Lesson video

Hi, I'm Mrs. Wheelhouse and welcome to today's lesson on checking and securing understanding of solving and interpreting linear equations.

This lesson is from our unit on algebraic manipulation.

Now algebra can be incredibly useful, so let's see how we're going to use it today.

By the end of today's lesson, you'll be able to solve complex linear equations involving brackets.

Now here are some of the words we're going to be using today in our lesson.

We've got additive inverse, reciprocal, and like terms. Now if those keywords are not familiar to you, I suggest you pause the video now and have a read through of them before you begin the lesson.

Now that you're ready to begin, let's get going.

Our lesson is broken into three parts today and we're going to begin by solving simple linear equations.

When we are solving an equation, we are looking to find a value for the unknown, which makes the equation balanced.

To do that, we need to isolate the unknown to see its value on its own.

So for example, 6x + 7 = 55.

We can add the additive inverse of positive 7 to find the value of 6x.

To maintain equality, you need to add the same value to the expressions on both sides of the = signs.

This results in 6x = 48.

So after undoing one operation, we have isolated the term with the unknown and now we have a 1 step equation to solve.

6x remember means 6 times x, so we can multiply by the reciprocal of 6 to find the value of x.

So multiplying both sides by 6 results in 6x/6 = 48/6.

Remember that multiplying by 6 is the same as dividing by 6 so you can write it this way if you prefer.

This leads to the solution, x = 8.

We can check this is correct by substituting 8 into the original equation.

So 6(8) + 7 = 55.

When we see equations with brackets, we can expand the brackets in order to solve.

So for example, 9(x - 1) = 33.

Well, using the distributive law, I can expand those brackets.

And I know that I can write this therefore as 9x - 9 = 33.

And from here of course, we can solve this by manipulating, so we can add 9 to both sides, giving us 9x = 42 and then divide both sides by 9, giving us the answer x = 14/3.

Now when the answer is not an in integer, it is easiest to leave your answer as a fraction.

Now you should try to write your fraction in the simplest form possible, and if you don't want to leave it as an improper fraction like I have here, then you could write it as a mixed number.

Let's solve this equation by expanding the brackets.

So 5(3x + 7) is 15x + 35.

I can then take away 35 from both sides.

This leads me with 15x = -30, dividing 3 by 15 means that x = -2.

I can check this by substituting.

so 5(3(-2) + 7) and I can see that I get 5 = 5, so I know I'm right.

It's your turn now.

Please solve this equation by expanding the brackets and then check your answer by substituting.

Pause and do this now.

Welcome back.

When you expand the brackets, you'll get 6x + 33.

You then need to take 33 away from both sides giving us 6x = -24.

Dividing through by 6 gives the solution x = -4.

We then check by substituting and indeed we should reach 9 = 9.

Where we see common factors dividing by the coefficient of the bracket first instead of expanding could be more efficient.

For example here, remember I have 3(2x + 11) so multiplying by the reciprocal of 3 or dividing 3 by 3, we'll find the value of what 2x + 11 is.

So we can see here if 3(2x + 11) = 9, then one lot of 2x + 11 = 3.

This is now in a form that I'm quite familiar with.

So by subtracting 11 from both sides and then dividing 3 by 2, I reach the solution x = -4.

Jun points out that it is useful knowing two methods because it means that you can check your answer with the other method to see that you've got it right.

When the coefficient of the bracket is a fraction, expanding the bracket can be time consuming and lead to some tricky calculations.

Let's consider this example.

I've got 3/5(x + 7) = 30.

If I expand the brackets, that gives me 3/5 of x + 21/5 = 30.

Well, okay, I need to subtract 21/5 from both sides, so I end up with 3/5 of x = 129/5, then into multiplied by the reciprocal of 3/5, which is 5/3 and eventually get to x = 43.

Now, I dunno about you, but that looked awfully complicated.

I mean I can do it because I've got a good knowledge of fractions, but just because I can do it this way doesn't it mean I should.

Let's consider an alternative method.

So this time, I'm going to divide both sides of the equation by 3/5.

Remember, dividing by a fraction is the same as multiplying by the reciprocal of that fraction.

In other words, I could multiply both sides of my equation by 5/3.

While that means that the left hand side results in x + 7, well that looks a lot nicer already, and I need to know what 5/3 of 30 is.

Well that's quite nice.

I can do 30/3 = 10x 5 = 50, so I've got x + 7 = 50, well, x = 43.

Izzy points out the more I practise, the easier it gets to spot the most efficient method.

I dunno about you, but I certainly prefer this method over expanding the brackets.

How could we solve this equation involving division? 3x + 4/5 is the same as 1/5(3x + 4) and the reciprocal of 1/5 is 5, so I can multiply both sides by 5.

This gives me 3x + 4 = 70.

Taking away 4 from both sides gives me 3x = 66 and therefore x = 22.

Let's try solving this one.

I'm going to start by multiplying both sides by 4.

This gives me 3(2x + 5) = 60.

I'm then going to divide both sides by 3.

Now remember I could have expanded the brackets, but because I noticed that 3 is a factor of 60, I've chosen to divide instead.

This leads me with 2x + 5 = 20, therefore 2x = 15 and x = 15/2 or you could have written 7.

5.

Now I'm going to check this by substituting and this is where leaving it as a fraction was really helpful because 2(15/2) = 15, so I end up with 15 + 5 inside the brackets, which is 20.

I then have 3(20)/4 and it is indeed 15.

It's now your turn.

Please solve this equation and then check your answer by substituting.

Pause and do this now.

Welcome back.

You should have started by multiplying both sides by 3.

Then you could have divided 3 by 2.

This will result in the equation 5x - 3 = 27, adding 3 to both sides and then dividing by 5 gets you to the solution x = 6.

When you substitute that in, inside the brackets we have 5(6 - 3) well that gives me 27.

I multiply that by 2, divide by 3, and I do indeed get 18.

So my answer of x = 6 is correct.

It's now time for your first task.

What I've got here are three equations and then the working to solve them, but in each I've made a mistake.

I'd like you please to spot the mistake in each set of working.

Pause and do this now.

And then the second part of question one, again, three equations.

Can you please spot where I've gone wrong in each of them? Pause and do this now.

Welcome back.

Question two.

Please solve each equation thinking about what the most efficient method may be.

Remember, it will vary from question to question, but it doesn't matter if you don't have the most efficient.

If you've got one that you really prefer and you feel really confident with, then it's fine to do that.

But you might want to consider how long it's taking you versus alternative methods.

Pause and do this now.

Welcome back.

It's time to go through our answers.

So question 1A, did you spot the mistake? 15/2 - 4 is not 11/2, so I didn't get that right.

In B, I was supposed to subtract 12x from both sides, not subtract it from 1 and add it to the other.

And then in 1C, I didn't expand the brackets correctly.

I should have multiplied the 5 and the -6 together.

So it should have been 5x - 30 = 44.

In 1D, did you spot I was supposed to multiply 3 by 3, not divide by 3, so I should have ended up with 36 on the right hand side.

In E, I was doing really, really well until the very final step of working.

I was supposed to multiply 20 by the reciprocal of 4/5 and so my answer should have been 25, not 16.

And then in F, you can see here I expanded the brackets on the left hand side, but the right hand side shouldn't have changed at all except I did change it.

Oops.

Another mistake.

Well done if you spotted all of these.

Question two, you had to solve each equation thinking about the most efficient method.

So for A, I suggest you divide 3 by 5 first.

In B, again, I suggested dividing through and this time by 4 and if you follow your working, you should get x = 7/3.

In C, I opted to expand the brackets because 5 is not a factor of 48.

Following the working through, I reached the solution x = -4/5.

For D, I suggested starting by multiplying by 4, following the working through leads to x = 17.

For E, simple rearrangement here, I added 7 to both sides and then I opted to divide 3 by 3 first before multiplying 3 by 4.

You could of course at this stage 3x/4 = 18 have multiplied both sides by the reciprocal of 3/4 but you'll get to the answer x = 24 either way.

For F, I opted to multiply both sides by the reciprocal of 7/10.

Following the working through leads to x = -4.

And then G, I opted to first divide both sides by 5 and then multiply both sides by 6.

Following the working through leads to x = -5/2.

It's now time for the second part of our lesson on further solving linear equations.

We can solve equations which have unknowns on both sides of the = sign.

For example, 6x + 3 = 3x + 15.

Often, the easiest first step is to manipulate the equation so there are only unknowns on one side of the equation.

In this case, I'm going to add -3x to both sides.

This will mean that I'll get no x terms on the right hand side.

So after I've done that, I'm left with 3x + 3 = 15.

We're now left with a two-step equation which we saw in the first part of the lesson, so I can solve this quite easily leading to the solution x = 4.

Now it is useful to substitute the solution back into the equation to check that the equation balances.

So 6(4 + 3) and on the right hand side 3(4 + 15).

Well that results in 24 + 3 on the left hand side and 12 + 15 on the right hand side or in other words, 27 = 27.

That's definitely true and so I know x = 4 as a solution.

Knowing this allows us to solve equations where there are brackets on both sides.

So for example, 5(x + 4) = -3(2x + 1).

We can start by expanding the brackets.

This gives us 5x + 20 = -6x - 3.

By adding 6x to both sides, I'm left with 11x + 20 = -3.

I can then subtract 20 from both sides and divide by 11.

When solving an equation, there are often options which order to complete the steps in.

You should always pick the way that is easiest for you.

You could then use a different method and see if you've got the same answer.

Let's explain each step of Sofia's working.

So what has she done? Well, first of all, she's expanded the brackets on both sides of her equation.

She's then collected like terms on the left hand side.

She's then added -18x and positive 15 to both sides of the equation.

She's then multiplied both sides by -1 and then divided by 15 and simplified.

Is there anything you would've done differently? Well, let's have a look here.

Jacob and Alex have both had a go at solving the same equations and you can see here they've reached the same answer, but their methods are different to Sofia's.

How do they differ? And of the three methods you've seen, which is your favourite? Well, Jacob chose to add -3x and positive 9, which meant that his coefficient of x was positive.

Remember, Sofia reached -15x = 6, but Jacob's method meant he never saw a -x coefficient.

And Alex started by dividing everything by 3.

Let's solve this equation.

So the first thing I'm going to do is I'm going to expand the brackets on the left hand side.

I've then expanded the brackets that are on the numerator of the fraction on the right hand side.

I've gathered the like terms on the left hand side and then I've multiplied both sides of the equation by 3.

Now at this point, I decided to add 6x to both sides.

I wanted to do that because I wanted the coefficient of x to be positive.

I've then taken away 8 from both sides and divided 3 by 10.

Now you might have written that as a decimal, which would be 4.

6 and it's absolutely fine if you did.

It's now your turn to solve this equation.

So think about the steps that I took.

You might want to choose different ones or do it in a different way.

It's absolutely fine if you do.

Just be careful and make sure you are maintaining the equality in the equation the whole way through your working.

Pause and do this now.

Welcome back.

Let's go through this then and see if you reach the same solution that I did.

So I'm going to start by expanding the brackets.

5(2x - 1) = 10x - 5.

And then -3(2x) = -6x and -3(4) = -12.

So on the left hand side I have 10x - 5 - 6x - 12.

On the right hand side, I end up with 5x + 30/4.

I then gather the terms on the left hand side so that I have 4x - 17.

I then multiply 3 by 4 giving me 16x - 68 = 5x + 30.

I then took 5x away from both sides, added 68 to both sides and then finally divided through by 11.

Well done if you reach the same answer as me.

It's now time for your second task.

The solutions to the equations are given.

Please fill in the steps to show how I got to these answers.

Pause and do this now.

It's now time for the second part of question one.

So three new equations, and again the three solutions.

Please show me the working for how I got there.

Pause and do this now.

Welcome back.

Now what you see on the screen are just examples of the way to get to the answer.

You might have chosen a different way to do this and it's absolutely fine if you did as long as you always maintained the equality in the equation.

So feel free to pause and compare your method to mine.

Same for D, E, and F here.

I've just shown you one way to get to the solution, but you may have chosen a different way.

Feel free to pause at this point so you can compare your method to mine.

It's now time for the final part of today's lesson and that's on forming and solving equations.

We can form and solve equations in a variety of context to solve problems. For example, Laura has four blocks of different heights.

She lines them up from smallest to largest.

Each block is 3 centimetres taller than the previous block.

If she stacks the two smallest blocks on top of each other, they're the same height as the largest block.

How tall would a tower of all four blocks be? Well, I dunno about you, but that's an awful lot of text there.

I'm not finding that easy to understand.

I wonder what I could do to make that a little bit easier.

Well, if I can form an equation, it should be quite easy to manipulate and solve it, but I think the drawing a diagram will be a good first step.

At the moment, I'm unsure how to start.

So I'm going to draw the four blocks that Laura talked about.

I've started with the smallest and then shown that each block is 3 centimetres taller than the previous block.

If I decide to call the height of the smallest block x, then what is the total height of the next block? That's right, it'd be x + 3.

And now I can label the heights of all the blocks in relation to the smallest.

So the first block has a height of x.

The next will be x + 3.

The next is x + 3 + 3, so that's x + 6.

And then the last one will be x + 3 + 3 + 3 or x + 9.

Now let's think about the information I was given in the question.

The sum of the height of the two smallest blocks is equal to the height of the largest block.

In other words, I can use this information to form an equation.

What am I going to write? That's right, x + x + 3 = x + 9.

Adding the heights of the two smallest blocks gives us the height of the largest block.

I can simplify and write 2x + 3 = x + 9 and now I can solve this subtracting x from both sides and then subtracting 3 from both sides tells me that the height of the smallest block must be 6 centimetres.

Well what does that mean in context of the problem? Exactly.

It means that the height of the smallest block is 6.

Hang on, Lucas, x = 6? That's not the solution.

Exactly, Laura.

We should check that just because I solved an equation doesn't mean I necessarily solved the problem.

I need to go back and check what I was actually asked to do.

I was asked how tall would a tower of all four blocks be? Well, 6 centimetres is only the height of the smallest block, so I definitely haven't answered the question yet.

So let's use that information though to write down the heights of the four blocks.

I know that the smallest has a height of 6, the next block has a height of 9, and then the next block is 12, and the next block is 15.

I had to sum these, remember, to find the height of all four blocks combined.

So that's 42 centimetres.

Exactly, Lucas.

That's my final answer.

And Laura points out I could even check my answer by making sure the two smaller blocks are the same height as the largest block or 6 + 9 is indeed equal to 15.

Let's do a check now.

Izzy has 4 cats and each cat is 2 years younger than the previous cat.

If the youngest cat has age c, write an expression for the age of each of the other cats.

And then for the second part, this year, their combined age is 32.

Form an equation from this information and work out what is the age of the oldest cat.

Pause a video and do this now.

Welcome back.

How did you get on? Well, if the youngest cat has age c and the other cats, remember we start with the youngest one, are all going to be 2 years older than the previous cat.

We'll have C + 2, C + 4, and C + 6.

Now this year they're combined age is 32.

So we had to sum the 4 ages and say that was equal to 32.

And you could of course have simplified by collecting like terms. To work out the age of the oldest cat, well, we'll start by solving the equation.

So I'll take away 12 from both sides and then divide by 4.

Now is C = 5 the final answer? Now, that's right, it's not.

This is the age of the youngest cat and I want the age of the oldest cat.

The oldest cat remember is C + 6.

So given that C is 5, I add 6, the oldest cat must be 11 years old.

Andeep is going to give us a number puzzle.

He's thinking of a number, multiplies it by 5, subtracts 2, multiplies that by 3, and then divides by 11, and his answer is 9.

So Andeep has set as a challenge.

Can we work out his original number? Of course we can.

We're going to start by forming an equation.

So let's follow through the steps Andeep took.

He thought of a number and he multiplied it by 5.

So that's 5a then subtracted 2 so that's takeaway 2.

He multiplied all of this by 3, which is why I've used brackets, and then divided the result by 11, and got to the answer 9.

So let's undo the steps of Andeep's working.

We'll multiply 3 by one, divide 3 by 3, add 2, and divide by 5.

So Andeep's original number was 7.

Remember, you could have used any letter to stand for Andeep's original number.

You didn't have to use A like I did.

We can of course check our answer by starting with 7 and following Andeep's calculations.

Let's do a quick check.

To convert degrees Celsius to degrees Fahrenheit, you multiply the value in degree celsius by 9, then divide it by 5 and then add 32.

Write a formula for this information and then write an equation given the information that the temperature in Andeep's hotel room was 59 degrees Fahrenheit.

And then in C, what was the temperature in Andeep's hotel room in degrees Celsius? Pause and work this out now.

Welcome back.

Let's start with the formula.

Remember, to convert to degrees Fahrenheit, we had to multiply degrees through Celsius by 9.

So that's 9c/5 + 32.

We're then told that the temperature in Andeep's hotel room was 59 degrees Fahrenheit.

So we can replace the F with a 59 because we now know that that is what it's equal to.

And we can now solve this equation to work out the temperature in Andeep's hotel room in degrees Celsius.

So I subtract 32 from both sides, multiply by 5, and then divide by 9.

The temperature in Andeep's hotel room was 15 degrees Celsius.

We can use shape properties to form and solve equations as well.

For example, a quadrilateral has these interior angles.

What type of quadrilateral could it be? Well, what do we know about interior angles in a quadrilateral? We know they sum to 360 degrees and we can form an equation with this information.

So those four angles must sum to 360 degrees.

If I collect the like terms, this will make it a lot easier to manipulate.

So when I collect like terms, I get to 46x + 130 = 360, and now I can solve for x.

x = 5, so now what I can do is substitute that in for each expression and I can work out the size of each angle.

So when I do that, I get the following.

So I have two angles that have a size of 115 degrees and two angles that are 65 degrees.

So what type of quadrilateral could this be? Well, two pairs of equal angles means it could be a parallelogram, in fact, it could even be a rhombus, and it could be a isosceles trapezium.

We would need to know the side lengths to work out which of these two it actually is.

Let's do a quick check now.

I have the following angles in this triangle.

I'd like you please to work out the value of x, then calculate the size of each angle, and then tell me what type of triangle this could be.

Pause and do this now.

Welcome back.

Well, if you form and solve the equation, you should find out that x = 8 and therefore the size of each angle is 26 degrees, 64 degrees, and 90 degrees.

So this could be a right-angled scaling triangle.

It's now time for your final task.

For question one, please work out the house numbers for these sets of houses.

So where I have 3 adjacent houses that sum to 57, what would the 3 houses be? And if 5 adjacent houses sum to 45, what would their 5 numbers be? In question two, be careful.

We're no longer increasing by 1 for each adjacent house, we're now increasing by 2.

So again, find me the 4 adjacent houses where the numbers on the houses sum to 40, and then 6 adjacent houses where the numbers on those sum to 174.

Pause and do this now.

For question three, work out the numbers that Laura and Sam are thinking of.

And then in question four, work out the size of the angles.

Pause and do this now.

In question five, a quadrilateral has the following interior angles.

Using the fact that interior angles in a quadrilateral sum to 360 degrees, form a simplified equation for x, then find that value for x, and then find the size of each angle, and tell me what type of quadrilateral this could be.

Pause and do this now.

Time to go through our answers.

For 1A, you should have discovered that the 3 adjacent houses were 18, 19, and 20.

In B, the 5 houses were 7, 8, 9, 10, and 11.

In question two, the 4 adjacent houses were 7, 9, 11, and 13.

And for B, the 6 adjacent houses were 24, 26, 28, 30, 32, and 34.

In question three, you had to work out Laura and Sam's numbers.

For Laura's number is 8 and Sam's number is 7.

Remember, you could check this by taking those numbers and following the steps that both Laura and Sam went through to check that you reached the final value that they did.

In question four, you had to work out the size of the marked angles.

You should have found that x was 20 degrees and therefore the two angles were both 100 degrees.

We knew they had to be equal to each other, remember? They're vertically opposite angles and that's how we set up our equation in the first place.

In part B, you know that these two angles are equal to each other 'cause they're the base angles in isosceles triangle.

That leads to x having a value of 15 degrees.

You can then substitute for x and discover that the two angles are 55 degrees.

You could even work out the remaining angle, which is why I asked you to do in the question, which was 70 degrees.

A quadrilateral had the following interior angles.

The simplified equation should have been 15x + 165 = 360.

Solving that leads to a value of 13 for x, and then you had to find the size of each angle.

Or when you substituted for x, you would've got 90 for each of these angles.

So this means I either have a rectangle or a square.

Without knowing the side lengths, I can't say which one it is.

Well done for having a go at all of these.

It's now time to solve what we've learned today.

With linear equations involving brackets, the brackets can be expanded before solving.

Where there are common factors, it can be more efficient to divide by the coefficient of the bracket first.

When the coefficient of the bracket is a fraction, it can sometimes be easier to multiply by the reciprocal instead of expanding the brackets.

And equations can be formed and solved in a variety of context to find unknown values.

Well done.

You've worked really well today.

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