A coffee shop sells lattes for $4 and

cappuccinos for $3. Last Friday, sales totaled

$252. If the number of lattes sold was 6 more

than 4 times the number of cappuccinos, how

many lattes were sold?

(If you can please write the system of equations and how you solve it)

54 lattes were sold

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations with the information given:

If each latte costs $4 , each cappuccino cost $3, and sales totaled $252:

4 x + 3y = 252

Where  

x = number of lattes

y = number of cappuccinos

The number of lattes sold was 6 more than 4 times the number of cappuccinos. Mathematically speaking:

x = 4y +6

Now, we have the system:

a)4 x + 3y = 252

b)x = 4y +6

Replacing the value of x in a) for b):

4 (4y +6) +3y =252

16y +24 +3y =252

16y +3y =252 -24

19y = 228

y= 228/19

y = 12

Replacing the value of "y” in b ) for the value obtained:

b)x = 4y +6

x = 4 (12) +6

x = 48+6 = 54

x =54

So , 54 lattes were sold.

step-by-step explanation:

d

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8