Question

A human 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 represent this situation. How many of each animal is up for adoption?

Answer 1

Total: 73
dogs: x
cats: 2x+10

2x+10+x=73
3x+10=73
3x=63
x=21

dogs: x
dogs: 21

cats: 2x+10
cats: 52

Check answer:
cats+dogs=73
52+21=73

There's no need for a system of linear equations if you were to solve a real life problem like this. In calculus, they don't give a rat about HOW you get your answer to an easy algebra 1 equation, as long as you used one of the correct methods and got the right answer.

Best of luck my friend. :)

Answer 2

A linear equation of one variable is expressed in the form of ax + b = 0, where a and b are the two integers, and x is a variable.

In the given question, the linear equation can be represented as:

2x + 10 + x = 73.

The number of dogs is 21, and the number of cats available is 52.

Given that:

Total number of dogs and cats = 73

Let the number of dogs be x.

The number of cats, as given is = 10 + 2x

Now, solving for the equation:

x + 10 + 2x  = 73

Solving for the variables:

3x + 10 = 73

3x = 73 -10

3x  = 63

x = $\dfrac {63}{3}$

x = 21

The number of dogs available is 21.

The number of cats available is:

2x + 10

Putting the value of x:

2 (21) + 10

52

Thus, the number of dogs is 21, and the number of cats available is 52.

To know more about linear equations, refer to the following link:

https://brainly.com/question/11897796