent to 17 over 12.

We then need to find common
denominators in order to be able to subtract each pair of fractions. On the left-hand side, weβll use
the common denominator of six, and three over two is equivalent to nine over six,
while five over three is equivalent to 10 over six. So we have nine-sixths minus
ten-sixths multiplied by π¦. On the right-hand side, weβll use
the common denominator 12, and three over two is equivalent to 18 over 12. So we have negative one-sixth π¦ is
equal to negative one twelfth. And by multiplying or dividing both
sides by negative one, we can cancel out the negatives, leaving one-sixth of π¦ is
equal to one twelfth. To solve for π¦, we multiplied both
sides of the equation by six, giving π¦ is equal to six over 12, which is equivalent
to one-half.

So, weβve found the value of one of
these two variables, and now we need to find the value of the other. To do this, we substitute the value
of π¦ weβve just found into any of the three equations, either equation one, two, or
three. Letβs use equation one. This gives two-thirds of π₯ plus
three-quarters multiplied by one-half is equal to 17 over 24. Three-quarters multiplied by
one-half is three-eighths, so we have two-thirds of π₯ plus three-eighths equals 17
over 24. We can then write all of the terms
in this equation with a common denominator of 24.

By multiplying both the numerator
and denominator of the first fraction by eight, we see that two-thirds is equivalent
to 16 over 24. And by multiplying the numerator
and denominator of the second fraction by three, we see that three-eighths is
equivalent to nine over 24. Multiplying the entire equation
through by 24 then will eliminate all of the denominators and give the simpler
equation 16π₯ plus nine is equal to 17. To solve this equation, we can
subtract nine from each side, giving 16π₯ is equal to eight. We can then divide both sides of
the equation by 16, giving π₯ equals eight over 16, which is of course equal to
one-half. So, we found the solution to this
pair of simultaneous equations. π₯ is equal to one-half, and so is
π¦.

We should always check our answers
where possible. And we can do this by substituting
the values we found for π₯ and π¦ into the second equation. Substituting both π₯ and π¦ equal a
half into the left-hand side of this equation gives four-thirds multiplied by a half
plus five-thirds multiplied by a half, which is four-sixths plus five-sixths. This is equal to nine-sixths, and
this fraction does indeed simplify to three over two, the value on the right-hand
side of equation two, which confirms that our answer is correct. Using the elimination method then,
we found that the solution to the given pair of simultaneous equations is π₯ equals
a half and π¦ equals a half.