Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Lovely of you to join me today, especially lovely because we're solving equations involving functions.

I know you love equations and I know you love functions.

So what are we waiting for? Let's get started.

Our learning outcome is that we'll be able to solve equations involving functions.

Some keywords we'll hear today, function and equation.

A function is a mathematical relationship that uniquely maps values of one set to the values of another set.

An equation is used to show two expressions that are equal to each other.

We've got two parts to our learning today, and we're going to begin with solving.

A function such as F of X equals three X minus five maps values from the domain to values in the range.

Evaluating F of eight shows that the value eight maps to the value 19.

When we evaluate F of eight, eight goes into the expression, three X minus five.

Three lots of eight minus five becomes 19.

The input eight from the domain maps to 19, the output, in the range.

There are times when we know a value in the range and we want to go backwards to find the value in the domain.

In such a case, we might be solving F of X equals 31.

What does this mean? Effectively it means the output is 31.

What was the original input? We can form and solve an equation to answer this question.

If F of X equals 31, then three X minus five equals 31 because three X minus five is F of X.

From here, we've got an equation and we can rearrange.

Let's add positive five to both sides.

Let's divide both sides by three.

Oh look, X equals 12.

That's our solution to the equation, F of X equals 31.

At least we think it is.

It would be nice to know it is.

Solving equations should be one of your favourite things in maths, because you can check your work for yourself.

12 is the value in the domain, that was the original input, we think, but let's check so that we know.

F of X equals three X minus five.

What's F of 12? Well, it's three lots of 12 minus five, which is 31.

Wonderful, we're right.

Our solution, X equals 12, that's an input that we think gives us the output 31.

After checking it, we know that to be true.

Quick check you've got that.

F of X equals three X minus five.

Can you solve F of X equals negative 11? Pause.

Give it a go.

I'll see you in a moment to run through a solution.

Welcome back.

Let's see how we did.

You start by writing F of X equals negative 11, and if F of X equals negative 11, then three X minus five equals negative 11.

Add positive five to both sides.

Divide three by three and we know X equals negative two.

That's the input into F of X.

That will give us an output of negative 11.

At least we think it is.

Of course we're going to check that.

If we input negative two into F of X, does it give us the output negative 11? Absolutely it does.

So at this point we know we're right.

In some cases we need to form and solve a quadratic equation such as this case.

F of X equals X squared plus three X.

Solve F of X equals four.

Our first step is very similar.

F of X equals four.

What's F of X? It's X squared plus three X.

So we've now got the equation X squared plus three X equals four.

Now if we want to solve a quadratic, we want to rearrange it so all the terms on the left hand side are just a zero term on the right hand side.

X squared plus three X minus four equals zero.

I'm gonna ask myself a key question.

Will that factorise? And you're gonna tell me, yes, Mr. Robson, it does.

If factorises to X plus four, X minus one.

If that's gonna be equal to zero, then either X equals negative four or X equals one.

Two inputs that map to four, or at least we think we found two inputs that map to four.

We will of course check this one.

If we input negative four into the function F of X, we get the output four.

We know that negative four is a valid solution.

If we input one into the function F of X, we also get the output four.

Hoo rah.

Both of our solutions worked.

Quick check you can do that.

I'd like you to form and solve a quadratic equation, to find the values of X such that F of X equals 10.

Pause.

Give it a go.

See you in a moment to run through a solution.

Welcome back.

Let's see how we did.

F of X equals 10.

If F of X equals 10, then X squared plus three X equals 10.

Let rearrange that.

We're solving X squared plus three X minus 10 equals zero.

There are lots of ways that we could solve this quadratic equation.

Often a good thing to ask yourself is will it factorise because that might be the quickest and simplest way.

In this case it did factorise, so our solutions must be X equals negative five or X equals two.

Did you get that? Well done.

Did you check that they were the right solutions? If so, super well done.

Two inputs that map to 10, at least we think they do.

Let's check.

F of negative five does indeed map to 10.

F of two does indeed map to 10.

We know we're right.

We can also form and solve an equation to find any moments where two functions share a common input and output.

For example, if F of X equals five X minus 17 and G of X equals two X plus seven, then we can solve F of X equals G of X.

If F of X equals G of X, then five X minus 17 equals two X plus seven.

It's a linear equation with unknowns both sides.

Let's manipulate.

We could add positive 17 to both sides.

Add negative two X to both sides, multiply both side by a third, and we get X equals eight.

Eight is the input, the value of X in the domain, that gives both functions the same output in their respective ranges.

Again, at least we think it is.

Let's check that it is.

What's F of eight? Five lots of eight minus 17.

That's 23.

What's G of eight? Two lots of eight plus seven, that's 23.

Wonderful.

Eight is an input that maps to the same output in both ranges.

Quick check you can do that.

I'd like to form and solve an equation to find the moment where these two functions share a common input and output.

Pause.

Give this one a go.

I'll be back in a moment to run through a solution.

Welcome back.

Hopefully you started by writing F of X equals G of X.

That's what we need to solve in order to find the moment where the two functions share a common input and output.

If F of X equals G of X, then two lots of X plus four equals 10 X minus 48.

Lots of ways we could manipulate to solve this.

One nice simple way is to divide both sides by two.

Then we can add negative X to both sides.

Add positive 24 to both sides.

Divide both sides by four, X equals seven.

I think you know what's coming next.

Let's check that we're right.

F of seven equals 22, G of seven equals 22.

Wonderful.

We found the moment when the input, seven, shares the same output, 22, for both functions.

We know we're right.

Practise time now.

Question one, F of X equals seven X minus 4, G of X equals eight minus three X.

There's four equations there for you to solve.

Pause and give them a go now.

Question two.

F of X equals X squared plus nine X.

Two equations for you to solve.

F of X equals 10, F of X equals negative 20.

Pause, find those solutions now.

Question three.

We're gonna deal with three functions this time.

F of X equals seven X plus six, G of X equals three lots of X plus 10, and H of X equals X squared minus 10 X.

In this occasion, for part A, you're gonna solve F of X equals G of X.

What's the moment where the same input has the same output for both functions? And then for part B, solve G of X equals H of X.

Pause.

Give these two a go now.

Feedback time now.

Let's see how we did.

Question one, Part A.

I asked you to solve F of X equals 31.

If F of X equals 31, then seven X minus four equals 31.

So seven X equals 35 and X equals five.

I hope you didn't stop there.

I hope you evaluated F of five to check that you're correct.

F of five equals 31.

We know we're right.

For part B, G of X equals two, therefore, eight minus three X must equal two.

So three X must equal six and X must equal two.

If this is correct, then G of two must equal two, and it does.

Part C, F of X equals negative four.

Seven X minus four equals negative four.

So seven X must equal zero, so X must equal zero.

If this one's right, then F of zero is equal to negative four or zero maps to negative four.

When put into the function F of X, it does.

We know that one's right.

Part D, G of X equals negative 34.

Eight minus three X equals negative 34.

So three X equals 42 and X equals 14.

Let's substitute to check that's right.

G of 14 is equal to negative 34.

We know we're right.

Question two, slightly trickier, 'cause we're forming and solving quadratic equations, but nothing to be afraid of.

Part A, F of X equals 10, X squared plus nine X equals 10.

Rearrange like so, and then lots of ways we could solve this.

Well, it factorises often.

The quickest and easiest way to resolve this one did factorise.

Do X plus 10, X minus one.

So our solutions are X equals negative 10 and X equals one.

Again, we can check that these are two inputs that map to 10.

F of negative 10 is indeed equal to 10, and one maps to 10, so we know we're right.

Part B, F of X equals negative 20, X squared plus nine X equals negative 20.

So X squared plus nine X plus 20 equals zero.

Lots of ways we could solve.

Will it factorise? It does, so we can quickly find two solutions.

X equals negative five, X equals negative four.

Two inputs that map to negative 20.

Let's show that that is absolutely true.

F of negative five is negative 20.

F of negative four is negative 20.

We're absolutely right.

Question three, Part A, solve F of X equals G of X.

If F of X equals G of X and seven X plus six equals three lots of X plus 10.

Let's expand that bracket on the right hand side.

Do a little bit of rearranging to get to four X equals 24, so X equals six.

Let's evaluate F of six and evaluate G of six.

Wonderful.

When the input is six, both functions share the same output of 48.

We have solved F of X equals G of X.

Part B, a little bit trickier, because H as a function was quadratic.

Three lots of X plus 10 equals X squared minus 10 X.

Let's expand the bracket on the left hand side.

Let's rearrange.

Lots of ways we could solve this.

It factorised, which is possibly the most efficient way to solve it.

Two solutions, X equals 15, X equals negative two.

We can check that they're correct.

G of 15 is 75, H of 15 is 75.

G of negative two is 24, H of negative two is 24.

When the input is 15, both functions share the output 75.

When the input is negative two, they share the output 24.

We've solved G of X equals H of X.

On to the second half of our learning now, where we're going to be solving with more complex cases.

Lovely.

Some of the equations we have to solve will involve fractions.

If we look at this function, G of X equals 10 over X.

If we try to solve G of X equals two, then we get 10 over X equals two.

I'm gonna think of that left hand side, rather than 10 over X, I'll think of it as 10 multiplied by one over X, because it helps me to understand the next step more easily.

I wanna multiply by X, because that's the reciprocal of one over X.

We multiply both sides of the equation by X.

We get 10 equals two X.

We now have only integer coefficients and terms. You'll see those steps written more efficiently as when we start with 10 over X equals two, multiply both sides by X and it becomes 10 equals two X.

From there we can see that X equals five, and if you were to substitute that back into G of X, G of five would be 10 over five, which is two.

Ah, wonderful.

That's what we were expecting.

We know that solution is right.

Quick check that you can do one of these equations that involves fractions.

I showed you how to solve G of X equals two.

I'd like you to have a go now at solving G of X equals five.

Pause, give it a go now.

Welcome back.

Let's see how we did.

Let's start with 10 over X equals five and then multiply 'em by X, the reciprocal of one over X.

To get 10 equals five X, therefore X equals two.

We can check that.

G of two equals 10 over two, which is five.

We know we're right.

We will see other uses of reciprocals.

Example if we try to solve this.

G of X equals 2/13.

What's different now? When we set up, 10 over X is not equal to an integer of value, it's equal to a fraction, 2/13.

So we could make the first step multiply both sides by X, and we'll get 10 equals two over 13 X.

From here we can multiply by the reciprocal of 2/13.

Let's multiply both sides by 13 over two.

Why would we do that? Because, on the right hand side, it leaves us with just X.

So if X equals 10 multiplied by 13 over two, then X equals 65.

We will of course put 65 into the function to check that we're correct.

G of 65 is 10 over 65, which is indeed equal to 2/13, so we know we're right.

Often in maths, we can use multiple methods to solve the same problem.

Same function, G of X equals 10 over X, and we're solving G of X equals 2/13.

We start by setting up like so, but from here I'm gonna create an equivalent fraction to 2/13 on the right hand side.

I'm gonna multiply both the numerator and the denominator by five to turn 2/13 into 10 over 65.

Why did I want to do that? Well done.

I can hear you shouting at the screen, because if we've got equal numerators, 10 equals 10, and these are equivalent fractions, 10 over X equals 10 over 65, then those denominators are equal.

X must equal 65.

Well done.

Quick check you can do that now.

F of X equals 15 over X.

I'd like to solve F of X equals three over 17.

Pause, give this a go now.

Welcome back, let's see how we did.

15 over X equals three over 17.

We can multiply both sides by X, and then we can multiply both sides by the reciprocal of three over 17, which is 17 over three.

That'll leave us with X on the right hand side.

From there, X is equal to 15 multiplied by 17 over three, which is 85.

X is equal to 85.

We will of course put 85 into our function F of X.

We'll get 15 over 85, which is indeed equal to three over 17.

We know we're right.

An alternative method for that.

We start by writing 15 over X equals three over 17 and then create an equivalent fraction to three over 17.

An equivalent fraction for three over 17 is 15 over 85.

Why did we want that exact equivalent fraction? Because the numerators are equal, they're both 15, which means the denominators are equal, X must be equal to 85.

Next, when we solve F of X equals G of X, something interesting happens.

F of X equals X plus three, G of X equals 10 over X.

Let's have a go at solving F of X equals G of X.

X plus three equals 10 over X.

Now you know that we're gonna multiply both sides by X.

Now we have a quadratic that needs to be solved.

We expand that bracket and rearrange.

We're gonna have to solve X squared plus three X minus 10 equals zero to find the solutions to F of X equals G of X.

Will it factorise? It does, giving us two solutions, X equals negative five and X equals positive two.

We'll check that.

We know we're right.

Quick check that you can do one just like that.

F of X equals X minus four and G of X equals 21 over X, and you solve F of X equals G of X.

Pause, give this one a go now.

See you in a moment to check your answer.

Welcome back, let's see how we did.

We start by writing X minus four equals 21 over X, multiplying both sides by X, expanding that bracket on the left hand side, and then rearranging to get X squared minus four X minus 21 equals zero.

Again, lots of ways we could solve this.

It will factorise, giving us two solutions, X equals negative three, and X equals seven.

Now I know you didn't stop there.

I know you wanted to know that these were indeed the right answers, so we check them.

We know they both work.

Well done.

We may see denominators with multiple terms. For example, F of X equals nine over X minus four.

We've got an X term and an integer term in that denominator.

So let's see what happens when we solve F of X equals three over 11.

Well, if F of X equals three over 11, then nine over X minus four equals three over 11.

There's lots of ways we could solve this.

One useful first step will be to multiply by 11, because 11 is the reciprocal of one over 11.

That will leave us with just three on the right hand side.

Now, we wanna multiply by the reciprocal of one over X minus four, which is, of course, X minus four, multiply both sides by X minus four.

We're left with just 99 on the left hand side, and we get three lots of X minus four on the right hand side.

Lots of ways we could go from here.

I think the simplest way is to divide both sides by three.

If X minus four equals 33, then X equals 37.

We'll check that.

F of 37 becomes nine over 37 minus four, which is indeed three over 11.

We know we're right.

Again, we've got multiple methods to choose from.

Same function, same equation, same starting setup.

From here, let's change that fraction on the right hand side.

Let's find an equivalent fraction to three over 11.

Three over 11 is equal to nine over 33.

You can see what I'm doing here, can't you? The numerators are equal, they're both nine, which means that the denominators are equal.

X minus four equals 33, therefore X equals 37.

Oh, that's lovely.

That's a far more efficient way to reach that answer.

Quick check you can do one just like that.

If G of X equals 14 over five minus X, can you solve G of X equals two over three? I believe you can, but pause and show me that you can.

Welcome back.

Hopefully you started by setting up like so.

14 over five minus X equals two over three.

Lots of ways we can go from here.

We can multiply both sides by three, so leave just two on the right hand side, then multiply both sides by five minus X, and we get 42 equals two lots of five minus X.

Divide both sides by two.

X must be equal to negative 16.

Let's check that's right.

G of negative 16 is indeed 2/3.

We know that's right.

We did of course have an alternative method.

Let's manipulate two over three, turn it into an equivalent fraction, 14 over 21, because now we've got matching numerators, so the denominators must be matching.

21 equals five minus X, X equals negative 16.

Same answer, different method, arguably more efficient this one.

Some functions will give us more complex equations to solve.

If F of X equals X over three, G of X equals five over two X plus one.

When we come to solve F of X equals G of X, we're gonna get something rather marvellous.

Let's multiply both sides by three and then multiply both sides by two X plus one.

Two X squared plus X equals 15, therefore two X squared plus X minus 15 equals zero.

We've got a quadratic where the X squared coefficient is not one.

That's a little more complex.

Again, we should ask ourselves, will it factorise? This one does.

So we've got two solutions.

X equals 2.

5 or X equals negative three.

Let's check that they work.

F of 2.

5 is five over six, G of 2.

5 is five over six.

X equals 2.

5 is our solution.

F of negative three is negative one.

G of negative three is negative one.

That means X equals negative three is another valid solution.

Quick check you've got that.

In this case, in solving F of X equals G of X, which of these rearrangements is valid? Pause.

Have a good, a good think, and pick out the right option.

Welcome back.

Let's see how we did.

If F of X equals G of X, then two over X equals five X plus 33 over seven.

From there, we can multiply both sides by seven and then multiply both sides by X.

Oh look, it's option B.

14 is equal to X multiplied by five X plus 33.

Next I'm gonna ask you to continue the good work we've done so far.

In solving F of X equals G of X, we've manipulated to get to the point where we've got 14 is equal to X multiplied by five X plus 33.

I'd like you to keep going and solve the quadratic in order to solve F of X equals G of X.

Pause, give this a go now.

Welcome back, let's see how we did.

Hopefully you expanded that bracket on the right hand side, rearranged to get that quadratic, and then spotted that it factorises, to give us two solutions, X equals 0.

4, or 2/5, and X equals negative seven.

We would of course check that they work.

F of 0.

4 is five, G of 0.

4 is five.

That solution works.

F of negative seven is negative two over seven.

G of negative seven is negative two over seven.

That solution works.

Practise time now.

If F of X equals three over X minus one, G of X equals 18 over X plus nine.

For part A I'd like to solve F of X equals 1/3, but I'd like you to solve it twice, using two different methods, demonstrating your mathematical fluency.

For part B, I'd like to solve F of X equals G of X, and again, can you solve that equation twice using two different methods to demonstrate your mathematical fluency? Pause, give this a go now.

Question two.

Three functions, F of X, G of X and H of X.

Can you solve F of X equals 21 over 39, H of X equals 33 over 18, F of X equals G of X, and F of X equals H of X? Pause and give those four a go now.

Feedback time, let's see how we did.

Question one, Part A, I asked you to solve F of X equals 1/3 twice using two different methods.

One way would be to set up the equation, three over X minus one equals 1/3, multiply both sides by three, then multiply both sides by X minus one to get X equals 10.

An alternative method would've been to manipulate the one over three on the right hand side, so that's the equivalent fraction three over nine.

Therefore, X minus one equals nine, X equals 10.

Two different methods, same result.

Part B, I actually solve F of X equals G of X and do it two different ways.

We could have multiplied both sides by X plus nine, multiplied both sides by X minus one.

Expand the brackets on both sides, rearrange, simplify and find the X equals three.

Alternatively, we could have started by finding an equivalent fraction on the left hand side, because if we make both those numerator 18, then the denominators are equal.

Six lots of X minus one equals X plus nine.

Expand that bracket on the left hand side, rearrange, and once again find that X equals three.

Question two, part A, solve F of X equals 21 over 39.

You should have done this like so.

And found that X equals 15.

You might have used an alternative method and arrived at the same answer.

As long as you got to X equals 15, that's absolutely fine.

For part B, solving H of X equals 33 over 18.

We get to X equals five.

Question two, part C, solving F of X equals G of X.

Well, we start like so, and lots of ways we can manipulate that, but however we manipulate it, we should arrive at that quadratic, X squared plus X minus 20 equals zero, which will factorise to give us two solutions, X equals four, X equals negative five.

Part D, F of X equals H of X.

We can set up like so.

When we manipulate, we end up at a quadratic, two X squared minus three X minus 44 equals zero.

That factorises, giving us two solutions, X equals 5.

5, X equals negative four.

We're at the end of the lesson now.

We have learned equations involving functions can be solved.

When solving equations such as F of X equals negative 11, we're finding the original input that maps to negative 11 under the function F of X.

When solving equations such as F of X equals G of X, we're finding the moment when both functions share the same input and output.

Hope you've enjoyed today's learning as much as I have.

I look forward to seeing you again soon for more mathematics.

Goodbye for now.