Lesson video

Hi, I'm Mrs. Dennett.

In this lesson we're going to be solving simultaneous equations by substitution, where one of the equations is quadratic and the other is linear.

We have two equations that we want to solve simultaneously.

Let's label them A and B.

Equation A is linear.

Equation B is quadratic.

Notice that both the x term and the y term is squared.

In order to solve, we will need to substitute the linear equation A into the quadratic equation B.

I could substitute x equals y minus three into equation B to eliminate x and form a quadratic equation for y, or I can rearrange equation A to make y the subject and put y equals x plus three into equation B.

This will eliminate y and make a quadratic equation for x.

As I've already rearranged to make y the subject, let's put this one into equation B.

I get x squared plus y squared, which is now x plus three all squared, equals 20.

I'm going to write the squared bracket containing x plus three out twice, as this will help me to expand it accurately.

I get x squared plus three x plus three x plus nine from my expansion.

And I can now simplify the entire equation and make it equal to zero.

As it's a quadratic, and in order to solve, I need to have the equation equal to zero.

So to summarise, I must firstly rearrange my linear equation, if required.

I then substitute into the quadratic.

I expand and simplify.

And finally, make my equation equal to zero so that it is ready to solve.

Here is a question for you to try.

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

Here are the answers.

Kaia has missed out the middle terms, three x and three x, when expanding the bracket x plus three all squared.

To avoid this, always write out square brackets twice.

We are now ready to solve our pair of equations simultaneously.

We have substituted in a rearranged equation A into the quadratic equation B.

We then expanded and simplified and then we made our quadratic equation equal to zero.

So we've now got two x squared plus six x minus 11 equal to zero.

I cannot factorise this equation, so instead I will need to use the quadratic formula to find the values for x.

Here is a reminder of the quadratic formula for you.

We need to work out the values for a, b, and c.

a is a coefficient of x squared, this is two.

b is positive six, the coefficient of x, and c is minus 11.

Remember to include the signs for each coefficient.

I substitute each of these values into the quadratic formula.

I can then simplify this and divide throughout by two to get minus three plus or minus root 31 all divided by two.

Some questions may ask you to leave your answer in surd form and then you would have to use these values for x, substitute them into equation A, or you could use equation B, and find the values for y.

However, I'm going to get my scientific calculator out now and work out the two values for x.

I get the first answer when I add root 31 I get 1.

28.

When I subtract root 31, I get minus 4.

28.

Now I must find the values for y.

I substitute x equals 1.

28 into equation A.

You can use B if you wish, but usually the linear equation is more efficient to work with in these equations.

I get 4.

28 as my answer for y.

I can now do the same with the second value for x.

So I get minus 4.

28 plus three, which is minus 1.

28 for y.

State your values for x and y clearly, being careful not to accidentally switch them around.

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

Here are the answers.

Each term in the quadratic equation is a square number.

The solutions are the square root of the number term.

The square root of 100 is 10, which is the number 10 in a linear equation.

So you may have spotted the solutions to the second pair of equations.

Can you think of any other pairs of equations that will produce similar solutions? Here is another question for you to try.

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

Here are the solutions.

We have to rearrange the linear equation here first.

I made x the subject.

x equals two y plus six, and substituted it into the quadratic, eliminating x.

When factorised, this gave me y bracket five y plus 24 equals zero.

So I found my y solutions first.

Y equals zero and y equals minus 24 over five, or minus four and 4/5.

You may have been tempted to write minus 4.

8 for the second solution, which is right, but the question specifically asked for mixed number solutions where appropriate.

You then find your solutions for x, which are six and minus three and 3/5.

Here is final question for you to try.

I'll give you a hint for this one, you will be needing the quadratic formula.

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

Here is the answer to the last question.

I've included some working out for you here where I have made y the subject of the linear equation and eliminated y when substituting into the quadratic.

This meant I found the x values first, but of course, it's fine if you eliminated x and found the y values first instead.

Notice I have written my solutions in surd form, as requested, and tried to save myself some time by using the plus minus notation before the surd in the x solutions.

If you do this though, be careful to invert the plus minus to minus plus for the y solutions, so that it's clear which pair of solutions belong together.

So x equals minus three plus root 41 all divided by two goes with y equals three minus three minus root 41 divided by two.

And for x equals minus three minus root 41 over two, this goes with y equals three minus three plus root 41 divided by two.

What a way to finish the lesson.

That was a tough question.

That's all for this lesson.

Thank you for watching.