Lesson video

Hello, Mr. Robson here.

Welcome to maths, Solving Simultaneous Equations Via Any Method.

If you're like me and you love being able to do the same problem using multiple different methods, then you are gonna love this lesson.

Let's go.

Our learning outcome is that we'll be able to determine which method of solving is most appropriate and interpret the solutions in context.

Some keywords we'll use today, substitute, substitution.

Substitute means to put in place of another.

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

Another keyword you'll see is elimination.

Elimination is a technique to help solve equations simultaneously where one of the variables in a problem is removed.

Look out for those keywords throughout our lesson.

Two parts to the lesson today.

And we're gonna start by comparing methods to solve.

In mathematics, it can be useful to know a variety of ways to solve a problem.

Here is a pair of simultaneous equations.

We can solve by substitution.

If we label them equation 1 and equation 2, we can substitute equation 1 into equation 2 to create the equation 5x minus 3 equals 2x plus 9.

From there we can rearrange to get 3x equals 12.

So x itself must be 4.

Substitute x back into one of the original equations and we find the y value of 17.

We could solve the same equation by elimination.

Again, label them equation 1 and equation 2.

This time we're gonna subtract equation 2 from equation 1.

That's gonna give us 0 on the left-hand side, 'cause y subtract y is nothing, and on the right-hand side we'll be left with 3x minus 12.

From there we can rearrange to 12 equals 3x, so x must be equal to 4, substitute back in and we would obviously get the same y value, 17.

Having two different methods allows us to use one method to check our answers from the other method.

This is really reassuring.

In selecting a method we should consider efficiency.

When we look at these two different methods, we should hopefully have noticed elimination required fewer steps.

For this pair of equations, elimination was the most efficient method.

We will encounter simultaneous equations where y will not be the subject.

That last example was rather a kind one.

This one's different.

We can solve it by substitution.

Again, we'll label it equation 1 and equation 2.

I'm gonna rearrange equation 2 to make y the subject.

Once I've got y as the subject, I can substitute that into equation 1.

And we get 5x plus what we know y to be, which is 60 minus 2x, and that's equal to 4.

Remove that bracket, rearrange, and solve.

We find that x equals -4.

Substitute that back in because we need to know the value of y, and we find y to be 24.

That's a solution.

The only solution that satisfies both equations simultaneously.

We weren't limited to solving by substitution.

We could solve by elimination.

Equation 1 and equation 2, and we're gonna subtract equation 2 from equation 1 this time.

We're left with 3x equals -12, because 5x minus 2x on the left-hand side leaves us with 3x and y minus y leaves us with no ys.

The ys are eliminated.

Therefore, we get x equals -4.

Substitute that back in, y equals 24.

Do note the exact same solution.

We must be right.

In this case, elimination was more efficient.

There were fewer steps when we solved by elimination.

It's lovely to have multiple ways to solve the same problem.

It confirms our answer.

Ultimately, we're looking for the most efficient way to solve a problem.

Quick check you've got what we've covered so far.

Which statements are true when solving a pair of simultaneous equations.

Statement a true? Is statement b true? Is statement c true? I'll leave that for you to decide.

Pause and have a think now.

Welcome back.

I wonder what you said.

Hopefully you said statement a is true.

Using multiple methods reassures us that our solution is correct.

It's nice to know when doing algebra we've got lots of checking mechanisms to confirm that we're right.

I hope you said option b is not true.

We have to use the same method each time.

No, we've got versatility, we have choice, we have multiple methods to solve the same problem when faced with a simultaneous equation.

That's nice.

C was also true.

We can choose which method to use and should consider efficiency when doing so.

We mathematicians, we love efficiency.

When coefficients don't match, we have even more to consider about which method to use.

One option here is to eliminate the y variables.

We'll start by labelling them equation 1 and equation 2, and then we're going to quadruple everything in equation 2.

Why would we want to do that? Because we wanna turn that negative y into -4y, because when we insert equation one underneath, oh look, -4y, positive 4y, wonderful.

I'll add those two equations together and the y variable will be eliminated.

If 13x equals negative 26, x equals -2.

Substitute that back in, and we find y is equal to 5.

X equals 2, y equals 5.

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

We weren't limited to just eliminating the y variable.

We could eliminate the x variable.

Same two equations.

I'm going to double equation 1 to turn 5x into 10x.

Then I'm going to quintuple, that's to multiply by five, quintuple, lovely word, hey, gonna quintuple equation 2 to turn 2x into 10x.

By doing that we can now subtract 5 lots of equation 2 from 2 lots of equation 1, 10x subtract 10x, it is no xs left.

They've been eliminated.

If 13y equals 65, y equals 5, substitute that back in to find the x value.

X equals -2.

Lovely.

Our second approach has verified our solution.

You get the same answer from two different methods, you're probably right.

Elimination of y in this case was slightly more efficient.

In eliminating y, we had y as a factor of 4y.

In the second case when we went to eliminate x, 2x was not a factor of 5x.

That meant we had to multiply both equations.

In terms of efficiency, multiplying just the one equation made it a little quicker.

So eliminating the y variable is the quickest route to elimination to find this solution.

But when not restricted to elimination, we could solve the same pair of equations using substitution.

We'll label them equation 1 and equation 2 again.

We'll rearrange equation 2 to make y the subject and then substitute that into equation 1.

From there, expand the brackets.

Simplify.

13x equals -26, so x must be -2.

And that would again give us the same y value of 5.

Funnily enough, it's the exact same solution.

There's very little difference in efficiency here.

So either method works well.

Quick check you've got this.

I'd like you to solve this pair of linear simultaneous equations using two different methods.

That will help you to confirm that your solution is correct.

Pause, give this problem a go now.

Welcome back, see how we did.

I'm gonna start by eliminating the y variable.

I'm going to treble equation 1 to turn y into 3y, because when I insert equation 2 underneath positive 3y, <v ->3y, wonderful.

</v> I'll add those together.

9x plus 2x equals 11x, 3y add -3y, no ys left, the ys are eliminated.

And on the right-hand side, 18 plus 15 equals 33.

So if 11x equals 33, x equals 3, substitute that back in and we find y equals -3.

There's our solution.

Now in this case, if we wanted to do eliminating by eliminating the x variable, we'd have to multiply both equations.

That would be less efficient.

So I'm gonna give substitution a go, see if that's got equal efficiency to eliminating the y variables.

So in substitution I label them equation 1 and equation 2.

Rearrange equation 1 to make y the subject and insert that into equation 2.

Okay, 2x minus 3 lots of bracket 6 minus 3x equals 15.

Expand the bracket.

Simplify.

Rearrange.

11x equals 33.

X equals 3.

Substitute that back in, y equals -3.

Same result.

That's verified our solution.

I'm pretty confident that we're correct here.

And very little difference in efficiency between elimination of y and substitution for this pair of simultaneous equations.

If the simultaneous equations arise from a context, any solutions should be interpreted in that context.

For example, we might be faced with a question, in which quadrant do the linear graphs 2x minus 3y equals 32 and y equals 3x minus 27 intersect? We can find the intersection by solving this pair of simultaneous equations.

Because y is already the subject, it'll be really efficient to substitute, so we'll take equation 2 and substitute it into equation 1.

From there, expand those brackets, rearrange and simplify.

If 7x equals 49, x equals 7.

Substitute that back in and we'll find the corresponding y value to be -6.

So we're finished, right? Wrong, we haven't interpreted what this means in context.

We were asked in which quadrant do the linear graphs intersect? We have to consider this coordinate, x equals 7, y equals -6.

We could plot that point on a graph, and it would be there in quadrant four.

So, we need to answer the question.

They intersect in the fourth quadrant.

If you're interested, there's the graphical representation of that intersection.

Quick check that you can interpret a solution in context now.

Sofia knows a taxi firm has a fixed charge per journey followed by a cost per mile travelled.

She records a 3-mile journey costs 10.

50, an 8-mile journey costs 18 pounds, then forms and solves two simultaneous equations.

That's a lovely (indistinct).

What I'd like you to do is write a sentence to interpret this solution in the context of the question.

Pause and write a sentence now.

Welcome back.

"You might have written the fixed charge is 6 pounds "and the cost per mile is 1.

50." That's what's meant by Sofia's solutions, m equals 1.

5 and f equals 6.

Oh, Sofia said thank you.

You're welcome Sofia.

Practise time now.

Question 1, I'd like to solve these pairs of simultaneous equations using two different methods.

You're going to do each pair of equations twice with two different methods.

Pause and do this now.

For question 2, I'd like you to solve this pair of simultaneous equations.

For part a I'd like you to solve by eliminating the y terms. Part b, do the same problem, but this time eliminate the x terms. And then part c, I'd like you to write a sentence comparing the efficiency of methods a and b.

Pause and try this now.

Question 3, Sofia and Alex are going to solve this pair of simultaneous equations using substitution.

Sofia says, "I'm going to substitute "2x minus 1 equals 7y into equation 2." Whereas Alex says, "I'm going to make y the subject "and use y equals 2/7 of x minus 1/7.

Part a, I'd like you to select the most efficient method and solve.

You're gonna choose Sofia's method? Or you're gonna choose Alex's method? For part b, I'd like you to write a sentence explaining why the other method, the one you didn't choose, was less efficient.

Pause and do those now.

Question four, it's a question in context.

Which pair of lines intersect closest to the origin? Is it pair a or pair b? A lot of interesting maths to do here.

You'll wanna pause and have a good think about this one.

See you in a moment.

Feedback time now.

For question 1, I am to solve pairs of simultaneous equations using two different methods.

For part a, I'm gonna start by using substitution.

When I substitute equation 1 into equation 2, I get 7x plus 5 equals 4x minus 16, rearrange, and I find that x equals -7, substitute that back in and y equals -44.

I wonder if I'm right.

I can find out by doing it again using a different method.

Elimination.

When I subtract equation 2 from equation 1, I get 0 on the left-hand side, 3x plus 21 on the right-hand side, and look, x equals -7.

Substituted back in, y equals -44.

I think I'm right.

Question one part b, by substitution.

Nice way to do it, because in equation 2 we've got y as a subject.

That means we can substitute it nice and easily into equation 1.

From there, remove the bracket, simplify, rearrange, x equals 6.

Substitute that back in, we find y equals 3.

I think we're right, but if I also solve by elimination, I'll know we're right.

We can eliminate by adding the two equations together, because on the left-hand side there, -y and positive y, that eliminates the ys.

We get to 0 equals x minus 6, so x must equal 6, substitute it back in, y must equal 3.

Wonderful.

I'm pretty confident that I'm right.

Question two.

I have to solve this pair of simultaneous equations firstly by eliminating the y terms, secondly by eliminating the x terms, and then writing a sentence to compare the efficiency of the two.

By eliminating y, we need to double equation 1 to turn -3y into -6y, because from there, we can add the two equations together and the y variable is eliminated.

23x equals 115, x must equal 5.

Substitute that back into one of the original equations and we find that y equals 12.

Let's contrast that with eliminating x, part b.

If we want to eliminate the x variables, we have to multiply by 8, equation 2, and multiply by 7, equation 1.

We need 56x, and 56x matching coefficients.

From there, I can take one from the other, leaving 69y equals 828, divide through by 69.

Y equals 12, substitute back in, and x equals 5.

In terms of efficiency, you might have written eliminating y was more efficient because 3y was a factor of 6y, so we only had to multiply one equation.

Eliminating x meant multiplying both equations and the numbers got much larger.

It's fine to work with large numbers.

We got to the same solution, but we didn't need to.

We kept it nice and simple when we eliminated the y variables.

For question 3, I asked you to think about Sofia's idea and Alex's idea for how we might solve this pair of simultaneous equations, select the most efficient method, and solve.

I hope you spotted that Sofia's method is more efficient.

3x plus 14y equals 5.

Well, rather joyously, 14y is two lots of 7y.

So when Sofia said "I'm going to substitute "2x minus 1 equals 7y into equation 2," aha, there we are, 7y.

We can substitute the 2x minus 1 there.

From there, expand the bracket, simplify, rearrange, we find X equals 1.

And when we substitute that back in y equals 1/7.

Let's contrast that with Alex's method.

That's how Alex wanted to go about it, which is fine, and it works, but it's less efficient.

Alex's method required more skills than Sofia's method.

Sofia's method was simpler, and therefore, more efficient.

For question 4, I asked you which pair of lines intersect closest to the origin.

To find the intersection of the two lines, we can solve simultaneous equations.

If I do equation 1 and subtract equation 2, we're left with 4y plus two equals -30.

If 4y equals -32, y equals -8, substitute back in, we'll find an x value of 6.

For part b, if we do equation 1 add equation 2, we eliminate those y variables.

If 8x equals -40, x equals -5, substitute back in, y equals 12.

But we're not done there.

They're solutions to the simultaneous equations, absolutely, but we haven't interpreted them in context yet.

We have to think about the intersections or the coordinates of the intersection and how far they are from the origin.

For part a, there's an intersection at (6,-8).

And for part b, there's an intersection at (-5,12).

We can calculate the distance of the origin by using Pythagoras.

We do that for part a, 6 squared add -8 squared, square root of that is 10.

We'll do the same for b, and we find the distance to the origin there to be 13.

We're ready to draw a conclusion.

Pair a intersects closest to the origin.

That's our answer, considering the context of the question.

On to the second part of the lesson now, using a calculator to solve.

You don't always have to use a written method to solve a pair of simultaneous equations.

If permitted, you can solve them using your calculator.

From the home screen of the Casio Classwiz, select Equation.

That's the option there.

Hit OK, and these options come up.

The one you want is the one called Simul Equation.

What do you think that's short for? From there, select 2 unknowns.

There is such a thing in maths as a simultaneous equation with three unknowns, four unknowns, but today we're just solving ones with two unknowns.

Your calculator display should look like that.

We're now ready to enter a pair of simultaneous equations.

We're gonna enter this pair, on account of we've already seen this pair.

We solved it earlier.

We know the solution is x equals 3, y equals -3.

We're just gonna check that we can enter this into a calculator to come up with the same solution.

In order to do so, we need to enter the coefficients.

So, for equation 1 I want it to read 3x plus 1y equals 6.

In order to do that on my calculator, I hit the button 3, then execute button, the button 1, and then execute and then 6 and then execute.

Curiously, you have to enter the coefficient of 1, even though we'd not typically write this when solving algebraically.

Your calculator needs to know that the coefficient of 7 is 1.

Entering the second equation now, 2x minus 3 equals 15.

I press 2 and then execute, <v ->3 and then execute, 1, 5, and execute.

</v> It's really important that you change the positive to a negative in the second equation to calculate to know that the coefficient of y is -3.

From here, hit execute.

That comes up, oh look, x equals 3.

Hit execute again, oh look, y equals -3.

Super.

I'd like to check you can do that now.

I'd like you to use your calculator to solve this pair of simultaneous equations.

Pause, see if you can do that now.

Welcome back.

How did we get on? Hopefully you were able to make your calculator screen look like so.

From there, hit execute, x equals 54/7 and y equals -12/7.

There's our solution.

Your calculator will only accept equations in the form axe plus by equals c.

So sometimes it's necessary to rearrange before inputting.

For example, if we add the pair of simultaneous equations, x plus y plus 8 equals 0 and y equals 4x minus 17 we can't put them in in that format.

We need to rearrange.

So x plus y plus 8 equals 0.

I can add -8 to both sides.

That's in the right form.

Y equals 4x minus 17.

I'm gonna add positive 17 to both sides, and then I'm gonna subtract y, and then if 17 equals 4x minus y, then 4x minus y equals 17.

That is in the right form to be input.

There are the two equations I'm gonna input.

Once I've done it the calculator screen looks like that.

It's execute x equals 9/5 and y equals -49/5.

That's our solution.

Quick check you've got that now.

True or false, you can enter any pair of simultaneous equations into your calculator without needing to rearrange.

Is that true or is it false? Once you've decided, you can you use one of the two statements at the bottom of the screen to justify your answer.

Pause and think about this now.

Welcome back.

I do hope you've said it's very, very false.

And justify that with we can only enter equations in the form axe plus by equals c.

Sometimes it is necessary to rearrange first.

We would need to rearrange these equations to read 3x minus y equals 8 and 2x plus y equals 7.

Practise time now.

Question 1, solve these pairs of simultaneous equations using your calculator.

Pause, give them a go.

For question 2, Alex is trying to solve these pairs of simultaneous equations with a calculator, but he's made some errors, which is okay, it's maths.

We're always going to make errors.

We just need to keep a keen eye out for them and not be afraid to try again when we do make an error.

I'd like you to spot Alex's errors and write a sentence explaining what he needs to correct.

Pause and do this now.

Feedback time now.

For the first pair of simultaneous equations, your calculator display should have looked like so when you hit execute x equals 45/34 and y equals 63/34.

There's your solution.

For pair b, your calculator display should look like so.

It's execute x equals -47/253 and y equals 1317/2530.

There's your solution.

Kinda glad we had the calculator to help us with that one.

For question 2, we were helping Alex out, spotting errors and giving them feedback as to what could be done better.

In this case, the error was there, 1x.

That should be -1.

In the second case, 2x plus 1y, oh no, the x and y coefficients are the wrong way round.

It should be 1x and 2y.

On equation 2 there was an error as well.

It's an incorrect rearrangement.

If you're rearranging 2y plus 3 equals 7x, the rearrangement to input would be 7x minus 2y equals 3.

Sadly, that's the end of our lesson now, but we've learned when solving simultaneous linear equations, we can determine which method of solving is most appropriate.

The features of the equations can help us with this.

We should interpret the solutions in context when required.

We can also use our calculator to solve simultaneous equations.

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

Goodbye for now.