### Video Transcript

Solve the following simultaneous
equations using algebra: π₯ is equal to two π¦ minus seven over four, π₯ squared
minus one over 16 is equal to two π¦ squared.

First, weβll number the two
equations as equation one and equation two so that we have some shorthand to use in
referring to them. Now, note that equation one is the
simpler equation. Itβs a linear equation, meaning
that the powers of π₯ and π¦ are both one, whereas equation two is a quadratic β the
powers of π₯ and π¦ are two.

Equation one gives us an expression
for π₯ in terms of π¦. So we can substitute this
expression for π₯ into the second equation. And this will give an equation in
terms of π¦ only, making this substitution gives two π¦ minus seven over four all
squared minus one over 16 is equal to two π¦ squared. And therefore, thereβre no π₯ terms
in this equation.

We want to solve this quadratic
equation for π¦. And the first step is we need to
expand the bracket on the left-hand side. Remember when weβre squaring a
bracket, this actually means that weβre multiplying the bracket by itself. We can use the FOIL method to help
us with this expansion.

Multiplying the first terms in the
brackets together, we have two π¦ multiplied by two π¦ which gives four π¦
squared. Multiplying the outer terms of the
brackets together, we have two π¦ multiplied by negative seven over four. The two in the numerator of this
product and the four in the denominator can be cancelled by a factor of two, leading
to negative seven π¦ over two.

Next, we multiply the inner terms
in the brackets together: negative seven over four multiplied by two π¦ which is the
same product as weβve just done, but in reverse. So it simplifies to the same thing,
negative seven π¦ over two. Finally, we multiply the last term
in each bracket together: negative seven over four multiplied by negative seven over
four which gives positive 49 over 16.

When we add the four terms in the
expansion together negative seven π¦ over two plus another lot of negative seven π¦
over two gives negative seven π¦. So the simplified expansion is four
π¦ squared minus seven π¦ plus 49 over 16.

Returning then to our solution to
the simultaneous equations, we now have four π¦ squared minus seven π¦ plus 49 over
16 minus one over 16 is equal to two π¦ squared. Now, 49 over 16 minus one over 16
is equal to 48 over 16. And our 16 times three is 48. This just simplifies to three. Weβll also group all the terms on
the left-hand side of the equation by subtracting two π¦ squared from both
sides. This gives two π¦ squared minus
seven π¦ plus three is equal to zero.

Now, this is a quadratic equation
in π¦ only. And to solve, we have a number of
possible methods: we could factorize, we could use the quadratic formula, or we
could complete the square. If a quadratic does factorize, then
itβs almost always quickest to use this method. So itβs always worth a quick check
first.

In this quadratic, the coefficient
of π¦ squared and the constant term are both prime numbers. So we donβt have a lot of options
that we need to try. The two π¦ squared must be the
product of two π¦ and π¦. So this gives the first term in
each bracket. The only factors of three are three
and one. And as the constant term is
positive, but the coefficient of π¦ is negative, this tells us that both values in
the brackets must be negative. So we need negative one in one of
the brackets and negative three in the other. And we can use trial and error to
see which gives the correct factorization.

Now, Iβve put the negative one in
the first bracket and the negative three in the second for now. And we need to check that when we
expand these brackets, we get the correct number of π¦s. We already know that weβll get two
π¦ squared and weβll get positive three. So we only need to check the middle
two terms in the expansion. The π¦ terms come from multiplying
two π¦ by negative three giving negative six π¦ and negative one by π¦ giving
negative π¦. In total then, we have negative six
π¦ plus negative π¦ giving negative seven π¦ which is what we wanted.

So we have the correct
factorization for this quadratic and weβre nearly there with solving for π¦. Remember if two brackets multiplied
to give zero, then at least one of the brackets must itself be equal to zero. So we have that two π¦ minus one is
equal to zero or π¦ minus three is equal to zero. These are linear equations that we
can solve to find π¦.

To solve the first equation, we add
one and then divide by two, giving π¦ is equal to one-half. To solve the second equation, we
just need to add three to each side giving π¦ is equal to three. So we found the values of π¦ that
satisfy the simultaneous equations. And the final step is we need to
find the values of π₯.

To do so, weβll substitute into the
more straightforward equation β equation one. When π¦ is equal to a half, π₯ is
equal to two multiplied by a half minus seven over four. Two multiplied by a half is
one. So we have one minus seven over
four. Now, one can be expressed as four
over four. So we have four over four minus
seven over four which is negative three over four.

When π¦ is equal to the other value
three, π₯ is equal to two multiplied by three minus seven over four which gives six
minus seven over four. Six can be written as 24 over
four. So we have 24 over four minus seven
over four which is 17 over four.

The solution then to the
simultaneous equations are π₯ equals negative three-quarters, π¦ equals a half and
π₯ equals 17 over four, π¦ equals three.