What is 1/5 (3g+4)-1/3 (g+1)>= 1-1/3 (g+5) ?

The solution to 1/5 (3g+4)-1/3 (g+1)>= 1-1/3 (g+5) is g>=-17/9

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1/5 (3g+4)-1/3 (g+1)>= 1-1/3 (g+5)

\frac{1}{5} (3 g + 4) - \frac{1}{3} (g + 1) \geq 1 - \frac{1}{3} (g + 5)

g \geq - \frac{17}{9}

Interval Notation

[- \frac{17}{9}, \infty)

Decimal Notation

g \geq - 1.88888 \dots

Solution steps

\frac{1}{5} (3 g + 4) - \frac{1}{3} (g + 1) \geq 1 - \frac{1}{3} (g + 5)

Expand

\frac{1}{5} (3 g + 4) - \frac{1}{3} (g + 1) : \frac{4 g}{15} + \frac{7}{15}

Expand

1 - \frac{1}{3} (g + 5) : - \frac{2}{3} - \frac{1}{3} g

\frac{4 g}{15} + \frac{7}{15} \geq - \frac{2}{3} - \frac{1}{3} g

Move

\frac{7}{15}

to the right side

\frac{4 g}{15} \geq - \frac{17}{15} - \frac{1}{3} g

Move

\frac{1}{3} g

to the left side

\frac{3 g}{5} \geq - \frac{17}{15}

Multiply both sides by

5

3 g \geq - \frac{17}{3}

Divide both sides by

3

g \geq - \frac{17}{9}

35 \cdot 5

\frac{n ^{2} + 7 n}{4 n ^{2} + 28 n}

∣ x + 7 ∣ \leq 7

lo g_{6} (6)

\frac{x}{1 - x} - \frac{y}{1 - y}

What is 1/5 (3g+4)-1/3 (g+1)>= 1-1/3 (g+5) ?
The solution to 1/5 (3g+4)-1/3 (g+1)>= 1-1/3 (g+5) is g>=-17/9