algebra prob?

Question

There's a cage with 10 pets in it - dogs and cats. There are 56 biscuits to be fed to them. Each dog eats 6 biscuits and each cats eats 5 biscuits. How many of each pet are there?

The algebraic equation to this problem is

6D + 5C = 56
6(10 - C) + 5C = 56
60 - 6C + 5C = 56
60 - C = 56
- C = -4
C = 4

but where did the 6(10-C) come from? Also, how did you came to having -C=-4?

wow. thanks a lot. now that i understand this problem, i was wondering if you guys would kindly give me some more of these problems so that I could practice. Please give the answer, but press enter many times before typing the answer in.

ok, one more prob. why do you minus 60 from both sides?

Answer 1

Snowy,

Read my answer more carefully. I gave you BOTH equations,

D + C = 10
6D + 5C = 56

I took the first equation. D + C = 10 and solved it for D.

D + C = 10
....-..C........-.C
--------------------
D = 10 - C

This expression, 10 - C is to be substituted into the other equation in place of D.

6D + 5C = 56 becomes
6(10 - C) + 5C = 56 The D was replaced by (10 - C).


When you had

60 - C = 56, you want to solve for C by getting it alone. To get C alone, start by subtracting 60 on both sides of the equation.

.60 - C = 56
-60.........-60 60 - 60 is 0 and cancels out. Also, 56 - 60 = -4
----------------
- C = -4

But, C is not alone yet because -C = -4 means -1C = -4. To get C alone divide both sides by -1.

-1C = -4
-----....----
-1.......-1

C = 4

I hope that explains it all to you!! Keep on plugging! :-)

Answer 2

The 6(10-C) came from 6 biscuits for the dogs and the 10 came from the total number of pets which is 10 and C is the variable. You get C=4 because you take 60-C=56 and you substitue like this 60- C = 56 so to get C by itself make the 60 negaitve on the left side carry it over to the 56 so it would be 56 - -60 and you left the variable C as a negative which you cannot so you have to take the inverse of that and that is how you get a negative 4 you switch the negative C with to the 4. It is hard to explain in words hopefully this helps.

Answer 3

You know that each dog (D) eats 6 biscuits, so all of the dogs together eat 6D biscuits. You also know that there are 10 animals total, so the number of dogs (D) is equal to 10 minus the number of cats (10-C). So the number of biscuits eaten by the dogs is 6(10-C).

60 - C = 56
You know the number of cats (C) by solving for C in the above equation, but you must isolate C.
First, you subtract 60 from both sides,
-C = -4
But you want the number of cats, not it's negative, so you multiply both sides by -1, and get
C = 4.

Now you know that the number of cats is 4. And the number of dogs is 10 - C = 10 - 4 = 6 dogs

So 6 dogs ate 6 biscuits each 6 X 6 = 36
and 4 cats ate 5 biscuits each 4 X 5 = 20
36 + 20 = 56

Answer 4

D+C=10
D=10-C

6D+5C=56
6(10-C)+5C=56
60-6C+5C=56
60-C=56
60-56=C
C=4
D=10-4=6

Answer 5

there are 10 pets in the cage so C + D = 10 or D = 10 - C

by subtracting 60 from both sides of the equation

Answer 6

This is a system of equations with 2 variables, so you need 2 equations. Your first line is one of the equations. The second equation you need to solve the problem is C+D=10 because the amount of dogs plus cats equals 10. Now you can substitute D for (10-C)

Answer 7

(10-C) is the total number of pets in the cage minus the number of cats otherwise you could not solve the equation with two "numbers" ( C and D ) which you don't know how much are worth

Answer 8

1) 6D + 5C = 56
2) D + C = 10 [ten pets right?]

Two independant equations
Two variables
Hence one unique solution - hopefully not complex.

Answer 9

a million) upload C to the two sides of the equation: 9 -c +c = -13 + c (-c +c) = 0 so left facet is purely 9 9 = -13 + c 2) upload 13 to the two sides: 9 + 13 = c 3) remedy for c: 22 = c wish this helps!