Math help trig. ?....?

Question

Solve 5/3(x+1) is less than or equal to x+1

Answer 1

Is 5/3(x+1)= (5/3) (x+1) or 5/[3(x+1)]?

in second case it solved as follows

5/3 < (x+1)^2

so x+1 < - 5/3 or 5/3 < x+1.

In the first case 5/3 > 1 and so (5/3)(x+1) > (x+1) when x+1 is +ve.

(5/3)(x+1) < (x+1) if x+1 < 0.

Where is trigonometry?

Answer 2

Solve 5/3(x+1) is less than or equal to x+1

whadupdd.. here it is.

5/ [3(x + 1)] ≤ (x + 1)

5 ≤ (3x + 3)(x + 1)

5 ≤ (3x² + 3x + 3x + 3)

5 ≤ 3x² + 6x + 3)

0 ≤ 3x² + 6x - 2

Quadratic

a=3  b=6  c=(-2)

x=[-b ±√b² -4ac]/2a

=[-6 ±√36 -4(3)(-2)]/2(3)

=[-6 ±√60]/6

x= [-6 ± 7.75]/6

x+ =[-6 + 7.75]/6

x+ = 0.29

x-  =[-6 - 7.75]/6

x- = - 2.29

where

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Ans:: x ≥ 0.29   &   x ≥ - 2.29

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hope this helps

Answer 3

x >= rt(5/3) -1

Answer 4

5/3 (x + 1) ≤ x + 1

5/3x + 5/3 ≤ x + 1 Distribute

5/3x -x + 5/3 ≤ x - x + 1 Subtract x from both sides

2/3x + 5/3 ≤ 1 Subtract 5/3 from both sides

2/3x + 5/3 - 5/3 ≤ 1 - 5/3

2/3x ≤ -2/3 Divide both sides by 2/3

x ≤ -1

Not trigonmetry, algebra.