Answer to Question #347206 in Algebra for Douglas

Question

Solve the inequalities. Give your answer in interval notation, and indicate the answer geometrically on the real-number line.

a) t + 6 ≤ 2 + 3t

b) 3(2 – 3x) > 4(1 – 4x) 

Answer 1

a) $t + 6 ≤ 2 + 3t,$

$t-3t\leq2-6,$ (we have got all the terms with the variable t on the left and the numbers on the right)

$-2t\leq-4,$ (we have simplified the both sides)

$t\geq\frac{-4}{-2},$ (We have divided by the negative number, so we had to change the direction of the inequality)

$t\geq2,$

in interval notation: $t\in[2, \infty),$

the interpretation on the real-number line:

b) $3(2 – 3x) > 4(1 – 4x),$

$6-9x>4-16x,$  (we have removed parentheses)

$-9x+16x>4-6,$ (we have got all the terms with the variable x on the left and the numbers on the right)

$7x>-2,$ (we have simplified the both sides)

$x>-\frac{2}{7},$ (We have divided by the positive number, so we hadn't to change the direction of the inequality)

in interval notation: $x\in(-\frac{2}{7}, \infty),$

the interpretation on the real-number line: