## Algebra: A Combined Approach (4th Edition)

$-3(1+2x)+x\geq-(3-x)$ $(-3-6x)+x\geq-3+x$ $-3-5x\geq-3+x$ $-3+3\geq x+5x$ $0\geq6x$ $\frac{0}{6}\geq x$ $0\geq x$
$-3(1+2x)+x\geq-(3-x)$ 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) $\frac{0}{6}\geq x$ $0\geq x$