# Unsolvable Maths Question

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Alex and Ben go to a cafe with some friends

Alex buys 4 cups of cofee and 3 cups of tea

He pays a total of £6.95

Ben buys 5 cups of coffee and 2 cups of tea

He pays a total of £7.20

Work out the cost of each cup of coffee and the cost of each cup of tea

I feel like its something really easy but at the same time i feel like its something really hard . Thanks for help

Alex buys 4 cups of cofee and 3 cups of tea

He pays a total of £6.95

Ben buys 5 cups of coffee and 2 cups of tea

He pays a total of £7.20

Work out the cost of each cup of coffee and the cost of each cup of tea

I feel like its something really easy but at the same time i feel like its something really hard . Thanks for help

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#3

(Original post by

Alex and Ben go to a cafe with some friends

Alex buys 4 cups of cofee and 3 cups of tea

He pays a total of £6.95

Ben buys 5 cups of coffee and 2 cups of tea

He pays a total of £7.20

Work out the cost of each cup of coffee and the cost of each cup of tea

I feel like its something really easy but at the same time i feel like its something really hard . Thanks for help

**AleksMM**)Alex and Ben go to a cafe with some friends

Alex buys 4 cups of cofee and 3 cups of tea

He pays a total of £6.95

Ben buys 5 cups of coffee and 2 cups of tea

He pays a total of £7.20

Work out the cost of each cup of coffee and the cost of each cup of tea

I feel like its something really easy but at the same time i feel like its something really hard . Thanks for help

and solve.

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#5

Form algebraic simultaneous equations

let t= price of one tea and c= price of one coffee

from Alex

4c+3t=6.95

from Ben

5c+2t=7.20

Do you know how to solve these?

let t= price of one tea and c= price of one coffee

from Alex

4c+3t=6.95

from Ben

5c+2t=7.20

Do you know how to solve these?

0

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(Original post by

Form algebraic simultaneous equations

let t= price of one tea and c= price of one coffee

from Alex

4c+3t=6.95

from Ben

5c+2t=7.20

Do you know how to solve these?

**glad-he-ate-her**)Form algebraic simultaneous equations

let t= price of one tea and c= price of one coffee

from Alex

4c+3t=6.95

from Ben

5c+2t=7.20

Do you know how to solve these?

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#8

Start by creating a set of two equations for the cups of coffee and the cups of tea. Let c = cups of coffee and t = cups of tea. So, for example, the first equation would be 4c + 3t = 6.95. Once you have both equations, solve them simultaneously for c and t.

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#9

(Original post by

if you could show me the working out i would be really greatful ,its been a while since i did them

**AleksMM**)if you could show me the working out i would be really greatful ,its been a while since i did them

4c+3t=6.95

5c+2t=7.20

lets choose t, we need to find the lcm of 2 and 3, smallest number divisible by both

can you think of it?

once you have it, subtract one equation from another, solve to find c then substitute back in to find t

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#10

**AleksMM**)

Alex and Ben go to a cafe with some friends

Alex buys 4 cups of cofee and 3 cups of tea

He pays a total of £6.95

Ben buys 5 cups of coffee and 2 cups of tea

He pays a total of £7.20

Work out the cost of each cup of coffee and the cost of each cup of tea

I feel like its something really easy but at the same time i feel like its something really hard . Thanks for help

So the question is based on the topic of simultaneous equations:

So if we let 1 cup of coffee= x

and 1 cup of tea= y

and create equations for both alex and ben e.g:

Alex's equations is going to be 4x ~+3y =£6.95

And Ben's will be 5x+ 2y= £7.20

you have two equations and solve them simultaneously. So e.g you could make the 'y's' both 6y so times the No.1 equation by 2 so you get 8x +6y =13.9

and for equation no2 times it by 3 so it will be : 15x + 6y =14.40

Then solve... Ill leave you to work it out

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#11

**AleksMM**)

if you could show me the working out i would be really greatful ,its been a while since i did them

Rearrange one equation to find an expression for or , and substitute it into the other, or

Multiply them such that you can subtract them to cancel one variable.

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#12

I'm really glad 7 of us said the same thing on this thread, hopefully we get an 8th just to make sure OP knows what system of equations they need to construct.

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#13

(Original post by

I'm really glad 7 of us said the same thing on this thread, hopefully we get an 8th just to make sure OP knows what system of equations they need to construct.

**RDKGames**)I'm really glad 7 of us said the same thing on this thread, hopefully we get an 8th just to make sure OP knows what system of equations they need to construct.

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(Original post by

choose either c or t and try to get coefficients equal in both equations

4c+3t=6.95

5c+2t=7.20

lets choose t, we need to find the lcm of 2 and 3, smallest number divisible by both

can you think of it?

once you have it, subtract one equation from another, solve to find c then substitute back in to find t

**glad-he-ate-her**)choose either c or t and try to get coefficients equal in both equations

4c+3t=6.95

5c+2t=7.20

lets choose t, we need to find the lcm of 2 and 3, smallest number divisible by both

can you think of it?

once you have it, subtract one equation from another, solve to find c then substitute back in to find t

do i only multiply the top equation?

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(Original post by

Hey,

So the question is based on the topic of simultaneous equations:

So if we let 1 cup of coffee= x

and 1 cup of tea= y

and create equations for both alex and ben e.g:

Alex's equations is going to be 4x ~+3y =£6.95

And Ben's will be 5x+ 2y= £7.20

you have two equations and solve them simultaneously. So e.g you could make the 'y's' both 6y so times the No.1 equation by 2 so you get 8x +6y =13.9

and for equation no2 times it by 3 so it will be : 15x + 6y =14.40

Then solve... Ill leave you to work it out

**Bluebell1234**)Hey,

So the question is based on the topic of simultaneous equations:

So if we let 1 cup of coffee= x

and 1 cup of tea= y

and create equations for both alex and ben e.g:

Alex's equations is going to be 4x ~+3y =£6.95

And Ben's will be 5x+ 2y= £7.20

you have two equations and solve them simultaneously. So e.g you could make the 'y's' both 6y so times the No.1 equation by 2 so you get 8x +6y =13.9

and for equation no2 times it by 3 so it will be : 15x + 6y =14.40

Then solve... Ill leave you to work it out

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#17

(Original post by

do i add or subrtact them?

**AleksMM**)do i add or subrtact them?

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(Original post by

Subtract to "cancel" one variable.

**_gcx**)Subtract to "cancel" one variable.

and got 7x= 0.5 im not sure where i went wrong and i feel really stupid

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#19

(Original post by

i did 15x+6y=14.4 - 8x+6y=13.9

and got 7x= 0.5 im not sure where i went wrong and i feel really stupid

**AleksMM**)i did 15x+6y=14.4 - 8x+6y=13.9

and got 7x= 0.5 im not sure where i went wrong and i feel really stupid

.

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#20

**Bluebell1234**)

Hey,

So the question is based on the topic of simultaneous equations:

So if we let 1 cup of coffee= x

and 1 cup of tea= y

and create equations for both alex and ben e.g:

Alex's equations is going to be 4x ~+3y =£6.95

And Ben's will be 5x+ 2y= £7.20

you have two equations and solve them simultaneously. So e.g you could make the 'y's' both 6y so times the No.1 equation by 2 so you get 8x +6y =13.9

and for equation no2 times it by 3 so it will be : 15x + 6y =14.40

Then solve... Ill leave you to work it out

0

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