# Lesson video

In progress...

Hi, I'm Mrs. Dennett and in today's lesson we're going to be solving a pair of simultaneous equations where one equation is quadratic and the other is linear.

We're going to be setting the equations equal to one another in order to do this.

Solve this pair of equations.

The first equation is a quadratic, you can see the x squared term.

And the second is linear.

I'm going to label them A and B.

As both equations have y as their subjects, both say y equals, we can easily eliminate one of the unknowns, in this case, y.

We set the equations equal to one another.

As both are equal to y, this eliminates y.

We can now rearrange to form a quadratic equation equal to zero.

We now need to solve this quadratic to find the values of x.

Can we find a pair of factors of negative 15, but also add to make negative two? Yes, positive three and negative five.

Put these into brackets.

Brackets are being multiplied together, so you get zero.

At least one of the brackets must be equal to zero.

So x plus three equals zero, or x minus five equals zero.

We can solve these two equations to find our values for x.

X is negative three or x is five.

We still need to find our y values, so we substitute each value of x into one of our original equations.

I have chosen equation A, as it looks quicker to do.

So, y equals negative three squared or y equals five squared.

Given y equals positive nine for the first one, remember you are multiplying negative three by negative three which gives us positive nine, and y equals 25 for the second solution.

Even though we have our two pairs of solutions, it's always a good idea to check them in the other equations.

So using equation B, we can see that when x equals negative three and y equals nine, we get two times negative three plus 15 equals nine.

So negative six add 15 is nine, which is correct.

And when x equals five and y equals 25, we get two times five plus 15 equals 10 plus 15 which is 25.

Both pairs of solutions have worked, so we can be sure that we haven't made any silly errors.

Write your solutions clearly taking care not to mix them up, and we're done.

Here are some questions for you to try.

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

For part a, we set the equations equal to one another.

This gives you x squared equals two x plus three.

You must rearrange this so that the quadratic equation is equal to zero.

Then factorise, x minus three in one bracket and x plus one in the other.

Work out the x solutions, we get three and negative one and put these into y equals x squared to find y.

So y equals nine and y equals one.

Similarly, with b, make the equations equal to one another.

We get x squared minus 14 equals 10, take away 5x.

Rearrange to make the equation equal to zero and then factorise.

We get x minus three and x plus eight in brackets.

This gives us x equals three and x equals negative eight as our solutions for x.

Use either of the equations to find the corresponding y values.

We get y equals negative five and y equals 50.

So, what do we do when we can't factorise our quadratic? Here is another pair of simultaneous equations, the first quadratic and the second linear.

First we label them and then equate them as we did before.

This eliminates y so that we can find x.

Rearrange to make the resulting quadratic equal to zero, just like we did in the previous questions.

You now have two options, solve using inverses or use the quadratic formula.

Here it's much quicker to use inverses 'cause we only have two terms. Add five to both sides and then square root, remembering that you have two solutions when you square root, one positive and one negative.

So we have positive square root of five and negative square root of five.

Now, substitute these two values for x into one of the equations.

To find the values for y, equation B looks like the nicer one and the faster one this time.

So, we get y equals eight root five and y equals negative eight root five.

Always check to make sure you haven't made any silly errors.

Use the other equation, equation A to do this.

Root five squared is just five, and this cancels with negative five.

So the first pair of solutions work perfectly.

Now for the next pair.

X equals negative root five and y equals negative eight root five.

Negative root five squared is just positive five, and we take away the five to get negative eight root five.

Perfect.

So we write our solutions clearly in their x and y pairs, and we're finished.

Here's a question for you to try.

Pause the video to complete the task and resume once you're finished.

Hopefully you are getting much quicker at following this method by now.

Set the equations equal to each other, we get x squared plus 4x, take away three equals 4x.

Rearrange to make the equation equal to zero, x squared minus three equals zero.

And this time we can't factorise, we have to solve using inverses.

We get x equals root three or x equals minus root three.

Finally, substitute these x values into an equation to find the y values.

Y equals 4x looks like the nicer option here.

Check your pairs of solutions work in both of the original equations, and then you've finished the question.

So now, I bet you can't wait to plot the quadratic formula and use it to help you solve this pair of simultaneous equations.

Again, labelling equations first.

Both say y equals, so put them equal to each other, eliminating y.

Can you remember what to do next? Make the equation equal to zero.

Now, we would try to factorise at this point, but we can't find factors of seven that add to make negative nine.

Rearranging to find x isn't an option here either, so we use the quadratic formula.

Here's a reminder.

Our value for A is the coefficient of x squared, which is one.

For B, we have negative nine.

Remember to include the sign.

And for C, we have positive seven.

Substitute these values into the formula.

A equals one, B equals negative nine, and C equals seven.

Once you're happy that you have accurately substituted into the formula, you can now simplify.

You can do this on a scientific calculator if you wish.

But remember, whichever way you simplify, there's two solutions.

One where we add root 53 and one where we subtract root 53.

But this question asks for answers to two decimal places, so you will need to use your calculator at this point.

I found the first solution for x is 8.

14 to two decimal places.

Now we need to find the y solution.

I'm going to substitute x into equation B.

Notice I'm using my simplified surd form to do this.

This is so that my answer for y doesn't lose its accuracy as it would if I use my rounded decimal, 8.

14.

Using my calculator, I can find that y is 73.

26 to two decimal places.

I now need to do the same for my other value of x.

Y minus root 53 divided by two.

The x solution is 0.

86, I then used my simplified surd answer to find y, and I substitute it into equation B.

A check to make sure that these two pairs of values work shows that I haven't made any silly errors.

Now I then write my two pairs of answers to two decimal places.

Here is a question for you to try.

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

I've included some working out for you here as it can be quite easy to make a mistake when you're substituting and simplifying into the quadratic formula.

Just take your time, maintaining accuracy and always check your answers in the other equation at the end.

Here is a final question for you to try.

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