What is 3/4 (4k-1)>= 4/3 (3k-2) ?

The solution to 3/4 (4k-1)>= 4/3 (3k-2) is k<= 23/12

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3/4 (4k-1)>= 4/3 (3k-2)

\frac{3}{4} (4 k - 1) \geq \frac{4}{3} (3 k - 2)

k \leq \frac{23}{12}

Interval Notation

(- \infty, \frac{23}{12}]

Decimal Notation

k \leq 1.91666 \dots

Solution steps

\frac{3}{4} (4 k - 1) \geq \frac{4}{3} (3 k - 2)

Expand

\frac{3}{4} (4 k - 1) : 3 k - \frac{3}{4}

Expand

\frac{4}{3} (3 k - 2) : 4 k - \frac{8}{3}

3 k - \frac{3}{4} \geq 4 k - \frac{8}{3}

Move

\frac{3}{4}

to the right side

3 k \geq 4 k - \frac{23}{12}

Move

4 k

to the left side

- k \geq - \frac{23}{12}

Multiply both sides by

- 1

k \leq \frac{23}{12}

3 d^{- 4} \cdot 4 d^{4}

- 4 (3 x + 5) > - 116

- \frac{1}{- 10}

0 = x^{2} - 4 x + 3

\frac{10 \cdot 8.2}{17.5}

What is 3/4 (4k-1)>= 4/3 (3k-2) ?
The solution to 3/4 (4k-1)>= 4/3 (3k-2) is k<= 23/12