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Unsolvable Maths Question

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

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simultaneous equation broski
Original post by 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


Simultaneous equations. If a cup of tea costs tt and a cup of coffee costs cc then:

4c+3t=6.954c+3t=6.95
5c+2t=7.25c+2t=7.2

and solve.
It's just simultaneous equations.

4c+3t=6.954c + 3t = 6.95
5c+2t=7.205c + 2t = 7.20
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?
Simple simultaneous equation bro
Reply 6
Avatar for AleksMM
AleksMM
OP
4
Original post by 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?


if you could show me the working out i would be really greatful ,its been a while since i did them
Reply 7
Avatar for Joshiee
Joshiee
11
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.
Original post by AleksMM
if you could show me the working out i would be really greatful ,its been a while since i did them


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
(edited 6 years ago)
Original post by 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


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
:smile:
Original post by AleksMM
if you could show me the working out i would be really greatful ,its been a while since i did them


Two approaches:

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

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


i think OP should use a system where they attempt to solve two equations at the same time, cant remember the name of the system
Reply 13
Avatar for AleksMM
AleksMM
OP
4
Original post by 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

6...
do i only multiply the top equation?
Original post by AleksMM
6...
do i only multiply the top equation?


no, get both as 6t so times top by 2 and bottom by 3 then subtract
Reply 15
Avatar for AleksMM
AleksMM
OP
4
Original post by 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
:smile:


do i add or subrtact them?
Original post by AleksMM
do i add or subrtact them?


Subtract to "cancel" one variable.
Reply 17
Avatar for AleksMM
AleksMM
OP
4
Original post by _gcx
Subtract to "cancel" one variable.


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
Original post by 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


No, the first equation is incorrect.

7.2∗3≠14.47.2 * 3 \neq 14.4.
Original post by 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
:smile:


Your second equation is incorrect, 7.2∗37.2 * 3 not 7.2∗27.2 * 2 :smile:

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