What is 8(t+2)+7(t+2)-3(t-2)<0 ?

The solution to 8(t+2)+7(t+2)-3(t-2)<0 is t<-3

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8(t+2)+7(t+2)-3(t-2)<0

8 (t + 2) + 7 (t + 2) - 3 (t - 2) < 0

t < - 3

Interval Notation

(- \infty, - 3)

Solution steps

8 (t + 2) + 7 (t + 2) - 3 (t - 2) < 0

Add similar elements:

8 (t + 2) + 7 (t + 2) = 15 (t + 2)

15 (t + 2) - 3 (t - 2) < 0

Expand

15 (t + 2) - 3 (t - 2) : 12 t + 36

12 t + 36 < 0

Move

36

to the right side

12 t < - 36

Divide both sides by

12

t < - 3

2 = \frac{7}{2} (5 x + 6)

12 = \frac{2 x - 4}{3}

4 v^{2} + 24 v = 0

\frac{1 + 2 i}{1 - 2 i}

x - \frac{1}{3} \geq - \frac{3}{2}