### Video Transcript

Solve the system of equations
algebraically. π₯ squared minus π¦ squared equals
seven and two π¦ plus π₯ equals 10.

Letβs start by labelling the
equations number one and number two. Now, we can take equation number
two and make π₯ the subject by subtracting two π¦ from both sides of the
equation. And this gives us that π₯ is equal
to 10 minus two π¦. And weβll call this equation
three. So now, we have an equation for π₯
in terms of π¦ and we can substitute this into equation number one. This gives us 10 minus two π¦ all
squared minus π¦ squared is equal to seven.

Now, we have a term of 10 minus two
π¦ all squared, which we can expand using the FOIL method. First, we can write 10 minus two π¦
all squared as 10 minus two π¦ times 10 minus two π¦. Now, we can use the FOIL
method. So we start by multiplying the
first terms in each bracket. So thatβs 10 times 10 which is
equal to 100. Next, we multiply the outside terms
β so that 10 timesed by minus two π¦ giving us minus 20π¦. Next, we multiply the inside
term. So thatβs minus two π¦ times 10
which gives us minus 20π¦. And then, finally, we multiply the
last term in each bracket. So thatβs minus two π¦ times minus
two π¦. And since a negative times a
negative gives a positive, this gives us four π¦ squared.

Now, we notice that we have minus
20π¦ minus 20π¦, which we could write as minus 40π¦. And we get 100 minus 40π¦ plus four
π¦ squared. And we can substitute this back
into the equation for 10 minus two π¦ all squared. Next, we can combine the terms with
the same powers of π¦. So we have four π¦ squared minus π¦
squared, giving us three π¦ squared. Then, we have minus 40π¦, which we
donβt have anything to combine with. So weβre just left with minus
40π¦.

Finally, we have 100 and seven. But the seven is on the other side
of the equation. So we need to subtract seven from
both sides of the equation, which gives us minus seven on the left of the
equation. And we are left with 100 minus
seven, which gives us 93. And then since we subtracted seven
from the left, we also subtract seven from the right of the equation. So this means that this is all
equal to zero. And we are left with a quadratic in
π¦, which we can solve to find π¦.

In order to factorize this
equation, we can start by drawing our brackets. And we know that in each bracket,
there will be some lot of π¦ plus some constant since we have a π¦ squared term. So letβs write the π¦s in each
bracket. Next, we notice that we have three
π¦ squared. In order to make this three π¦
squared term, we will multiply the two π¦ terms from the brackets. So therefore, there must be a
coefficient in front of one of the π¦ terms in order to give us the three. And since three is a prime number,
this number has to be three. So we know that one of the π¦ terms
in one of the brackets is three π¦. And when multiplying this by the π¦
in the other bracket, this will give us the three π¦ squared.

Next, we can work out the signs of
the constants. Since we have a positive constant
in the equation and a negative π¦ term in the equation and since the constant term
comes from multiplying the last term in each bracket, this means that the constants
must have the same symbol β so either be positive or both be negative. Now since we have a minus π¦ term,
this is how we know that both of them are negative or else we would have no
negatives in our equation.

In order to work out what the two
constants are, letβs look at the possible factor pairs of 93. We either have one and 93 or three
and 31. We need to find a factor pairing
such that when we multiply one of the factors by three and add it to the other one,
we end up with 40. So if we tried multiplying this one
by three and adding it to 93, this gives us 96. So therefore, it canβt be this
pairing. Next, letβs try one plus three
times 93. This gives us 280. So therefore, this canβt be a
factor pairing either. So we know that the first one is
not right. Next, we can try three timesed by
three plus 31 and this gives us 40. Therefore, we found our factor
pairing.

Now, since we know that we need
three times three plus 31, therefore, the three has to go in the opposite bracket to
the three π¦. Since then when we expand it, we
will end up multiplying the three π¦ by the three to give us the nine, which we then
add to the 31 multiplied by the π¦ and then this is what gives us the negative
40π¦.

So now that we have factorized our
π¦ equation, we can find the solutions of π¦. And since we have two things
multiplied by one another equal to zero, this means that either one of them has to
be equal to zero. So three π¦ minus 31 equals zero or
π¦ minus three equals zero. Now, we can solve the equation on
the left for π¦ by adding 31 to either side, giving us that three π¦ is equal to
31. And then, we can divide both sides
by three and this gives us that π¦ is equal to 31 over three. Now, we can solve the equation on
the right by simply adding three to both sides and we get another solution of π¦ is
equal to three.

So now, we have two possible
solutions of π¦ and we can substitute them into equation three, which is π₯ is equal
to 10 minus two π¦, in order to find the corresponding π₯-values. If we substitute in π¦ is equal to
31 over three, we get π₯ is equal to 10 minus two timesed by 31 over three. Then, two timesed by 31 over three
is equal to 62 over three. And then, we know that we can write
10 as 10 timesed by three over three since three over three is equal to one and 10
timesed by three over three is equal to 30 over three. And now, since 30 over three and 62
over three have a common denominator, we can combine the fractions and we get 30
minus 62 all over three. And this simplifies to give us an
π₯-value of minus 32 over three.

Next, we can find the corresponding
π₯-value when π¦ is equal to three. Substituting π¦ into the equation,
we get that π₯ is equal to 10 minus two times three. Then, two times three is equal to
six. So we get 10 minus six, which is
the same as four.

So now, we have found a solution to
the system of equations. And the solution is π¦ is equal to
31 over three and π₯ is equal to minus 32 over three or π¦ is equal to three and π₯
is equal to four.