Question

Which of the equations below have infinitely many solutions? Select all that apply.

a. 3x - 1 = 3x + 1
b. 2x - 1 = 1 - 2x
c. 3x - 2 = 2x - 3
d. 3(x - 1) = 3x - 3
e. 2x + 2 = 2(x + 1)
f. 3(x - 2) = 2(x - 3)

Answer 1

Equations d and e have an infinite number of solutions, since both sides of the equations are identical (so any value of x will work).

Step-by-step explanation:

a. 3x - 1 = 3x + 1 Equation is false (not possible): 0 = 2

b. 2x - 1 = 1 - 2x One solution: x = (1/2)

c. 3x - 2 = 2x - 3 One solution: x = -1

d. 3(x - 1) = 3x - 3 Infinite solutions: 3x-3 = 3x-3 valid for ANY x value

e. 2x + 2 = 2(x + 1) Infinite solutions: 2x+2 = 2x+2 valid for ANY x value

f. 3(x - 2) = 2(x - 3) One solution: x = 0

Answer 2

(d) and (e)

Step-by-step explanation:

To identify which of these equations have infinitely many solutions, you need to understand that for an equation to have infinitely many solutions, both sides of the equation must be identical, meaning that they are essentially the same mathematical statement.

Let's process these options:

(a) 3x - 1 = 3x + 1: Subtracting 3x from both sides we obtain -1 = 1, hence, no solutions.

(b) 2x - 1 = 1 - 2x: If we add 2x and 1 to each side, we get 4x = 2 ⇒ x = 1/2, thus it is not an identity and has a unique solution.

(c) 3x - 2 = 2x - 3: Rearranging we have x = -1, it’s not an identity, hence, this equation has a unique solution.

(d) 3(x - 1) = 3x - 3: Simplifying we obtain 3x-3 = 3x-3, an identity, hence, infinitely many solutions.

(e) 2x + 2 = 2(x + 1): Simplifying we obtain 2x + 2 = 2x + 2, an identity, hence, infinitely many solutions.

(f) 3(x - 2) = 2(x - 3): This simplifies to 3x - 6 = 2x - 6 ⇒ x = 0, thus it is not an identity and has a unique solution.

Thus, the correct options are (d) and (e).