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2 At Bea's Pet Shop, the number of dogs, d, is initially five less than twice the number of cats, c. If she decides to add three more of each, the ratio
of cats to dogs will be – Write an equation or system of equations that can be used to find the number of cats and dogs Bea has in her pet shop. Could Bea's Pet Shop initially have 15 cats and 20 dogs? Explain your reasoning. Determine algebraically the number of cats and the number of dogs Bea initially had in her pet shop.​

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ianwall

Let D be dogs and C be cats

dogs, d, is initially five less than twice the number of cats, c

D + 5 = 2C

If she decides to add three more of each, the ratio  of cats to dogs will be

D + 8 = 2C + 3

Could Bea's Pet Shop initially have 15 cats and 20 dogs?

Simply plug in the numbers

20 + 5 = 2(15)

This is clearly not true: 25 does not equal 30

akposevictor

1. The system of equations would be:

d = 2c - 5 ---> eqn. 1.

(c + 3)/(d + 3) = ¾ ---> eqn. 2.

2. Bea's shop cannot initially have 15 cats and 20 dogs because:

20 = 2(15) - 5

20 = 25 (false)

3. Initially, Bea's Pet shop had 9 cats and 13 dogs.

1. To solve algebraically, we would represent the problem as a system of equations as shown below:

d = number of dogs

c = number of cats

Thus:

"number of dogs is 5 less than twice the number of cats initially" can be represented as: d = 2c - 5 ---> eqn. 1.

Adding three more of each to give a ratio of cats to dogs of ¾ will be represented as: (c + 3)/(d + 3) = ¾ ---> eqn. 2.

  • The system of equations would be:

d = 2c - 5 ---> eqn. 1.

(c + 3)/(d + 3) = ¾ ---> eqn. 2.

2. To know if Bea's Pet shop could initially have 15 cats and 20 dogs, substitute c = 15 and d = 20 into d = 2c - 5.

  • Thus:

20 = 2(15) - 5

20 = 25 (false)

Bea's shop cannot initially have 15 cats and 20 dogs.

3. Solve the system of equations to find the number of cats (c) and the number of dogs (d) Bea initially had in her pet shop.

Substitute d = 2c - 5 into eqn. 2.

(c + 3)/(2c - 5 + 3) = ¾ ---> eqn. 2.

(c + 3)/(2c - 2) = ¾

  • Cross multiply

4(c + 3) = 3(2c - 2)

4c + 12 = 6c - 6

  • Combine like terms

4c - 6c = -12 - 6

-2c = -18

  • Divide both sides by -2

c = 9

To find d, substitute c = 9 into eqn. 1.

d = 2c - 5 ---> eqn. 1.

d = 2(9) - 5

d = 13

Thus, initially, Bea's Pet shop had 9 cats and 13 dogs.

Learn more about system of equations on:

https://brainly.com/question/13476446

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