Answer
$-3(1+2x)+x\geq-(3-x)$
$(-3-6x)+x\geq-3+x$
$-3-5x\geq-3+x$
$-3+3\geq x+5x$
$0\geq6x$
$\frac{0}{6}\geq x$
$0\geq x$
Work Step by Step
$-3(1+2x)+x\geq-(3-x)$
Expanding both the brackets by multiplying the outer term with the inner term, we get:
$(-3-6x)+x\geq-3+x$
Simplifying, we get:
$-3-5x\geq-3+x$
Using the addition/subtraction property of inequalities, we add 5x on both sides and add 3 on both sides.
$-3+3\geq x+5x$
Simplifying, we get:
$0\geq6x$
Dividing both sides by a positive number (6) means we don't have to change signs. (Multiplication/Division rule of inequalities)
$\frac{0}{6}\geq x$
$0\geq x$