In the cafeteria all sandwich are priced the same in all drinks are priced the same a customer ordered three sandwiches and two drinks for $14.70 another customer bought two sandwiches and four drinks for $13.80 find the cost of one sandwich and the cost of one drink

Cost of 1 sandwich = $3.9

Cost of 1 drink = $1.5

Step-by-step explanation:

Let us assume the cost of 1 sandwich  = $ m

And the cost of 1 drink   = $ n

Now, according to the question:

Cost of 3 sandwich + 2 drink = $14.7

⇒ 3 m  + 2 n  = $14.7 (1)

Cost of 2 sandwich + 4 drink = $13.8

⇒ 2 m  +  4 n  = $ 13.8 (2)

Solving both equation, we get:

Multiply (1) with -2 and add both equations, we get:

- 6 m - 4 n + 2 m + 4 n = 13 .8 -29.4

or, -4 m =  -15.6

or, m = 3.9  

Put m = 3.9   in  3 m  + 2 n  = $14.7, we get:

11.7 +  2 n = 14.7

2 n = 3 , or n = 1.5

Hence, cost of 1 sandwich  = $3.9

and the cost of 1 drink  = $1.5

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!