How do I solve 2x + 3_> 1?

2x + 3 ≧ 1
2x ≧ 1 - 3
2x ≧ -2
x ≧ -2/2
x ≧ -1

2x + 3 > 1
2x > - 2
x > - 1

Answer: x > - 1

2x + 3 ≥ 1
2x ≥ - 2
x ≥ - 1

2x + 3 -> 1

2x -> 1 - 3

2x -> - 2

x -> - 1

You should solve it in the same way as an equation.

Subtract 3 from each side so it's
2x_>1-3
Which is
2x_>-2
Then divide both sides by 2 so it's
x_>-1
Then it's solved.

2x _> 1-3
2x _> -2
x_> -1

2x+3=1
-3 -3

2x = -2

divide each side by 2

x > -1

SImple if x_>1 then
-1_>x_>1
therefore -1_>2x+3_>1
~ -1-3_>2x_>1-3
~ -4/2_>x_>-2/2
~ -2_>x_>-1
Therefore x value lies between -2 and -1or may be equal to -2 or -1

Do you propose {(2x + a million) / (x + 3)} > 0? this is, the numerator is 2x+a million and the denominator is x+3, then the fraction is larger than 0. If confident, then the perfect answer ought to be (-infinity, -3) U (-a million/2, +infinity).

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