**a humane society has 73 dogs and cats to be adopted. The number of cats is 10 more than twice the number of dogs. Write a system of linear equations that represents this situation. How many of each animal is up for adoption?**

**Answer:**

Cats = C

Dogs = D

Equations:

73 = Cats & Dogs

Cats = 10 + 2(Dogs)

We then solve the set of linear eqution set/system.

73=c+d

c=10+2d

Lets multiply the first eq. by 2.

146=2c+2d

c=10+2d

We rearange.

146=2c+2d

-10=-c+2d

Subtract to get solution.

146-(-10) = 156

2c+c = 3c

2d-2d = 0 = nothing to put in final Eq.

So, we add the numbers and get:

3c = 156

Divide 156 by three and get 52.

Therefore, there is 52 cats.

We can then substitute the 52 for c.

Lets do the first one to make it easier.

73 = 52 + d

73-52=21

21 = d

Turn around to simplify.

d = 21

Therefore, there is 21 Dogs and 52 Cats.