Lesson video

Hello, Mr. Robson here.

Superb choice to join me for maths today, especially seeing as we're solving a quadratic and linear pair of simultaneous equations using substitution.

Awesome! Our learning outcome is that we'll be able to solve two, one linear, one quadratic, simultaneous equations algebraically using substitution.

Substitution, substitute, the keywords today.

Substitute means to put in place of another.

In algebra, substitution can be used to replace variables with values, terms or expressions.

Two parts to today's lesson.

We're going to begin by solving using substitution.

It's possible to solve a pair of simultaneous equations, where one is quadratic and one is linear using elimination.

x squared plus y equals negative 4.

y equals 7x plus 6.

Let's call the top one equation one, the bottom one, equation two.

And then let's subtract equation two from equation one, giving us on the left hand side x squared, because y minus y is no y's left, and on the right hand side we get negative 7x, negative 10.

Subtracting equation two from equation one eliminates the y variable and leaves a quadratic, which we can solve to find the x values.

We'll rearrange that so that our quadratic is equal to 0.

That one factorises, delightful.

That'll give us the solutions x equals negative 5, x equals negative two, and of course we're not finished there.

We need to substitute in both x values to find the corresponding y values, and we'll get y equals negative 29 and y equals negative 8.

So a pair of solutions.

x equals negative 5, y equals negative 29, and x equals negative 2, y equals negative 8.

It's also possible to solve a pair of simultaneous equations using substitution.

I'm gonna show you how to solve that exact same pair of simultaneous equations using substitution instead of elimination.

Did you notice x squared plus y equals negative 4? And equation two told us exactly what y is, it's 7x plus 6.

So we can substitute that 7x plus 6 into the y position of equation one.

Once we've done that, we can make a quadratic which we can solve to find the x values.

Removing those brackets and rearranging gives us x squared plus 7x plus 10 equals 0.

And look, it's the exact same quadratic, so we would get the exact same solutions.

On the right hand side there, we see how we get to that quadratic by substitution rather than elimination.

Let's check we've got that skill.

I'm going to do an example where I solve by substituting, and then I'm gonna ask you to repeat with your own example.

I need to start by labelling my equations one and two, and I'm substituting equation two into equation one.

Equation one read x squared plus y equals 5, and I know what y is, y is 2x minus 10, so I substitute that expression into the y position of equation one.

I can remove those brackets and rearrange to get the quadratic x squared plus 2x minus 15 equals 0.

It factorises, lovely, x plus 5, x minus 3, giving me the solutions x equals negative 5 and x equals positive 3.

And you know I'm not finished there.

I need to substitute those x values back into one of the original equations to find the respective y values.

Once I've done that, I've got my pair of solutions.

x equals negative 5, y equals negative 20.

And when x equals positive 3, y equals negative 4.

Over to you.

I'd like you to solve this pair of simultaneous equations by substituting.

Pause and give this a go now.

Welcome back.

Let's see how we did.

Hopefully you labelled them as equation one and equation two, and then substituted equation two into equation one.

From there, you would rearrange to get the quadratic x squared minus 8x plus 15 equals 0, and yours factorised to two, giving you x solutions, x equals 5, x equals 3, and I hope you didn't stop there.

I hope you substituted them back into one of your original equations to get your corresponding y values.

There's two solutions therefore, when x equals 5, y equals negative 19, and when x equals 3, y equals negative 3.

Well done.

Sometimes it's helpful to rearrange before substituting.

If I label those equation one and equation two, if I can make y the subjects of equation two, I can substitute.

Rearranging equation two will leave us with y being equal to 3x plus 16.

We're now ready to substitute it into equation one.

When we do, it looks just like that.

And remove that bracket and rearrange, and we get the quadratic 2x squared minus 3x minus 20 equals 0.

That one will factorise again.

When 2x plus 5 equals 0, x equals negative 5 over 2.

The corresponding y value when I substitute that x solution back in is y equals 17/2.

The other x solution is x equals 4.

And the y value there is y equals 28.

We could have rearranged equation one and then substituted.

It's one of those wonderful moments in mathematics where there's multiple methods to solve the same problem.

You just watched me substitute equation two into equation one.

We could substitute equation one into equation two.

Let's see what that looks like.

We need to rearrange equation one, so that y is the subject.

y in this case is 2x squared minus 4.

Let's substitute that into equation two.

And then when you remove that bracket and simplify, oh, look, 2x squared minus 3x minus 20 equals 0.

Did that look familiar? Of course.

The same quadratic which would give us the exact same pairs of solutions.

I just wanna show you what this looks like graphically now, so you can see what's going on here.

The equation 2x squared minus y equals 4 is that beautiful parabola.

The equation y minus 3x minus 16 equals 0 is the straight line.

And look at the two points of intersection, 4, 28, that's our solution x equals 4, y equals 28.

And my graphing technology has described the other intersection as negative 2.

5, 8.

5.

That's my solution x equals negative 5/2, y equals 17/2.

Our solutions are the two intersections of the straight line and the curve.

Quick check that you've got this now.

I'd like to know which rearrangement correctly makes y the subject for equation two.

Is it option A, option B or option C? Pause.

Have a think.

Welcome back.

Well done.

It's option A.

When we rearrange equation two, y equals 2x plus 3.

We can substitute that in to equation one to get 3x squared plus, brackets, 2x plus 3 equals 8.

From there, which rearrangement correctly produces a quadratic that is ready to be solved? How would you rearrange that? Would it be into option A, option B, or option C? Pause.

Have a think.

Well done for saying it's not option A.

The right hand side must be 0 if we're going to solve this quadratic, so it was option C.

We now have 3x squared plus 2x minus 5 equals 0.

From there we can factorise, but how? I'd like you to select the correct factorization of 3x squared plus 2x minus 5.

Three options there.

Pause and take your pick.

Welcome back.

I hope you said it's option A.

That would factorised to 3x plus 5 and x minus 1.

Now that we've factorised, we're ready to find our solutions.

Four options there.

Pick the correct solutions.

Notice I said solutions plural.

There's more than one.

Pause and do that now.

Welcome back.

Let's see how we did.

A was indeed a solution.

If x minus 1 equals 0, then x equals positive 1.

Substitute that back into either equation and you'll find the corresponding y value y equals 5.

Our other solution, when the bracket 3x plus 5 equals 0, then x has to be negative 5/3 for the corresponding y value, y equals negative 1/3.

Sometimes when substituting, we end up with a term involving brackets.

Let's have a look at what that means.

I'll call that equation one, and that equation two, and rearrange equation two to make y the subject.

And when I substitute it in, oh, of course, x squared plus 5y becomes x squared plus 5 lots of, bracket, 2x plus 5.

Now you've got some brackets that need to be expanded but that's not a problem.

We've got 5 lots of 2x, 5 lots of 5.

It expands like so.

And we're ready to rearrange that, and we leave ourselves with x squared plus 10x plus 24 equals 0, which factorises, lovely.

From here, x equals negative 6, and when we substitute that back in, we find the corresponding y value, negative 7, and for the right hand bracket, x equals negative 4 with a y value y equals negative 3.

Again, just to show you the graphical representation of what's going on here, x square plus 5y equals 1 is that beautiful parabola.

y minus 2x minus 5 equals 0 is the straight line.

Two intersections at negative 6, negative 7, negative 4, negative 3.

Notice our solutions are those coordinates.

Let's check we're getting this now.

Sofia has made an error in trying to solve this pair of simultaneous equations.

That's fine, Sofia.

We're always going to make mistakes when we do maths.

It's important that we're able to spot them, reflect on them and correct them.

What I'd like you to do is spot Sofia's error and finish solving this pair of simultaneous equations.

Pause, see if you can spot the error.

Welcome back.

Did you say that's the error, positive 30? It's not positive 30.

We need negative 6 of everything inside the bracket.

Negative 6 lots of x is indeed negative 6x, but negative 6 lots of 5 is negative 30.

Errors regarding negatives are common.

It's really important that we double check our work, especially around brackets and negatives.

Once you're wrong at one stage of your working, unfortunately, the error carries on.

So if we correct that error, and instead of reading positive 30, it now reads negative 30, we can then rearrange.

We're left with x squared minus 6x minus 27 equals 0.

That factorises and gives us solutions x equals negative 3.

Substitute that back in and you'll find y equals 2.

x equals positive 9.

Substitute that back in and y equals 14.

You're welcome, Sofia.

We won't always be able to easily factorise to solve, and it won't always be efficient to make x or y the subject.

If we look at this pair, call the top one equation one, the bottom one equation two.

Rearranging equation two, we get 3y equals x minus 5, therefore y equals x/3 minus 5/3, and we substitute that back into equation one.

And we can absolutely definitely solve from here, but there's a more efficient approach.

And as mathematicians, we should always be on the lookout for efficiency.

If we go back to this moment, can you spot something? 3y equals x minus 5.

Look at equation one, 5x squared plus 6y.

Well, 6y rather conveniently is 2 lots of 3y.

So let's call equation one 5x squared plus 2 lots of 3y equals 9, because now we can substitute in what we know 3y to be.

We know it's equal to x minus 5.

So substitute that in and we get 5x squared plus two lots of bracket x minus 5 equals 9.

Remove the bracket, rearrange.

Hurrah, we've got a quadratic that we can solve.

Didn't factorised this one, so I'm gonna use the formula.

I use the formula, it simplifies to so, and I get a positive solution, x equals 1.

75.

When I substitute that back in, y equals negative 1.

08.

My negative solution, x equals negative 2.

15, y equals negative 2.

38.

Quick check we've got that.

True or false, to solve this pair of simultaneous equations, we have to make y the subject of equation two before we do any substituting.

Is that true or is it false? Once you've decided, can you use one of those two statements at the bottom of the page to justify your answer? Pause, have a think about this one now.

Welcome back.

I hope you said false and justified that with the statement you could substitute 4y equals 11 minus x into equation one.

What does that look like? Well, let's call equation one 3 lots of 4y minus 7x squared equals 3, and then substitute in what we know 4y to be, 11 minus x.

From there, we can rearrange to get a quadratic 7x squared plus 3x minus 30 equals 0.

We could solve that, and we didn't need to make y the subject before substituting.

Practise time now.

Question one, I'd like you to solve these pairs of simultaneous equations using substitution, rounding to one decimal place if necessary.

Pause and give them a go.

Question two, Sofia has tried to solve these pairs of simultaneous equations using substitution.

I'd like you to mark her work, looking for errors and correcting any errors you find.

Pause and do this now.

Feedback time now.

Part A was about solving a pair of simultaneous equations using substitution.

Hopefully we called the top one equation one, the bottom one equation two, and substituted equation two into equation one to get that.

Remove the bracket.

Rearrange.

x square plus 3x minus 28 equals 0.

It factorises which is lovely, nice and efficient.

Giving us solutions x equals negative 7.

Substitute that back in and you'll find y equals negative 41, and x equals positive 4.

Substitute that back in, you'll find y equals negative 8.

There's your pair of solutions.

For question one part B, we need to start by rearranging equation two to make y the subject, and then we can substitute that into equation one and it will look like so.

Remove the bracket, rearrange.

x squared plus 7x plus 5 equals 0.

Doesn't factorised that one.

I'm gonna use the formula.

Substitute into the formula.

Gives us x equals negative 7, plus or minus the root 29 over 2.

So to one decimal place, we get a solution x equals negative 0.

8, y equals negative 4.

7.

And our second solution x equals negative 6.

2, y equals negative 42.

3.

For question two, part A, we're looking for any errors in Sofia's working.

There's one right there, right at the beginning, and rearranging to make y the subject of equation two.

And unfortunately, once you've made an error there, that error continues and we're led to the wrong solutions.

So to correct that error, we should have rearranged equation two to y equals 3x plus 5, and then we substitute it into equation one, we get that.

Negative 2 lots of 3x is negative 6x.

Negative 2 lots of 5 is negative 10.

Rearrange from there the quadratic x squared minus 6x minus 16 equals 0, which factorises and gives us a pair of solutions, x equals negative 2, y equals negative one, x equals positive 8, y equals positive 29.

For part B, the error was there.

So close.

So much effort's gone into that work, but we need to be really careful around negatives.

Negative 2 lots of 5 is indeed negative 10, but negative 2 lots of negative x would mean that should be positive 2x there.

Unfortunately, once you make an error there, your solutions will be wrong at the very end.

If we go back and correct that line of working, 3x squared plus 2x minus 10 equal 6.

We can rearrange that and it'll factorise, and give us this pair of solutions.

Onto the second half of lesson now, further solving using substitution.

We can use substitution where elimination is not possible.

Call that equation one and the second one equation two.

If we were to try and eliminate by subtracting equation two from equation one, the left hand side looks like this, x squared plus y squared subtract y minus x, and that bracket expands to so, and then nothing simplifies.

We often see pupils make mistakes here when they think that you can add x to x squared, or you can subtract y from y squared, but that's not the case because they're not like terms. Because they're not like terms, we haven't eliminated anything.

We're left with x squared plus x plus y squared minus y equals 18.

We're no closer to solving.

We can't eliminate any variables by adding and subtracting in this case.

But we can use substitution.

So, let's rearrange equation two to make y the subject.

y equals x plus 7.

We can substitute that into equation one.

Instead of x squared plus y squared, we now have x squared plus bracket x plus 7 squared.

We now have the pair of brackets to expand.

x plus 7 squared is x plus 7 multiplied by x plus 7.

We can expand that to x squared plus 14 x plus 49.

Remove that bracket, rearrange, 2x squared plus 14x plus 24 equals 0.

Divide through by 2, and we've got a quadratic, which factorises.

Super.

I can find those solutions.

x equals negative 3, substitute that back into the original equations.

I find the corresponding y value of 4.

And for the second bracket, x must be equal to negative 4.

Substitute that back in and I find the corresponding value y equals 3.

We could have solved this same pair by making x the subject.

Look at equation two.

We could rearrange it to be x equals y minus 7.

We can then substitute that into equation one, but instead of writing x squared, we're gonna write y minus 7 squared.

There's a pair of brackets to be multiplied together, which will give you that.

Remove those brackets, simplify.

2y squared minus 14y plus 24 equals 0.

Divide 3 by 2 and our quadratic factorises.

Our solutions therefore, y equals positive 4, substitute it back into the original equation to find x equals negative 3.

y equals positive 3, substitute that back in, x equals negative 4.

Did you notice? The exact same pair of solutions.

When y equals 4, x equals negative 3, or when an x equals negative 3, y equals 4, and when y equals 3, x equals negative 4.

By making x a subject, we created a quadratic equation in terms of y.

This means we find the y values first, then use those to find the x values.

Either way round is absolutely fine.

Quick check you've got that now.

True or false, in order to solve this pair of simultaneous equations, we must start by making y the subject of equation two.

Is that true? Is it false? Once you've decided, could you use one of the two statements at the bottom of the page to justify your answer? Pause, have a think about this now.

See you in a moment.

Welcome back.

I hope you said false, and I hope you justified that with the statement "There are multiple ways to solve this problem.

We could make x subject of equation two and substitute." When we do that, we get x equals three minus y.

Substitute that into equation one.

Expand that pair of brackets.

Simplify, 2y squared minus 6y minus 21 equals 0.

We could find those y values, then use those to find the corresponding x values.

There are multiple ways to solve simultaneous equations.

The most efficient method is determined by the nature of the equations.

Making x the subject in this case, we get x equals 7 minus 3y.

When we substitute that into equation one, 7 minus 3y squared, we can expand and solve.

If we try to make y the subject, for equation two, y would be equal to 7/3 minus x/3.

When we substitute that back into equation one, we get x squared plus, bracket, 7/3 minus x/3, close bracket, squared equals 17.

It's not impossible.

We could certainly solve by multiplying out that pair of brackets and simplifying, but it's less efficient than the version on the left hand side, where we made x the subject.

If we continue from here, expanding the bracket 7 minus 3y squared gives us 49 minus 42y plus 9y squared.

Remove the bracket and simplify, and we've got a quadratic that we can solve.

Using the formula.

By the time we've done all that, we get y equals 3.

2, and y equals one.

We use those two y values to find the corresponding x values.

So y equals 3.

2, x equals negative 2.

6 is one solution, and y equals one, x equals 4 is another solution.

Quick check you've got this.

Could you choose the most efficient method from the two below and solve this pair of simultaneous equations? That's your pair of simultaneous equations, and the methods to choose from are making x the subject, or making y the subject.

Which of those two is going to be the most efficient? Once you've decided that, see if you can find the solutions.

Pause, do that now.

Welcome back.

I do hope you said making x the subject is not impossible but it's certainly less efficient, and we mathematicians are always on the lookout for efficiency.

So making y the subject is more efficient in this case.

When we substitute that into equation one, we get this.

Multiply the brackets.

Remove the brackets, simplify.

28x squared minus 70x plus 36 equals 0.

We can solve from here, but we can't factorised, so I'm using the formula.

And we end up with x equals 1.

8, y equals negative 1.

9, and x equals 0.

7, y equals 3.

4.

Practise time now.

For question one, I'd like you to solve this pair of simultaneous equations by substitution.

Pause and do this now.

Feedback time.

Part A, solving by substituting.

Lots of ways we could do it.

I'm gonna go for making y the subject of equation two, and substituting that into equation one, multiplying out the pair of brackets, simplifying, and from there it factorises, x minus 8, x plus 6 equals 0.

That'll give us a pair of solutions.

When x equals positive 8, y equals negative 6, and x equals negative 6, y equals positive 8.

For part B, it was more efficient to make x the subject of equation two this time, rather than making y the subject.

When I've rearranged equation two into x equals 1 minus 3y, substitute that back into equation one.

Expand the brackets, simplify, and from here, we have a quadratic we can solve, giving us y equals 2.

5.

Use that y value to find the corresponding x value of negative 6.

6, and y equals negative 1.

9.

Using that to find the corresponding x value 6.

8.

What we've learned is that we can algebraically solve two simultaneous equations, one linear and one quadratic, using substitution.

Sometimes this will involve multiplying out a bracket or a pair of brackets.

Setting out your work clearly and labelling the equations can help to eliminate errors and make your work easier to check.

I hope you've enjoyed this lesson as much as I have, and I look forward to seeing you again soon for more mathematics.

Good bye for now.