# Unsolvable Maths Question

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#1
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
2
3 years ago
#2
simultaneous equation broski
1
3 years ago
#3
(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 and a cup of coffee costs then:  and solve.
0
3 years ago
#4
0
3 years ago
#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?
0
3 years ago
#6
Simple simultaneous equation bro
0
#7
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
0
3 years ago
#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.
0
3 years ago
#9
(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
0
3 years ago
#10
(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 0
3 years ago
#11
(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 or , and substitute it into the other, or

Multiply them such that you can subtract them to cancel one variable.
0
3 years ago
#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.
8
3 years ago
#13
(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
2
#14
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?
0
3 years ago
#15
(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
0
#16
(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 do i add or subrtact them?
0
3 years ago
#17
(Original post by AleksMM)
do i add or subrtact them?
Subtract to "cancel" one variable.
1
#18
(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
0
3 years ago
#19
(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. .
0
3 years ago
#20
(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 Your second equation is incorrect, not  0
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