Lesson video

Hello, Mr. Robson here.

Great choice to join me for maths, "Solving two step equations" how exciting, let's get going.

Our learning outcome is that we'll be able to see that when an additive step and a multiplicative step are required to solve an equation, the order of operations will not affect the solution.

The word for today will be equation.

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

Two parts of today's lesson and we're gonna begin by solving using an additive step first.

Firstly, I'd like you to begin with something that should be very familiar, some one step equations.

Could you pause this video and solve, 2X equals 10, and X plus four equals 10.

Do that now.

I hope you wrote something along the lines of divide by two on the left hand side, divide by two on the right hand side and we get a solution of X equals two.

And then to undo the add four, we add a negative four to the second equation and we get a solution X equals six.

But what's different about this equation? 2X plus four equals 10.

What's different about that one? Pause and tell the person next to you.

I hope you said something along the lines of, it's a two step equation.

We have two operations that we need to undo to get to a solution.

What does it look like to undo those operations? It looks fairly similar to what you did with the one step equations.

I see 2X and four, I'm gonna undo that plus four, I'm gonna do that by adding a negative four to both sides of the equation.

It's best to write it like this.

You may see lots of textbooks that present this in a different way, but if we write add negative four on the left hand side expression and on the right hand side, just like so.

You can see 2X plus four add negative four simplifies to just 2X on the left hand side.

We have 10X, 10 and negative four on the right hand side giving us six, we now have a one step equation to solve.

What happened? We undid one operation and we isolated the term with the unknown.

Now we've got a one step equation to solve, divide through by two on both sides and we get the solution X equals three.

So it always good to check your solution.

If we substitute the X value of three back into the original equation, do lots of three plus four, does it indeed make 10.

We know that solution could be correct.

We can see these steps undoing on a balance model.

Balance model is a really powerful visual representation when we're solving equations.

If I start by putting 2X plus four on the left hand side and 10 on the right hand side, I've represented this equation.

I can then represent the addition of negative four on both sides like so.

Now you see positive four, negative four, they become a zero pair.

They cancel that's this will leave us with just 2X on the left hand side and six on the right hand side and I a half what I have on both sides of the equation.

We have our solution X equals three.

Andeep and Jun is solving this equation, 3X plus eight equals 17.

Andeep says, "I'm gonna start by adding negative three to both sides." Whereas Jun says, "I think we should add negative eight to both sides." Let's see who's right.

If we add negative three to both sides, we end up with 3X plus five equals 14.

We didn't undo an operation so we haven't isolated the term with unknown.

We've still got a two step equation to solve.

By contrast, if we add negative eight to both sides, that will undo the add eight on the left hand side.

We'll be left with 3X equals nine.

Look at the difference between those two methods.

Jun has isolated the term with the unknown and this should always be our goal from solving these two step equations.

From there, divide both sides through by three and we get this solution X equals three and it says we need to check that solution when we substitute three back into the original equation.

Three lots of three plus eight does indeed make 17.

It's right.

The good thing about solving the equations is that you can check your solution is correct.

Quick check you've got everything I've said so far.

Which first step will isolate the term with the unknown when solving the equation 7X plus 10 equals 31.

Is it add negative seven, add negative 10 or add negative 31? Pause and tell the person next to you.

I hope you said add negative 10, option B.

When we add negative 10 to both sides, you'll write it like that and you'll see we have 7X equals 21 as our new equation from there divide both sides three by seven and our solution is X equals three.

Seven lots of three plus 10 is indeed 31.

I can check that is correct.

Slightly different question now.

Which first step will isolate the term with the unknown when solving the equation, 2X minus 10 equal 24? Is it add 24, add negative 10 or add 10? Pause and tell the person next to you.

I hope you said option C, add 10.

Option B was tempting and I imagine a few people suggested it might be that but it's not right.

It's a frequent error though.

When people say 2X minus 10 and they want just 2X, they try to take away the 10 by adding negative 10.

That's not right, why? 2X minus 10 add another negative 10 and the left hand side will end up with 2X minus 20.

We've still got a two step equation, we haven't made it any more simple.

By contrast, we add 10 to both sides.

Negative 10 positive 10 becomes zero pair are left with just 2X on the left hand side, 2X equals 34, divide through by 2, X equals 17 is our solution.

Andeep and Jun are now solving this equation.

4X plus 13 equals 30.

And Andeep proficiently says, "We start by adding negative 13 to both sides." And this happens, we end up with a 4X equals 17.

Jun says, "But now we can't solve it because 17 is not invisible by four." Do you agree with Jun? Pause and have a little think.

Andeep says to Jun, "17 might not be a multiple of four.

I think we can divide it by anything.

Let's try." We wanna divide both sides of our equation by four.

We'll be left with X on the left hand side and 17 divided by four on the right hand side, which we write as 17 over four.

Jun says, "That's the solution! We did it! And it's right.

You could substitute 17 over four back into the original equation and you'll find that to be the correct solution.

You may see this written as a mixed number four and a quarter was a decimal 4.

25 but typically you'll see it written as a fraction.

It's more efficient to just write 17 over four.

Quick check that you can solve these now.

I'm gonna do two on the left hand side and then present you with two to solve (indistinct).

(indistinct) solve to start 6X minus 13 equals six.

I'm gonna add 13 to both sides to get 6X equals 19 and then divide both sides through by six and there's my solution ,X equals 19 over six.

6X plus 24 equals six.

I'm gonna add negative 24 to both sides.

Leave me with 6X equals negative 18.

Dividing through by six leads me with a solution X equals negative three.

Your turn now.

I'd like you to solve 5X minus 11 equals seven and 5X plus 12 equals seven.

Your working out should look a lot like mine.

Pause and do that now.

First step, add 11 to both sides leave us with 5X equals 18.

Divide through by 5X equals 18 over five.

How did you do? You might just wanna pause and just check out what you wrote is the same as what I've got.

5X plus 12 equals seven.

Let's add negative 12 to both sides to isolate the term with the unknown.

5X equals negative five.

Divide three by 5X equals negative one.

Practise time now, question one.

I'd like you to manipulate the balance scale to solve the equation 4X plus five equals 13.

On your balance scale, you should be drawing extra things, scribbling a few things out as we manipulate.

On the left hand side, whatever you are doing to that balance scale, I'd like you to record it so that your algebra runs alongside your manipulations with the balance scale.

Pause and do that now.

Question one part B, a similar problem but a different equation.

The equation 3X minus 10 equals two.

Manipulate the balance scale and make sure you write the correct algebraic notation on the left hand side as you work with balance scale.

Pause, do this problem now.

Question two, solve the following equations, six equations for you.

Make sure you write your steps correctly.

You probably want to check your solutions by substituting them back in as well.

Pause and have a go at those equations now.

Question three.

Jacob attempts to solve an equation X plus five equal 29.

You can see his method and his solution there.

Part A, I'd like to substitute Jacob's solution back into the original equation to show him that he is wrong and then for part B, I'd like you to write some feedback on his work to show him where he went wrong and explain his error to him.

You should write at least two sentences for that part.

Pause and do that now.

Let's see how it did.

Question one part A, 4X plus five equals 13.

Your first step should be to add negative five to the balance scale and the add the algebraic notation.

From there, positive five and negative five becomes a zero pair and they disappear of our balance scale leaving us with 4X equals eight.

Just gonna rearrange my eight on the right hand side there because that makes this next step look very easy.

I've got 4X is equal eight, but I can see in each row that X equals two and I divide both sides through by four, I get the solution X equals two.

Question B, 3X minus 10 equals two.

The first step would be to add positive 10 to both sides, which will look like that in your algebraic notation.

Once the negative 10 and positive 10 cancel each other out, you have 3X on the left hand side, 12 on the right hand side.

I rearrange my 12 like so because now I can see a solitary row of X equaling four.

That's our solution.

For question two part A.

9X plus seven equals 25.

The first step is to add negative seven to both sides.

9X equals 18, so X equals two.

For part B, 9X minus seven equals 25.

Undoing that minus seven would be to add positive seven to both sides.

9X equals 32, divide through by nine.

We get X equals 32 over nine.

For about C, four equals 5Y plus 19.

The unknown's on the right hand side of the equation this time, but that doesn't matter.

We're just operating exactly the same way.

We want to isolate it.

We'll add negative 19 to both sides.

To do that, we'll get negative 15 equals 5Y.

Divide through by five, negative three equals Y.

Just be aware if you've got that written, you want to turn it around if you will.

We would write Y equals negative three.

When we're writing solutions, we write the letter on the left hand side and the numerical solution on the right hand side.

D, half of X plus seven equals 25.

A tricky looking one but no different.

We'll add negative seven to both sides and we're left with half of X equals 18.

We multiply both sides by 2, X equals 36.

Part E, negative 10X minus 30 equals 20.

We'll add 30 to both sides, leaving us with negative 10X equals 50, divide through by negative 10 and we get X equals negative five.

For part F, 31 equals seven minus 4f, lots of ways you might solve this.

Adding negative seven to both sides will isolate the unknown term.

If 24 is equal to negative 4f, we can divide through by negative four and we'll get f equals negative six.

For question three part A, substitute Jacob's solution back in to show that he is wrong.

We'll put eight back into the original equation.

Eight lots of eight plus five equals 69, not 29.

So we know that X equals eight is not the solution to this equation.

The feedback that we should give Jacob would be, on the left hand side, you didn't just subtract five, you subtracted 5X and subtracted five.

However, on the right hand side we only did one of those two things.

If we don't do the same thing to both sides, we'll lose a quality in that equation.

It would've been good if Jacob had written the negative five step like this and you see the difference when we write it like that.

It's clearer and we can see the like terms. The positive five and the negative five become a zero pair leaving us with 8X equals 24, a solution of X equals three.

We can check that by subsiding it back in eight lots of three plus five as indeed make 29.

We know that it's the solution.

(indistinct) the second half of the lesson.

Now we're gonna look at multiple approaches to solving the same equation.

Alex is explaining to Izzy how he did this equation that we saw earlier, 2X plus four equals 10.

"First I added the additive inverse to isolate the variable term." "Then I multiplied by the multiplicative inverse to find the value of X." X equals three.

Izzy says," I don't think we always have to use that order.

I think we could do the multiplicative step first." Izzy's now explaining how she solved it.

"Do the multiplicative step first, like this." That's interesting.

Do you think Izzy's idea will work? Pause, have a think.

It does work if we half the left hand side of the equation, half of 2X is X, half of four is two.

Half the right hand side we get five and we get the solution X equals three.

It's the exact same answer, just a different route to getting there.

That's awesome Izzy? I think sometimes this might be the most efficient method.

You might see it written in a factorised form.

If we start with the expression 2X plus four left hand side, you'll see a factor of two.

The same factor exists on the right hand side, so we can factorised.

From there, if we divide both sides through by two, we're left with X plus two equals five.

You may see it written in that form.

You might also see balance scales representing problems like this.

You can see exactly what's happening, exactly what Izzy's doing.

2X plus four on the left hand side, 10 on the right hand side.

When we half what we have on both sides of our balance model, you can see X plus two equals five and then have negative two to both sides, X equals three.

Quick check that you've got that, we do the multiplicative inverse first to start solving this equation.

What will we write next? X equals 12, X minus 10 equals 12 or X minus five equals 12.

Pause and have a think about it.

I hope you wrote option C, X minus five equals 12, half of 2X is X, half of negative 10 is negative five, half of 24 is 12 and there we can solve X equals 17 being a solution.

Another question for you, which is a sensible first step to solving this equation? 3X plus 15 equals six.

Is it to divide by three, to divide by x, multiply by three or multiply by a third? What do you think? Pause this video and consult the person next you.

Option A, divide by three works.

Divide by X does not work.

Multiply by three, mm-mm, multiplied by a third does work.

Why does divide by X not work.

If I divide all those terms by X, I get 3X over X plus 15 over X equals six over X which simplifies to three plus 15 over X equals six over X, which is solvable but tricky from there to multiply by three, not a sensible first step either, three lots of what we have on the left hand side.

Three lots of what we have on the right hand side would give us 9X plus 45 equals 18.

We've still got two step equation to solve.

We whereas if we divide three by three, providing all the terms left hand side by three, will leave us with X plus five, dividing six by three leaves us with two and we can quickly get to our solution and that X equals negative three.

To multiply by a third also worked.

If we take our equation, multiply both sides by a third, you could expand that bracket.

One third of 3X, one third of 15, one third of six, X plus five equals two.

A solution X equals negative three.

And you see how those two things are the same thing, to divide by three is to multiply by a third.

You are welcome to write this in either way.

Izzy and Alex are comparing methods.

The same equation that they're solving 5X plus 35 equals 20, but Izzy says, "I'll do the multiplicative inverse first." Alex says, "I'll do the additive inverse first.

Let's see who's more efficient." Izzy starts by dividing through by five.

Alex starts by adding negative 35 and Izzy solves like so, Alex solves like so.

For this equation you can see that both methods are really efficient.

Izzy says," I think I'll always do the multiplicative step first.

Because she spotted in this equation, 8X plus 40 equals 96 as a common factor eight.

She divides through by eight.

She turns that equation into X plus five equals 12, which is really simple to solve.

X equals seven, lovely.

However, 8X plus seven equals 11 and Izzy assists.

She divides through by eight and she gets X plus seven eighths equals 11 eighths and then she adds negative seven eighths to both sides and gets X equals four eighths, which cancels to a half.

It can be awkward if all the terms don't share the same factor.

Still doable, we just have to use our fraction skills.

By contrast for the same equation.

If she'd added negative seven to both sides, and then divided through by eight, it would've been more simple to solve.

So Izzy concludes, "I should take time to think which method will work best." Well done Izzy.

Great, check you've got that.

For which of the following equations are you most likely to start with a multiplicative step and divide by three first? Is it A, 3X plus 21 equals 33? B, 3X plus seven equals 13? or C, 3X plus 23 equals 31? Pause and have a think.

It was option A.

The equation 3X plus 21 equals 33, why? Because all three terms have a factor of three.

By contrast equations B and C do not have a common factor in all the terms, you could divide by three, but that would introduce fractions.

How might you begin to solve this equation? And what an equation? X over 10 plus 23 over five equals five over two.

Beautiful, what do you think? We've got options.

The additive step, we could add negative 23 over five.

We've got the multiplicative step.

We would use the multiplicative step, multiply both sides by 10, why 10? I wonder if you can spot why 10.

Anyway, what do you think? You gonna add negative 23 over five or you're gonna multiply by 10 and we indeed solve that equation.

Pause and give it a little go.

You may or may not have been able to solve that equation, if you couldn't, our problem, well done for giving it a go.

It is a tricky one.

If we went down the route of adding negative 23 over five to both sides we'll look like so, we'll be left with X over 10 on the left hand side and on the right hand side, negative 21 over 10.

By that time we've had those two fractions.

From there we've got equal denominators, so we must have equal numerators, X equals negative 21.

This seems like a lot of steps.

Let's see if the multiplicative step first, makes it more efficient.

Multiply both sides by 10, why 10? Because the denominators 10, five and two have a common multiple 10 or 10 is the lowest common multiple with those three denominators.

So 10 lots of what we've got on the left hand side.

10 lots of what we've got on the right hand side becomes X plus 46 equals 25.

Well that's an awful lot more simple.

From there we can see X equals negative 21.

Starting with a multiplicative step here was far more efficient.

I'm gonna solve an equation just like that one and then ask you to have a go yourself.

I'm gonna solve X over 20, 17 over 10 equals three over five.

Lowest common multiple of 20, 10 and five, that's 20.

So I'm gonna multiply both sides by 20.

When I expand out those brackets, I get 20X over 20, plus 340 over 10 equals 60 over five.

I can simplify all of those X plus 34 equals 12.

From there, add negative 34 to both sides and I will get a solution.

X equals negative 22, your turn now give it a go, be brave.

It's okay to make mistakes in maths, we all do.

Solve X over 12 minus a sixth equals two thirds.

Pause and give it a go.

How did we get on? Did we spot and it would be efficient to multiply both sides by 12? Did we expand the brackets to get that? If we simplify from there to get X minus two equals eight and find a solution therefore of X equals 10.

Practise time now, question one, solve these equations in two different ways.

Pause and do that now.

For question two, Laura is trying to solve 4Y minus eight equals 44 by doing the multiplicative step first.

She's dividing through by four or rather multiplying by a quarter.

I'd like you to check Laura's work and if necessary, write a few sentences of feedback.

Pause and do that now.

Question three, I'd like to solve 13 over four minus X over eight equals seven over two.

Pause and give it a go.

All right, let's see how we did.

Part A to question one, 2X plus 18 equals 30.

Starting with a multiplicative step would be to divide both sides through by two.

Turning that equation into X plus nine equals 15, giving us a solution of X equals six.

Alternatively, the same method written slightly differently.

You could factorised two out of both sides of that equation.

And we then divide through by two, we're left the end with X plus nine equals 15, the solution X equals six.

Starting with an additive step would've been to add negative 18 to both sides, leaving it with 2X equals 12 and a solution X equals six again.

You should notice all of your solutions are the same.

There's many different methods.

As as your method is correct, you'll always get the same result.

For part B, 3D minus 27 equals nine.

We'd start with multiplicative step.

We could divide both sides by three or multiply them by a third.

That will turn the equation to D minus nine equals three, but the solution D equals 12.

Alternatively, you might have factorised both sides taking out the factor of three and you divide three by three.

Again, D minus nine equals three, so the solution is D equals 12.

If you start with a additive step, 3D minus 27 equals nine.

Let's add 27 to both sides to create the equation 3D equals 36 with a solution again, B equals 12.

Again, multiple methods, as long as your method is correct, you'll always arrive at the same solution.

Part C, multiplicative step because outcome factor, divide through by seven, turn it into five minus X equals three, add negative five to both sides and negative X equals negative two.

So positive X equals positive two.

Alternatively and probably more efficiently, factorised through by negative seven.

If you take out a factor of negative seven, you're left with negative five plus X equals negative three.

We then add five to both sides and get the solution X equals two.

I think that method is slightly more efficient.

For the additive step, we could add negative 35 to both sides and we'll have the equation negative 7X equals negative 14, divide through by negative seven and X equals two.

Same solution, multiple correct methods.

For question two, check Laura's work and if necessary, write a few sentences of feedback.

Well, we can check it by substituting the solution Y equals 19.

Four lots of 19 minus eight, that's 68, that's not 44.

So we know Y equals 19 is not the solution.

So where's it on wrong? Multiplying by quarter is a good idea, but it'll be better to write it like this.

One quarter of what's on the left hand side, one quarter of what's on the right hand side.

When we do that, we're less likely to miss any terms. If you now compare the second line of working out in Laura's work, she got Y minus 8 equals 11 because she didn't multiply negative eight by a quarter, she missed that term.

Whereas when we've expanded out our bracket, a quarter of 4Y is Y, a quarter of negative eight is negative two.

For question three, we're solving 13 over four minus X over eight equals seven over two.

Look at those denominators, four, eight, two.

Lowest common multiple of eight will be the most efficient thing that we can multiply both sides by.

Let's start with that.

Eight lots of what's on the left hand side, eight lots of what's on the right hand side.

Expand those brackets and we'll get 26 minus X equals 28.

The solution being X equals minus two or X equals negative two.

We're at the end of the lesson now sadly.

To summarise, we can solve two step equations by undoing the two steps, sometimes by undoing the additive step first, sometimes by undoing the multiplicative step first.

The order of operations does not affect our answer as long as we are accurate in our working.

We have multiple methods to get to the solution.

Some are more efficient for some equations than others.

You can always check your solution is correct by substituting it back into the original equation.

I hope you've enjoyed learning about solving equations as much as I have.

I hope to see you again soon for more mathematics.

Bye for now.