How do you solve $2 (t + 3) < 3 (t + 2)$ $2( t + 3) < 3( t + 2)$ ?

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Answer 1 · 2018-05-14T09:40:26

$t > 0$ $t > 0$ graph{2 ( x + 3 ) < 3 ( x + 2 ) [-10, 10, -5, 5]}

$2 t + 6 < 3 t + 6$ $2t+6 < 3t + 6$
$- t < 0$ $-t < 0$
$t > 0$ $t > 0$

Answer 2 · 2018-05-14T09:40:55

$t > 0$ $t>0$

First lets simplify the inequality:

$2 (t + 3) < 3 (t + 2)$ $2(t+3) < 3(t+2)$
$2 t + 6 < 3 t + 6$ $2t + 6 < 3t + 6$ -----> opening the brackets

$2 t - 3 t < 6 - 6$ $2t-3t < 6-6$

$- t < 0$ $-t<0$

When multiplying or dividing by a negative number, the inequality sign changes, from $<$ $<$ to $>$ $>$ .

Hence we get:

$t > 0$ $t>0$

Check the answer:
Take $t = 2$ $t = 2$

$2 (2 + 3) < 3 (2 + 2)$ $2(2+3) < 3(2+2)$ #

$(2 \times 5) < (3 \times 4)$ $(2xx5) < (3 xx 4)$

$10 < 12$ $10 < 12$

How do you solve 2(t+3)<3(t+2)2( t + 3) < 3( t + 2)?