Question

The school is having a bake sale and the math club is bringing cupcakes. They have 180 total cupcakes, both chocolate and strawberry. The number of chocolate cupcakes is eighteen more than twice the number of strawberry cupcakes. Which system of equations can be used to find the number of chocolate cupcakes, c, and the number of strawberry cupcakes, s? A. c + s = 180 s = 2c - 18 B. c + s = 180 c = 2s + 18 C. c - s = 180 s = 2c + 18 D. c = 2s c - 180 = s

Answer 1

Let $s$ be the number of strawberry cupcakes, and $c$ the number of chocolate cupcakes.
From our problem we know that the total number of cupcakes is  180, so $c+s=180$. We also know that the number of chocolate cupcakes is eighteen more than twice the number of strawberry cupcakes, so $c-18=2s$, or solving for $c$: $c=2s+18$.
Know we have our system of equations:
$\left \{ {{c+s=180} \atop {c=2s+18}} \right.$
We can conclude that the correct answer is B. c + s = 180 c = 2s + 18.

Lets solve our equations to find how many cupcakes of each they have:
from our second equation we know that $c=2s+18$; lets replace that value in our first equation to find the number of strawberry cupcakes:
$2s+18+2=180$
$3s=162$
$s= \frac{162}{3}$
$s=54$
Now that we know the number of strawberry cupcakes, lets replace that value in our second equation to find the number of chocolate cupcakes:
$c=2(54)+18$
$c=108+18$
$c=126$
As a bonus, we can conclude that the math club have 126 chocolate cupcakes and 54 strawberry cupcakes. 

Answer 2

The correct answer is B) c+s=180, c=2s+18

The number of chocolate and strawberry together make 180.  c+s=180.

The number of chocolate is 18 more than twice the number of strawberry; c=2s+18.