# Lesson video

In progress...

Hi, I'm Mrs. Dennett.

And in this lesson, we're going to be solving simultaneous equations where one of the equations is linear and the other is of the form xy equals a constant a.

In this pair of equations, we have a linear equation, y equals two x plus one, and a second equation, xy equals 45, which looks like it should be linear, but, in fact, it isn't.

It's a reciprocal function, but you'll see why we've included equations of this form in this section on quadratic and linear equations shortly.

Let's label the equations.

So when we have equations like this, we substitute the linear equation into the other equation.

We get x multiplied by y, which is two x plus one.

So make sure you put the two x plus one in brackets, as we want some multiply x by all of y.

And this is equal to 45.

Expand the bracket on the left-hand side to get two x squared plus x equal to 45, and rearrange to make a quadratic equation, which is equal to zero.

So now we have a quadratic equation to solve.

That's why we include these types of equations in this section of solving simultaneous equations.

Now, does this equation factorised? Yes, it does.

We can find two solutions for x.

So x is nine over two or x is minus five.

We put these into equation A to find the y-values.

Two times nine over two plus one gives us 10 and two times minus five plus one equals minus nine.

You can check that these solutions work in equations B and then, state your pairs of solutions clearly.

Here is a question for you to try.

Pause the video to complete the task and restart when you are finished.

We substitute x minus eight into the first equation, expand and rearrange to get x squared minus eight x plus 16 equals zero.

When we factorise, we get two brackets, which are the same.

We call this a repeated root.

So there is only one solution for x, x equals four.

We use this to find y using either the first equation or the second equation.

We get y equals minus four.

And so we only have one pair of solutions for these simultaneous equations.

Here is a question you to try.

Pause the video to complete the task and restart when you are finished.

I have put y equals two x plus seven into the first equation.

Expanding and rearranging gives us a quadratic, which factorises, and we have our two pairs of solutions.

Here is a question for you to try.

Pause the video to complete the task and restart when you are finished.

Tom and Jim have both correctly rearranged the linear equation in order to substitute it into the equation A.

However, the second rearrangement contains a fraction, and it is usually better to avoid this, as I see many students making errors with fractions, but I would definitely recommend trying both methods to solve these equations and decide which you personally prefer.

Sometimes, even awkward looking factions, can turn out to be easier to work with than you think they will be.

So whilst we may say the first equation looks easier to work with, it doesn't really matter which equation you use, as long as you're accurate.

And if you get stuck with one method, you can try the other method instead.

If you use the first equation, you will get the x-values first, as this eliminates y.

If you use the second equation, you get the y-values first and you can see the pairs of solutions.

When x equals minus three over two, y is minus two and when x is a quarter, y is 12.

Here is a question for you to try, Pause the video to complete the task and restart when you are finished.