Question

find the solution set of 1/3x-4≤ 2+x​

Answer 1

1/3x*3-4*3≤ 2*3+x*3

x-12≤6+3x

-12-6≤3x-x

-18/2≤2x/2

-9≤x

OR

1÷3x-4≤ 2+x

x-12≤6+3x

-12-6≤3x-x

-18/2≤2x/2

-9≤x

Answer 2

Answer:

[-9 ≤ x] or [x ≥ -9]

Step-by-step explanation:

QUESTION :-

Find the solution set of →  $\frac{1}{3} x - 4 \leq 2+ x$

SOLUTION :-

• Simplify the L.H.S.

$=> \frac{x-12}{3} \leq x + 2$

• Multiply both the sides by 3

$=> \frac{x - 12}{3} \times 3 \leq 3(x + 2)$

$=> x - 12 \leq 3x +6$

• Substract 'x' from both the sides

$=> x - 12 - x \leq 3x + 6 -x$

$=> -12 \leq 2x + 6$

• Substract 6 from both the sides.

$=> -12 - 6 \leq 2x + 6 - 6$

$=> -18 \leq 2x$

• Divide both the sides by 2.

$=> \frac{-18}{2} \leq \frac{2x}{2}$

$=> -9 \leq x$  or  $x \geq -9$