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5 votes
Please help me! im having trouble​

Please help me! im having trouble​-example-1
User Academia
by
6.1k points

2 Answers

5 votes

Answer:

Step-by-step explanation:


\left \{ {{c+s=4} \atop {1.75c+1.3s=6.55}} \right.

1.75c + 1.75s = 1.75 × 4 ........ (1)

1.75c + 1.3s = 6.55 ........ (2)

(1) - (2)

0.45s = 0.45 ⇒ s = 1

c + 1 = 4 ⇒ c = 3

[ 3 ] chocolate [ 1 ] strawberry

User Rashidah
by
6.9k points
4 votes

Answers:

3 chocolate cones

1 strawberry cones

===========================================================

Step-by-step explanation:

Let's isolate the variable c in the first equation

c+s = 4

c = 4-s

we subtract s from both sides to get c all by itself. This will then be plugged into the second equation. I'm using the substitution property.

1.75c + 1.3s = 6.55

1.75(4-s) + 1.3s = 6.55 ..... replace c with 4-s

1.75*4 + 1.75*(-s) + 1.3s = 6.55

7 - 1.75s + 1.3s = 6.55

-0.45s + 7 = 6.55

-0.45s = 6.55 - 7

-0.45s = -0.45

s = -0.45/(-0.45)

s = 1

So he bought 1 strawberry cone. Use this value of s to find c

c = 4-s

c = 4-1

c = 3

This says he also bought 3 chocolate cones.

-------------------

As a check, we see that c+s = 3+1 = 4, showing that he bought 4 cones total.

Also,

1.75c + 1.3s = 1.75*3 + 1.3*1 = 6.55

indicating he spent $6.55 total for the four cones. This matches with what the instructions tell us, so the answer is confirmed.

User Kirlich
by
7.1k points
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