Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Today we're going to be solving a quadratic and linear pair of simultaneous equations using elimination.

That sounds awesome.

Let's take a look.

Our learning outcome is I'll be able to solve two, one linear, one quadratic simultaneous equations algebraically using elimination.

Elimination's a key word for today.

"Elimination is a technique "to help solve equations simultaneously "where one of the variables in a problem is removed." Two parts to today's lesson.

And given that we're gonna be solving a quadratic and linear pair of simultaneous equations, we're gonna start by reviewing how we solve quadratic equations.

In maths we can solve equations with simple rearrangement at times.

For example, 8x minus 20 = 0, we could add positive 20 to both sides, divide through by 8, and we'll get the solution x = 20 over 8, and you'd simplify that to 5 over 2.

Some might even write it as a decimal 2.

5.

This is a linear equation.

It's got just one solution.

When does the line 8x minus 20 pass 0? It's that point there, when x = 5 over 2 or 2.

5.

Jun tries to solve this equation by simple rearrangement; x squared + 8x = 0.

And Jun says x is a factor of the terms on the left-hand side.

So I'll divide through by that.

Let's divide every term by x, and we get to having 8, x + 8 = 0.

Let's add -8 to both sides: x = -8.

Do you agree with Jun's solution or is something a little amiss here? Pause, have a think.

See if you can spot what's going on.

Welcome back.

I wonder if you notice something? Well, I hope you notice that x = -8 satisfies the equation.

It's a solution.

When we substitute -8 into the exposition, we do indeed have satisfied equation.

But this is a quadratic equation.

The nature of that line is different, it's a parabola.

Therefore, x = -8 is only one of the solutions.

Jun's realised this? "I remember now!" "Because x is a factor of both terms on the left-hand side "I can factorise." So we factorise x squared + 8x to become x bracket x + 8.

Now Jun's realised something else.

"I can use the fact that if the product "of two expressions is 0, "one of the expressions must be 0." So in this case, either x has to be 0 or x + 8 has to be 0 in order for the product of the two to be equal to 0.

So we've got two solutions now; x = 0, x = -8.

If we look at the graph, we can see x = 0, x = -8 are the two solutions to our quadratic.

For some quadratic equations, factorising can be the quickest and easiest way to solve x squared + 8x minus 20 = 0.

Well, that factorises beautifully: x + 10 x minus 2 equals to 0.

So either x + 10 = 0 or x minus 2 = 0.

So we get two solutions; x = -10 or x = positive 2.

If we look at the graph of that, we can see our solution x = -10 and our solution x = positive 2.

It's useful to know more than one way to solve the same problem in mathematics.

We can check and confirm our answer when we do that.

So we just solve by factorising this equation: x squared + 8x minus 20 = 0.

And we know the solutions are x = -10, x = 2.

If I had another way of doing this, I could check whether I'm right.

Enter the quadratic formula.

We can solve that same quadratic by using the quadratic formula.

In case you've forgotten what the quadratic formula is, it's simply x = -B plus or minus the square root of B squared minus 4ac, all over 2a.

So we need our quadratics to be in the form, axe squared + bx + c = 0.

So we can substitute a, b, and c into the formula.

Thankfully this one's already in that form.

We've got an x squared coefficient of 1, so a is 1.

We've got an x coefficient of 8, so b is 8.

We've got a constant of -20.

Let's substitute those into the quadratic formula.

It should look like that.

I'm now in a position where I'm going to have two solutions: x = -8 plus or minus 12 over 2.

I'll find the positive one first.

<v ->8 + 12 over 2 gives me 2.

</v> I'll go back into my calculator and find the negative one: x = -8 minus 12 over 2 = -10.

Oh look, the exact same pair of solutions.

So we must have been right when we factorised.

Some quadratics don't factorise easily.

In which case let's use the formula to solve it: x squared + 8x + 2 = 0.

I'm gonna use the formula when I substitute an a value of 1, a B value of 8, and a c value of 2 into the quadratic formula.

It looks like that: x = -0.

258, and x = -7.

741.

Using the formula wasn't the only way to solve this quadratic, we could have completed the square instead.

When completing the square, we need this quadratic in the form: bracket x + p squared + q = 0.

In order to get that form, what I'm gonna do is arrange x squared + 8x + 2, like so, because it's starting to look like a square.

I'm trying to arrange it into x + 4 squared, as in 2 lengths of x + 4.

But have you noticed I haven't quite got the whole square, I haven't got x squared + 8x + 16, I'm 14 short.

So x squared + 8x + 2 becomes bracket x + 4 squared minus 14.

Why would I want it in this form? Because I can add positive 14 to both sides, take the square root, remembering to take both the positive and the negative root, and then I've got a solution x = -4 plus or minus the root of 14.

Does that look familiar? So our two solutions become x = -0.

258 and x = -7.

741.

Oh look, the same pair of solutions.

Quick check you've been listening.

Which of these are methods to solve a quadratic equation? Pause, select the right options now.

Welcome back.

I do hope you said, "By factoring." Yes please.

"Using the quadratic formula." Yes please.

"And completing the square." Yes please.

You may have said "Trial and error." It could work, but in most cases you'll find this method very inefficient.

So stick to the top three.

Let's check we can use those methods.

I'm gonna solve a couple of quadratic equations by factorised and then ask you to repeat that skill.

So, solved by factorising: x squared minus 5x minus 14 = 0.

That'll factorise: so you x minus 7 x + 2.

If x minus 7 is going to be 0, then x has to be positive 7.

And if x + 2 is going to be 0, x has to be -2.

There's my two solutions to that quadratic.

Second quadratic's a little trickier 'cause the x squared coefficient's not 1.

So the factorization is a little trickier but not impossible.

I know I'm gonna have 3x as the first term, in the first bracket, and x is the first term in the second bracket, that factorises to 3x minus 14 and x + 1.

If 3x minus 14 is 0, then x is positive 14 over 3.

The second solutions nice and simple: x = -1.

Over to you now.

Pause.

Try this now.

Welcome back.

Let's see how you did: x squared minus 8 x + 15 factorised to x minus 3 x minus 5, giving you two solutions: x = positive 3, x = positive 5.

3x squared minus 17 x + 10 = 0 factorise to: 3x minus 2 x minus 5, your solution's: x = 2 over 3, and x = positive 5.

The next skill to check is that we can solve using the formula: 6x squared + 17x minus 30 = 0.

I tried to factorise it and couldn't.

I'm gonna use the formula.

And I'm asked to round my answers to two decimal places.

So there's the formula.

That's my a, b and c value substituted in.

That's what it looks like in my calculator display.

I can type the whole thing into my calculator display, and I'll find that solution first.

The positive root: x = 1.

23, it's two decimal places.

And I'll use the left arrow key on my calculator to go and change that positive root to a negative root, and I found my second solution: x = -4.

06, two decimal places.

Your turn.

Solve this quadratic equation using the formula and give your answers to two decimal places.

Pause.

Try that now.

Welcome back.

Let's see how you did.

Hopefully it look like that when you substitute the values in, and your calculator display looked like so.

And your positive root looked like that giving you x = 1.

26 to two decimal places, and your negative root looked like so, giving you x = -0.

16.

Well done.

Last thing to check.

Can we complete the square? I do enjoy completing the square.

Give your answers to three significant figures.

And I have to solve: 3x squared minus 7x + 16 = 2x square + 3x minus 1.

Wow! I need to do some rearranging, I need the right-hand side to be 0.

So I'll add -2x squared to both sides.

Then I'm gonna add -3x to both sides, and then I'm gonna add positive 1 to both sides.

So I've rearranged my complicated looking question into x squared minus 10x + 17 = 0.

I do rely on my visual representation to put that into the correct form.

I need it in the form: bracket x + p squared + q = 0.

So I'm gonna arrange my x squared minus 10x + 17 as if I was trying to arrange it into a square.

Have I got x minus 5 squared? No, I've not got the whole thing, I'm 8 short from having the whole thing.

So, I can rearrange it into: bracket x minus 5 squared minus 8 = 0.

Add positive 8 to both sides.

Take the square root, remembering to take the positive and the negative root of 8, and then add positive 5 to both sides.

I'm ready to find my two solutions: x = 7.

83, and x = 2.

17, they're rounded to three significant figures.

Your turn.

Pause.

Try this one now.

Welcome back.

How did we go? Hopefully you rearranged all that to read; x squared minus 18x + 71 = 0 and then is it x minus 9 squared? Not quite.

You're 10 short.

We can add positive 10 to both sides.

Take the square root, both positive and negative root of 10.

Then you'll get the solution x = 12.

2 and x = 5.

84.

Those are your solutions to three significant figures.

Practise time now.

For question one, I'd like you to solve both of these quadratic equations using all three methods.

Pause, and do this now.

For question two, I'd like to solve these three quadratic equations, but in this case I'd like you to select the most efficient method for each equation.

Pause, and do that now.

Feedback time.

I asked you for question one to solve these quadratic equations using all three methods.

x squared + 6x minus 40 = 0.

Remembering we needed that right-hand side to be 0 in order to factorised.

By factorising you get x + 10 x minus 4.

Our solution's x = -10, x = positive 4.

You could also solve it by using the formula and you'll notice x = -10, x = 4.

Same result.

You could have solved it by completing the square: x = -10, x = 4.

You should have all three of those methods written down and you should have achieved the same pair of solutions each time.

For part B, the quadratic equation: 5x square + 42x + 16 = 0.

Using all three methods.

That's by factorising.

That's by using the formula And that's by completing the square.

Notice again, each method led to the same pair of solutions.

I'll ask you to pause and just check that your working out looks exactly like mine for this one.

For question two, it's about efficiency this one.

Choosing the right method for each quadratic.

I hope you noticed in the case of a it factorised, and it's incredibly quick and simple to get those two solutions: z = -7, x = 1.

For b, but I quite like the fact that 6x I can half that to three, so I can put it immediately into the form of completing the square.

You didn't have to, you could use the formula.

I just want to complete the square for this one.

So I get those two solution: x = -1.

58, x = -4.

41.

Wasn't specified whether we want that to two decimal places, three decimal places, so I've just truncated those answers there.

For part c, I use the formula 'cause I know if I put all that into my calculation in one fell sweep, I'm gonna get these two solutions: x = -0.

158 truncated, and x = -0.

441 truncated.

You may have used slightly different methods, just check that your method was as simple and efficient and that your solutions are nice and accurate like mine.

Onto the second half of the lesson now where we're gonna solve simultaneous equations whereby one is a quadratic and one is a linear.

We can solve a pair of linear simultaneous equations by elimination.

In this example, if I call that equation one and that's equation two, and then I subtract equation two from equation one, we get 2x + 0 = 10.

Why? Because 3x minus x is 2x, 6y minus 6y is nothing.

Lovely.

The y variable has been eliminated.

We're left with 2x = 10, so x = 5.

Super, we've got our solution.

Well done.

You are screaming at the camera now.

No we haven't.

We need to find the corresponding y value.

Well done, we do.

So we substitute back in x = 5 into either equation.

I've chosen equation two this time, I will find 5 + 6y = 11, 6y = 6, y = 1.

There we are.

That is the solution: x = 5, y = 1.

It's the only pair of values that satisfy both equations simultaneously.

In this case we had a pair of linear simultaneous equations, two straight lines, and look at that point of intersection.

Can you see now why we required that y value? X = 5 wouldn't have been a valid solution.

We need that precise coordinate: x = 5, y = 1.

We absolutely have to have that y value as well.

We can also use elimination to solve a pair of simultaneous equations when one is quadratic.

Wow! x squared + y = 25, y + 8x = 5, one of them is quadratic.

Hmm, that looks like it's going to be difficult.

Not really.

It's no more difficult than what we've just seen.

I'll call that equation one, I'll call that equation two, and then I want to subtract equation two from equation one.

Firstly, I'm going to lay the equations out like this.

Instead of writing it as x squared + y, equation one, I'm gonna call it x squared + y + 0x.

And then equation two, I've got 0x squareds, one y and 8 x's.

I'm aligning the terms there.

I've aligned the x squared terms, I've aligned the y terms and I've aligned the x terms. Because when I come to subtract equation two from equation one, I can now appreciate x squared subtract no x squareds, leaves me with x squared.

y minus y, that's no y's left at all, they've eliminated.

And then 0x minus 8x is -8x.

We've subtracted equation two from equation one.

We didn't have to do it that way, we didn't have to align those terms. We could have written: x squared + y and I want to subtract bracket y + 8x.

But we have to be super careful.

I need to subtract the y and subtract the 8x.

That will simplify to x squared minus 8x, which you'll notice is the exact same left-hand side we got earlier.

Either method is fine.

Just make sure you double check your answers.

Now that we're at this stage, you'll notice when I rearrange that x squared minus 8x minus 20 = 0, we've eliminated the y variable and we've created a quadratic equation which we can solve.

In this case it factorised beautifully.

I get the solutions x = 10, x = -2.

It was a quadratic equation, so no surprise that we got two solutions.

Finished? Well done.

We're certainly not.

We need to substitute back both x values to find the corresponding y values.

I'm gonna substitute back into equation one and find when I substitute in x = 10, I get y = -75.

So we've got one solution x = 10, y = -75.

Then I'll substitute x = -2 back into the same equation and find that y = 21: x = -2, y = 21.

Two pairs of values that satisfy both equations simultaneously.

I'm gonna leave those two solutions there because I want to show you something.

This is the graphical representation of what's actually happened here.

x squared + y = 25 gives us that beautiful parabola.

y + 8x = 5 is the linear graph.

And look at the two intersections: - 2, 21.

Ah, that's one of our solutions.

10, -75.

That's the other solution.

So we absolutely needed that y value to pair with the x value.

Those are accurate solutions.

Now I don't want you thinking, "Mr. Robson said we always have to substitute "back into equation one." I don't care.

It can go back into either equation.

In fact substitute it back into both equations.

Double check your work.

We could substitute back into equation two instead, those x values.

It doesn't matter which equation we substitute back into, we will arrive at the same solutions.

Watch.

When we substitute x = 10 and x = -2 into equation two.

Look the exact same solutions.

I also don't want you thinking, "Mr. Robson said we have to subtract "equation two from equation one." We could have subtracted equation one from equation two.

Let's look at what that would've looked like.

equation two subtract equation one would've given us 8x minus x squared = -20.

Look familiar? Absolutely.

Once we've rearranged that we get 0 = x squared minus 8x minus 20.

Oh, the exact same quadratic.

I've solved this pair of simultaneous equations and then blanked out some moments in my working.

I'd like you to think what's in those blank spaces.

See if you can copy this down, fill in those spaces and arrive at a pair of solutions.

Pause.

Do it now.

Welcome back.

Let's see how we did.

Hopefully you wrote -6x there.

The y has been eliminated.

Lovely, we're ready to solve.

Or we're ready to rearrange so that we can solve.

That's gonna be x squared minus 6x + 5 = 0.

Wonderful, a quadratic that factorises 2x minus 5x minus 1, giving us two x solutions: positive 5 and positive 1.

When we substitute those back in and I just chose to substitute into equation one, either would've been fine.

We get 5 squared + y = 5, same, x = 5, y = -20.

And for the x = 1 solution, when x = 1, y = positive 4.

Sometimes we cannot immediately eliminate the linear term.

if we try to subtract equation two from equation one in this example, it would look something like that, which simplifies to that, and we're left with x squared minus 2x + 6y = -2.

Hmm, we haven't eliminated the y variable.

What if we add the equations together? It would look like this.

And it'll simplify the same.

Again, we haven't eliminated anything so we're a little bit stuck here.

So what do we do? Sometimes when we can't immediately eliminate a linear term, we can multiply one of the equations, so there's a common coefficient in both equations.

What's meant by that? Let me show you.

I'm gonna multiply equation two by 5.

Why would I want to do that? It's because 2x multiplied by 5 is 10x, <v ->y multiplied by 5 is -5y,</v> 13 multiplied by 5 is 65.

Spot something? Well done.

When I align that with equation one, we've got positive 5y and -5y, this is delightful.

What we can do now is add those two together.

x squared + 10x = 76.

What happened to the y's? Well, positive 5y add -5y is nothing.

We've eliminated the y variable.

We've now got a quadratic to solve.

Super.

We can solve that in lots of ways.

I'm gonna complete the square on this occasion.

I get the solution x = -5 plus or minus square root of 101, x = 5.

04, that's truncated, and x = -15.

04, that's truncated.

Our calculator's memory function is super useful now.

It's gonna help us preserve accuracy when we substitute those x values back into our equation to find the y values.

That's the x solution, x = 5.

049875621.

In fact, that's not the whole story, there's loads more decimal places to this answer.

And our calculator knows them, and we're gonna use the memory function to store it.

That's where I find the memory function on my calculator.

When I press that button, this comes up, and I'll store that solution under A.

So you can see my calculator now knows that the value A is -5 + the square root of 101.

I didn't have to type in 5.

04 et cetera, I'm just gonna type A into one of our equations.

I need to store the negative solution as well.

So -5 minus the root of 101.

When I press the memory button, I get to store that somewhere else.

I'm storing that under B.

So when I come to substitute these x values back in: 2x minus y = 13, I'll rearrange that: y = 2x minus 13.

So I need to substitute those x values into 2x minus 13.

It looks like that on the calculator.

That's two lots of a minus 13.

That's our first x solution under A.

I accessed it by pressing Shift and 4, you'll see a tiny little A next to the 4 button on my calculator.

I press Shift and 4, I can type the A in and it will substitute in the x value of 5.

04 et cetera, and it gives me the corresponding y value, <v ->2.

900248, et cetera.

</v> You'll notice I've just truncated that and I've written it down there.

I'm gonna do the same for the solution I stored under B.

I had to press Shift 5 on this occasion, you'll see a tiny little B next to my 5 button and that gives me the solution -15.

04, y = -43.

09.

So we found two solutions for this pair of simultaneous equations.

And look at the graphical representation.

The solutions give us the coordinates at the points where the graphs intersect.

You might be staring at the screen saying "They're not exactly the same: 5.

05, -2.

9.

"That's not the same as what you've got written." I've left mine truncated as accurately as possible, whereas Desmos, the graphing technology I use to make this visual representation, does round.

So the Desmos solutions you see there are rounded solutions.

By what factor do we need to multiply equation two by in order to eliminate the y variable? Pause, and have a think.

Welcome back.

I hope you said, "One multiplier by 4," because 4 lots of equation two becomes 12x -4y = 28.

Wonderful.

We've got matching coefficients.

From here, I'd like you to fill in those blanks to produce a quadratic to solve.

Pause, and work that out.

Welcome back.

When we're subtracting 12x -4y from 4x squared minus 4y we get 4x squared minus 12x.

That's the left-hand side.

19 minus 28 is -9.

From there we'll rearrange that to 4x squared minus 12x + 9 = 0.

Now I'd like you to solve that quadratic to find their x values.

Pause, and solve now.

Welcome back.

Lots of ways you could do this.

I've gone for completing the square because I want to show you something.

When I complete the square, I get to this point: x minus 3 over 2 close brackets squared = 0.

Oh look x = 3 over 2 is the one and only solution.

We only get one x solution for this quadratic.

That happens, quadratic equations sometimes have two solutions, sometimes have one solution, sometimes have no solutions.

In this case we've got a quadratic with just one solution.

I'd like you to take that one solution, substitute it in to find the y value.

Pause, work it out now.

Welcome back.

I hope whichever equation you substitute back into you found that the y value is -5 over 2.

So substituting it to either equation gives you the same y value, and we've got our solution.

Can you see what's different this time? Why have we only got one solution for this pair of simultaneous equations? It's because for this pair the linear equation was tangent to the parabola, therefore there's only one intersection and only one solution.

Practise time now.

For question one, I'd like to solve these pairs of simultaneous equations, and if necessary round to one decimal place.

Pause.

Give those a go now.

Question two.

I'd like you to show using simultaneous equations that the line 10x + y = 17 is a tangent to the curve: 3x squared minus 4x minus y = -20.

Tricky on this, but a lovely bit of maths.

Pause, and give it a go.

Feedback time.

Question one.

Solve these pairs simultaneous equations rounding to one decimal place if necessary.

For part a, we could add those two equations together and we create a quadratic x square + 4x minus 21 = 0, which we'll factorise.

Wonderful, x = 3, x = -7.

Substitute back into either equation and we'll find our y values: y = 2.

7 to one decimal place, and y = -10.

7 to one decimal place.

Pair of solutions: x = 3, y = 2.

7; x = -7, y = -10.

7.

For part b, we couldn't immediately eliminate, we had to do some scaling up.

I multiplied equation one by 2, multiplied equation two by 3, because I got a common coefficient for y.

When I subtract one from the other and create the quadratic 2x squared minus 12x minus 22 = 0, I can divide through by 2 on both sides there.

I'm left with x squared minus 6x minus 11 = 0.

Put that into the formula 'cause it wouldn't factorise: two x solutions 7.

47 and -1.

47.

When I substitute those back in, I get x = 7.

5, and y = -12.

9.

And when x = -1.

5, y = 4.

9.

They're rounded to one decimal place.

For question two, show using simultaneous equations that a line 10x + y = 17 is a tangent to that curve.

So we can add the two equations together, and we come up with this quadratic: 3x squared + 6x + 3 = 0.

Divide through by 3.

Oh look, it's a perfect square.

Only one x solution: x = -1.

Substitute back in, y = 27.

We only found one solution, therefore it is tangent.

Sadly it's end of lesson now, but we've learned that we can use elimination to solve a pair of simultaneous equations algebraically, and one is linear and one is quadratic.

In this example, subtracting equation two from equation one leaves us with a quadratic which can be solved to find the x values.

I enjoyed that lesson, I hope you did too.

And I look forward to seeing you for more maths very soon.