3D + 4C = 52

subtract 4C from both sides

3D = 48

48/3

D=16

3(16) + 4C = 52

48 + 4C = 52

sybtract 48 from both sides

4C = 4

4/4

C = 1

check

3(16dogs) + 4(1cat) = 52 on the ranch

48dogs + 4cats = 52

52dogs&cats= 52 on the ranch

wrong

- Thread starter humakhan
- Start date

3D + 4C = 52

subtract 4C from both sides

3D = 48

48/3

D=16

3(16) + 4C = 52

48 + 4C = 52

sybtract 48 from both sides

4C = 4

4/4

C = 1

check

3(16dogs) + 4(1cat) = 52 on the ranch

48dogs + 4cats = 52

52dogs&cats= 52 on the ranch

wrong

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- Feb 4, 2004

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Hint: Look up the word "exceeded". Once you know what it means, you'll know what operation to use.humakhan said:A total of 52 dogs and cats were on the ranch. if three times the number of dogs exceeded the number of cats by four, how many of each were on the ranch?

3D + 4C = 52

Since 1 + 16 does not equal 52, the above cannot be the correct solution.humakhan said:cats = 1

dogs = 16

Eliz.