Lesson video

Hi, I'm Mrs. Wheelhouse.

Welcome to today's lesson on checking and securing understanding of substitution.

This lesson comes from our algebraic manipulation unit.

I love algebra.

It's so useful.

So let's get started.

By the end of today's lesson, you'll be able to substitute particular values into a generalized algebraic statement to find a sense of how the value of the expression changes.

Now in our lesson today, we're going to be using the word substitute.

This may be a word that's very familiar to you.

The important thing to remember is that in algebra, substitution can be used to replace variables with values.

So when we do that process, we call this substitution.

Our lesson has three parts and we're gonna begin by reviewing substitution.

We can evaluate an expression for given values of its variables by substituting.

Let's consider these expressions.

What would their values be if a was equal to 3? So if a had the value 3, what would these expressions evaluate to? Pause the video now while you have a go.

Now good practice is to write the substituted value in brackets and evaluate the expression for that value.

So let's have a look at how it works with these.

I would write 4, then in place of the a, I would write the 3, which is the value I'm substituting in, and I placed it in brackets to show that I've replaced the a with a value of 3.

Now the reason I do this, of course, if you remember this from having seen it before, is so that my calculator, if I'm using one, can deal with negative values correctly.

And you'll see that later in our lesson today.

So I have 4 lots of 3, add 1, which is just 12 add 1, or in other words, 13.

What about with a squared plus 1? Well, that's 3 squared plus 1, which gives me 10.

And then 9 subtract 2 lots of a would be 9 subtract 2 lots of 3, which gives me 3.

And then the last one, I have 3 lots of 3 add 1, and the result is divided by 2, which means I have 10 divided by 2, which is just 5.

When variables are being substituted for negative values, the process is identical, but this is where putting brackets around the substituted value has real merit.

Let's have a look.

What would the values of these expressions be if a was negative 3 this time? Well, I would have negative 3 squared plus 1.

Now this is really important.

If I was typing this into my calculator and I didn't have the brackets, my calculator may not evaluate this correctly.

Remember, I'm doing negative 3 squared, so that's negative 3 multiplied by negative 3, and the result is 9.

Meaning that the result when I evaluate this expression is 10.

And then for the second expression, I've got 9 subtract 2 lots of negative 3.

Well, that's 9 subtract negative 6.

Remember, subtracting a negative value is the same as adding the positive value.

In other words, 9 subtract negative 6 is equivalent to 9 add 6, which gives us 15.

Now it is important to remember the priority of operations when we substitute.

So for example, let's evaluate these expressions when x is equal to negative 1.

For the first expression, I have 6 lots of negative 1, and then subtract 5.

Well, 6 lots of negative 1 is negative 6, then subtract 5 gives me negative 11.

But what about in my second expression? This time round, I've said that I want to do the subtraction before the multiplication, and I've shown that because I've included brackets.

So negative 1 take away 5 means I have negative 6, and it's that I'm multiplying 6 by, which gives me a result of negative 36.

So those brackets had a big effect there on what happened when I evaluated the expression.

How did these two expressions differ? How would you say them perhaps? So for the first expression, you can see I have 3 lots of x squared, and we can show it visually with our algebra tiles.

But in the second expression, remember, it is the value 3x which is being squared.

And again, I can show that with my algebra tiles.

So you can see visually, there's quite a big difference here.

Evaluate both of these expressions when x is equal to negative 2.

Well, for the first one, that means I have 3 lots of negative 2 squared.

Well, that's 3 lots of 4, which is 12.

Remember, I'm expecting something different here.

So I've got 3 lots of negative 2, and then the result is squared, or 3 lots of negative 2 is negative 6, and negative 6 squared is 36.

So as expected, I did indeed see something different.

Ah, Andeep, I wonder if there is a value of x that will make the value of 3x squared greater than all of 3x being squared.

Hmm.

Interesting.

I'm wondering if there's a value where they're the same.

Well, let's investigate.

I'm gonna explore using substitution.

So here, I've got a table of values, and I've gone for the values where x is negative 2, negative 1, 0, a half, 1, and 2.

And I'm evaluating for these values of x what my two expressions are.

So when x is negative 1, 3x squared gives me a result of 3 and 3x all squared gives me a result of 9.

When x is 0, I get 0 for both of them.

Ah, so there is a value of x.

That means the two expressions evaluate to the same result.

What about when there are half? Well, in that case, I get the two results of 0.

75 and 2.

25.

Oh, that 3x all squared is still the greater value.

Oh, and it's true for 1 and for 2 as well.

Hmm.

So substitution hasn't helped me find a value where 3x squared is greater than 3x all squared.

But it did give me an idea though.

Let's look at their graphs.

Oh, that's interesting.

I can see that one graph looks to be steeper than the other graph.

We can see from our graph that they have the same value when x is naught.

We notice that in our table of values.

But otherwise, 3x all squared, which is the green graph, the steeper one, is always greater than 3x squared.

So for example, if you look on the x-axis where x is one and go up the y-axis, you'll notice that you can see that y is 3.

But if I go up to the green line, I can see that's where y is 9.

I can always see that the green line has a greater value, which means that 3x all squared is gonna give me a larger value when I substitute than 3x squared will.

And in fact, by expanding those brackets, so writing 3x times 3x, and then simplifying it to be 9x squared, I can see 9x squared is always gonna be bigger than 3x squared.

Apart from when it's zero, that's when they get to be equal.

Where the variable appears more often than once in the same expression, the value should be substituted each time.

So for example, if I have the expression x squared plus 4x subtract 3, and I want to evaluate it, I'd write it as follows.

Remember, using the brackets to show where I've substituted, I have 1 squared plus 4 lots of one subtract 3, which gives me 2.

Aisha points out that 1 squared and negative 1 squared have the same value.

So will I still get the value 2 if I substitute in x equals negative 1? What do you think? Did you spot that when I substitute in the negative 1, I get the value negative 6 as the result.

Now that's because the first term had the same value, but the second term did not.

And therefore, the value of the overall expression was indeed different.

We could explore equations and formula by substituting different values for the variables.

So for example, a curve has equation y equals x squared plus 4x subtract 3.

We've seen the value of this expression for two values of the variable.

1 and negative 1.

A table of values can help us see how the value of y varies as x varies.

So for example, when x is negative 2, our expression evaluates to negative 7.

When X is 0, the expression evaluates to negative 3.

And when x is 2, it evaluates to 9.

Can you see what's happening here? If I substitute more values or I draw the graph, I'll get even more information about how this expression changes as the value of x changes.

Quick check now.

Andeep is copying a method for plotting a curve from the whiteboard.

He thinks the final answer looks wrong.

Where has his teacher made a mistake on the board? Pause the video and see if you can work out where the mistake is.

Welcome back.

Did you find it? That's right, it's here.

When I'm squaring the value of negative 2, that means negative 2 times negative 2, which is positive 4, not negative 4.

So what I end up with is negative 4 take away 4, which is negative 8.

It's time for your first task.

For question one, please evaluate the expressions when b is 5.

And for question two, evaluate the same expressions, but for when b is negative 5.

Pause and do this now.

Welcome back.

Question three.

Aisha is checking her brother's revision homework.

Help her identify any mistakes he has made.

Pause the video and do this now.

Welcome back.

Time now for parts C and D.

Pause the video while you identify any mistakes.

Time to go through the answers now.

For 1A, you should have 9.

For 1B, it should be negative 9.

1C, 50.

1D, 100.

1E, 5.

1F, 1/2 or 0.

5.

And 1G, 14.

For question two, A should be negative 21.

2B should be 21.

2C should be 50.

2D should be 100.

2E should be negative 5 over 2 or negative 2.

5.

2F should be negative 1.

And 2G should be 34.

For question three, you had to work out where any mistakes were in Aisha's brother's revision homework.

Oh, here's the first mistake in 3A.

Her brother has written 4.

50 times 5.

Now following the priority of operations, only the 3 should be multiplied by the 5, and then we add 1.

50.

For part B, did you spot? Her brother completely forgot about the pi.

It just disappears.

In other words, it's there in line two and not at all there in line three.

For 3C.

Here's our mistake.

The value for x should be the same everywhere in the expression.

If we're substituting in negative 2, then it needs to be negative 2 in both places.

And then over here, what Aisha's brother has done is done 2 multiply by 2, and then squared the result.

But only that second 2 should be being squared, it's 2 squared and then multiply the result by 2.

And then for D, by evaluating the term separately, Aisha's brother has ended up subtracting the wrong way round.

So in other words, it should have been 100 take away 7,500, not 7,500 take away 100.

It's now time for the second part of our lesson and that's on further substitution.

We can substitute into expressions or formula with multiple variables.

So here's a formula for working at the area of a trapezium.

How can the formula be used to calculate the area of this trapezium? Well, we need to substitute.

a and b, remember, stand for the two parallel sides.

So in place of a, I've written 13, and in place of B, I've written 8.

We then multiply by the perpendicular distance between these two parallel sides, and that's 5.

I can then evaluate this expression, remembering my priority of operations, and that gives a result a 52.

5 centimeter squared.

"Ah," says Izzy, "I remember you can find the length of the hypotenuse of a right-angle triangle using a formula." Do you know what formula she's talking about? "Ah," says Laura, "I do.

I've got notes on this.

It's c is equal to the square root of a squared plus b squared." And she's right.

Although you might have been saying to the screen, a squared plus b squared equals c squared.

But Laura has gone straight for the form that allows her to find the length of the longest side.

Well, given that a is 5 and b is 12, what is that value of c? Well, we can replace or substitute a with 5 and b with 12.

This means I have 5 squared, which is 25, add 12 squared, which is 144.

Adding these together gives us 169, which is what I have to square root.

So c is 13.

Izzy says, "Hang on a second though.

I thought when you square root a number, there are two possible answers, a positive value and a negative value.

So can't c be negative 13?" What do you think about Izzy's statement? Is she right? Well, she is correct because the square root of 169, it could indeed be negative 13, because negative 13 times negative 13 is 169.

However, in this context, c is the length of the hypotenuse, so it's not gonna be a negative value.

And therefore, the answer is just 13.

When we see the square root symbol, it implies the positive route unless otherwise stated.

Please have a go at evaluating all these expressions when a is equal to 16 and b is equal to 9.

Pause the video and do this now.

Welcome back.

Let's see how you got on.

For the first one, 16 squared plus 9 squared gives us a total of 337.

For a plus b all squared, we have to add the a and the b, which gives us 25, and 25 squared gives us 625.

For the next one, square root of 16 add the square root of 9 means 4 add 3.

And then for the final one, the square root of 16 add 9 is a square root of 25, which is five.

So what's the value with c squared to subtract 3ab when a is negative 2, b is 5, and c is 3? Well, as long as I'm careful about making sure I substitute the correct value for each of my variables, I should get this right.

So I'll have 3 squared subtract 3 lots of negative 2 multiplied by 5.

Well, that's 9 subtract 3 lots of negative 10, or 9 subtract negative 30, which gives me 39.

So your turn now.

What is the value of 4bc subtract 2a when a is negative 2, b is 5, and c is 3? Pause the video and work this out now.

Welcome back.

So if you substituted correctly, you'll have 4 multiplied by 5 multiplied by 3, and then subtract 2 multiplied by negative 2.

Remembering our priority of operations will lead us to 60 subtract negative 4 with a result of 64.

What's the value of the square root of 2a plus b and then subtract 3c, if a is negative 2, b is 5, and c is 3? Well, first of all, I have to substitute correctly, then I need to remember my priority of operations.

I'm going to deal with what's inside the square root sign first, which means I end up with a square root of 1 and then I subtract 9 from the result, which leads me with negative 8.

Your turn now.

What's the value of the square root of 2b subtract 2c and then adding 5a to the result, if a is negative 2, b is 5, and c is 3? Pause the video and work this out now.

Welcome back.

Remember, substitute correctly first.

So make sure you are replacing each of the variables with the correct value.

This will lead you to the square root of 4 subtract 10.

So negative 8.

Look, same result, even though it looked different.

It's time now for your second task.

For question one, you are given the formula for the area of a trapezium and asked to find the area if a is 11, b is 14, and h is 10.

In c, you're given the formula for finding the length of the hypotenuse on a right-angle triangle and asked to calculate the length of hypotenuse if a is 6 and b is 8.

Pause the video and do this now.

Welcome back.

Question three, evaluate these expressions given that x is equal to 3, y is equal to negative 4, and z is equal to negative 1.

And then question four, Izzy has evaluated the following expression for set values of a, b, and c.

Now use her working to suggest what values she substituted for each variable.

Pause the video and do this now.

Welcome back.

Time to go through our answers.

For question one, if you substituted correctly, you'll reach the area of the trapezium is 125.

Question two, if you've substituted correctly, you'll reach that the length of hypotenuse is 10.

Feel free to pause this video if you want to just check the substitution and working against what you have.

Question three, evaluating each of these expressions.

3A will be negative 6.

3B evaluates to negative 35.

3C to 41.

3D to 4.

And 3E evaluates to 143.

Question four asked you to suggest some values that Izzy has substituted in for each variable.

Now I've given you two suggestions here.

When I look at this, I can see that negative 3b evaluated to six.

And that says that given that negative 3b evaluates to positive 6, I must have substituted a negative value.

And what do you multiply negative 3 by to get positive 6? Well, that's negative 2.

So b has to be negative 2.

Now that I've got that, I can look at the first term.

So that's ab all squared.

Well, what do you square to make 36? Well, I either squared 6 or I squared negative 6.

If I squared 6, then a had to be negative 3.

And if I squared negative 6, then a had to be 3.

And that's why you can see two sets of possible values to the side on the screen.

If a is negative 3, then the third term tells me that c has to be negative 10.

And if a was 3, it tells me that c had to be 10.

Well done if you got this right.

It's now time for the final part of today's lesson on substitution using a calculator.

Now substitution can be a useful way at times to check our answers.

Jacob has tried to solve the equation 2 lots of 3x plus 5 equals 20.

"I think there's a solution when x is 5 over 3," says Jacob.

Now how can we check that using a calculator? Well, here we are.

If you type the left-hand side of the equation into your calculator, putting brackets around the value substituted.

In other words, if you can make your calculator display look like mine does, we can then press the answer key, or the execute key, depending on your calculator, to see if we get the same value of 20 that the expression should evaluate to.

And we can see we do.

So, Jacob has indeed found a solution.

Well done, Jacob.

Alex has tried to solve this equation.

He thinks there's a solution when a is equal to negative 3.

So, Jacob is going to check Alex's answer using a calculator.

"Oh no," says Jacob, "I got this far, but it's supposed to be a fraction.

Ugh, am I gonna have to clear the screen and start again?" Do you think he needs to? No, he doesn't.

He can put brackets around the whole expression for his numerator using the cursor keys.

He can then press the fraction button.

And all of this will appear as the numerator for his fraction.

Have a go yourself.

If you've done it correctly, your calculator display will look like mine, assuming you're using the Casio ClassWiz like I am.

We can now type the denominator, which is 5.

And remember, press the cursor key to move to the right before we add 1.

If we then press the execute button, we will find that the value is 10, so that Alex has indeed found a correct solution.

If you know you need fraction notation, it's easier to press the fraction button first before typing into your calculator.

And you can use your cursor keys to move between numerator and denominator.

Have a go at evaluating this expression.

When a is equal to 4 and b is equal to negative 1.

Pause and do this now.

You can see that I used the right cursor button at this point to navigate away from the square root of the operation.

When I press the execute button, I should get the value 3.

Time for a check now.

Andeep is filling in a table of values for the graph with equation y equals 3x squared plus 5x subtract 3.

He checks his first coordinate using his calculator.

You can see there he's checking when x is negative 2, but he doesn't get the answer negative 1.

He gets the answer negative 25.

So which answer is correct? His calculator or the one in his table of values? What I'd like you to do, please, is pause the video and work out which one's right, and explain why he might have gone wrong with the other one.

Pause and do this now.

Welcome back.

Did you see? The answer in the table is correct.

It's the answer from his calculator that's wrong.

He didn't put in the brackets around the negative 2.

Remember at the start of the lesson, I did say, if I don't put brackets around the value I'm substituting in, especially when it's negative, my calculator may go wrong.

And that's exactly what's happened here.

Try typing this expression in again, but putting the brackets in where they should be.

Can you see if you do, do that? You get the correct answer of negative 1 for the equation.

Time for your final task now.

For question one, Alex is using a textbook to revise solving equations.

And the values for x are the answers in the back of the textbook.

Check if they work for each of the equations you can see.

Don't forget to use your calculator.

Question two.

Jacob wants to plot the curve with equation y equals x squared subtract 2x plus 5.

Check his table of values and correct any mistakes you find.

Pause and do this now.

Question three, use your calculator to find the value of these expressions when a is negative 7, b is 4, c is 3/8, and d is negative 3.

Pause and do this now.

Welcome back.

Let's go through our answers.

So for question one, x equals four is indeed a solution, but x equals negative 3 does not work.

For part C, x is 3 over 2 also doesn't work, but for D, x is 5/8 does.

Question two, you should have found that there was a mistake right at the beginning.

When x is equal to negative 2, y is equal to 13.

Remember, use the cursor keys at this point to change the value of the variable so you don't have to type the whole expression in again.

You'll also discover that when x is negative 1 and when x is 2, we have mistakes again.

And this is what the correct table of values should look like.

Question three, I'm gonna suggest you pause at this point so you can check your working with mine and check that you reached the correct final value.

Pause and do this now.

It's time to sum up what we've learned today.

Expressions and formula in context can be evaluated for particular values.

Expressions may have multiple variables and you can substitute for some or all of these values.

Substitution can be a good way to check solutions to equations.

And being able to efficiently and accurately use a calculator can help when checking calculations.

Well done.

You've done a great job today.

I look forward to seeing you for more lessons in our algebraic manipulation units.