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10-6x<=2(5-3x)
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10-6x<=10-6x 0=0
havent solutions
\[10-6x \le 2(5-3x)\]divide both sides by 2 \[5-3x \le 5-3x\]add 3x to both sides \[5-3x+3x \le 5\]simplify\[5 \le 5\]Now 5 (on the right side) is always less than or equal to 5 (on the left side) so the solution is all real numbers, or \[(-\infty, \infty)\]
Actually if I had just distributed the 2 on the right hand side first step\[10-6x \le 2(5-3x)\]\[10-6x \le 10-6x\]You could see that the left and right sides are equal.
plexx are you good? cuz i'm going to leave for a while to eat
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btw your previous post 5(x+1) < 5x - 2 has no solution since 5x + 5 < 5x - 2 5 < -2 which is not true, thus no solution
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