Question

Indicate whether each of the following equations is sure to have a solution set of all real numbers. Explain your

Answer 1

Answer:

a) $0 = -2$ is not a valid equality. So a) will not have a solution set of all real numbers.

b) $0 = 0$ is a valid equality for all real numbers. So b) will have a solution set of all real numbers.

c) $0 = 0$ is a valid equality for all real numbers. So c) will have a solution set of all real numbers.

d) $24x^{3} = 72x^{3}$ is only valid for $x = 0$. So d) will not have a solution set of all real numbers.

Step-by-step explanation:

a)

$3(x+1) = 3x + 1$

$3x + 3 = 3x + 1$

$3x - 3x = 1 - 3$

$0 = -2$

$0 = -2$ is not a valid equality. So a) will not have a solution set of all real numbers.

b)

$x + 2 = 2 + x$

$x - x = 2 - 2$

$0 = 0$

$0 = 0$ is a valid equality for all real numbers. So b) will have a solution set of all real numbers.

c)

$4x(x+1) = 4x + 4x^{2}$

$4x^{2} + 4x = 4x + 4x^{2}$

$4x^{2} - 4x^{2} = 4x - 4x$

$0 = 0$

$0 = 0$ is a valid equality for all real numbers. So c) will have a solution set of all real numbers.

d)

$3x(4x)(2x) = 72x^{3}$

$24x^{3} = 72x^{3}$

This is only valid for $x = 0$. So d) will not have a solution set of all real numbers.