Indicate whether each of the following equations is sure to have a solution set of all real numbers. Explain youranswers for each.a. 3(x + 1) = 3x + 1b. x + 2 = 2 + xc. 4x(x + 1) = 4x + 4x²d. 3x(4x)(2x) = 72x³

Answer:a) [tex]0 = -2[/tex] is not a valid equality. So a) will not have a solution set of all real numbers.b) [tex]0 = 0[/tex] is a valid equality for all real numbers. So b) will have a solution set of all real numbers.c) [tex]0 = 0[/tex] is a valid equality for all real numbers. So c) will have a solution set of all real numbers.d) [tex]24x^{3} = 72x^{3}[/tex] is only valid for [tex]x = 0[/tex]. So d) will not have a solution set of all real numbers.Step-by-step explanation:a)[tex]3(x+1) = 3x + 1[/tex][tex]3x + 3 = 3x + 1[/tex][tex]3x - 3x = 1 - 3[/tex][tex]0 = -2[/tex][tex]0 = -2[/tex] is not a valid equality. So a) will not have a solution set of all real numbers.b)[tex]x + 2 = 2 + x[/tex][tex]x - x = 2 - 2[/tex][tex]0 = 0[/tex][tex]0 = 0[/tex] is a valid equality for all real numbers. So b) will have a solution set of all real numbers.c) [tex]4x(x+1) = 4x + 4x^{2}[/tex][tex]4x^{2} + 4x = 4x + 4x^{2}[/tex][tex]4x^{2} - 4x^{2} = 4x - 4x[/tex][tex]0 = 0[/tex][tex]0 = 0[/tex] is a valid equality for all real numbers. So c) will have a solution set of all real numbers.d)[tex]3x(4x)(2x) = 72x^{3}[/tex][tex]24x^{3} = 72x^{3}[/tex]This is only valid for [tex]x = 0[/tex]. So d) will not have a solution set of all real numbers.