30 points and !

1. solve the rational equation x divided by 4 equals x squared divided by quantity x plus 2 end quantity, and check for extraneous solutions.

x = two thirds; 0 is an extraneous solution
x = 0; x = two thirds is an extraneous solution
x = 0 and x = two thirds
no solution

2. solve the rational equation 3 divided by x equals quantity 4 times x plus 3 divided by x squared, and check for extraneous solutions.

x = 0; x = 3 is an extraneous solution
x = 3; x = 0 is an extraneous solution
x = 0; x = −3 is an extraneous solution
x = −3; x = 0 is an extraneous solution

3. find the domain for the rational function f of x equals quantity x minus 1 end quantity divided by quantity x plus 3 end quantity.

(−∞, 1) (1, ∞)
(−∞, −1) (−1, ∞)
(−∞, 3) (3, ∞)
(−∞, −3) (−3, ∞)

Part 1) x = 0 and x = two thirds

Part 2) x = −3; x = 0 is an extraneous solution

Part 3) (−∞, −3) ∪ (−3, ∞)

Step-by-step explanation:

Part 1) we have

Solve for x

Multiply by 4(x+2) both sides to remove the fractions

Solve the quadratic equation

Factor 3

Complete the square

Rewrite as perfect squares

square root both sides

Part 2) we have

Multiply by x^2 both sides to remove fractions

Part 3) we have

we know that

The denominator cannot be equal to zero

so

The value of x cannot be equal to -3

so

The domain of f(x) is all real numbers except the value of x=-3

therefore

(−∞, −3) ∪ (−3, ∞)

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.