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How do you solve $\frac{1}{3} x + 2 > 3$ $1/3x + 2 >3$ and $- 3 x > 12$ $-3x > 12$ ?

1 Answer

Sep 19, 2016

$3 < x < 4$ $3 < x <4$

Explanation:

Treat inequalities the same as equations, except if you multiply or divide by a negative value, in which case the inequality sign changes around.

$\frac{1}{3} x + 2 > 3 \times \times \times \times \times \times \times - 3 x > 12$ $1/3x+2 >3color(white)(xxxxxxxxxxxxxx)-3x >12$

$\frac{1}{3} x > 1 \leftarrow \times 3 \times \times \times \times \times x 12 > 3 x \leftarrow \div 3$ $1/3x > 1" "larr xx 3color(white)(xxxxxxxxxxx) 12>3x" "larr div3$

$x > 3 \times \times \times \times \times \times \times \times \times \times \times 4 > x$ $x > 3color(white)(xxxxxxxxxxxxxxxxxxxxxx)4 >x$

Combine the two inequalities,

$3 < x and x < 4$ $3 < x and x < 4$

$:. 3 < x <4$

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