Question

Please i need help with these questions

Answer 1

Answer:

S)No solution

T)Infinite solution

U)x=0

V)No solution

W) -1.2

X) No solution

Step-by-step explanation:

For solving linear equations of one variable, take variable on one side and constants on one side.

S)$\frac{-1}{3}(x) - \frac{1}{2} = \frac{-1}{3}(x) +12$

  $\frac{1}{2} = 12$ , this is false.

  thus, for what ever value of x, this is false, thus, no solution.

T)$\frac{-5}{6}(x) +40 = 40 + \frac{-5}{6}(x)$

   $40 = 40$ , this is true.

thus, for what ever value of x, this is true, thus, infinite solutions.

U)$(-10)(x) = 0$

Thus, x= 0.

V) $-6 = 6$,

  this is false.

  thus, for what ever value of x, this is false, thus, no solution

W)$\frac{5}{6}(x) = -1$

   $x = \frac{-6}{5} = -1.2$

X)$8x -5 = 6x+2+2x = 8x +2$

$-5 = 2$ , which is false.

thus, for what ever value of x, this is false, thus, no solution.

Answer 2

Answer:

ok so lets do this step by step ok?

Step-by-step explanation: