How do you solve $4 x + 8 \leq 6 x - 20$ $4x +8 <= 6x -20$ ?

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Answer 1 · 2017-11-08T08:42:09

$x \geq 14$ $x >=14$

Pretend that your $\leq$ $<=$ here is just a plain old $=$ $=$ . Balance it like you always would :)

$4 x + 8 \leq 6 x - 20$ $4x+8<=6x-20$ Subtract $6 x$ $6x$ from both sides

$- 2 x + 8 \leq - 20$ $-2x+8<=-20$ Subtract $8$ $8$ from both sides

$- 2 x \leq - 28$ $-2x<=-28$ Divide both sides by $- 2$ $-2$ and do not forget to flip the sign because you're dividing by a negative number!

$x \geq 14$ $x >= 14$

How do you solve 4x+8≤6x−204x +8 <= 6x -20?