-4x+3>= 7

Solution

- 4 x + 3 \geq 7

x \leq - 1

Interval Notation

(- \infty, - 1]

Solution steps

- 4 x + 3 \geq 7

Move

3

to the right side

- 4 x \geq 4

Multiply both sides by

- 1

4 x \leq - 4

Divide both sides by

4

x \leq - 1

\frac{x ^{2} ( x + 5 ) ^{3}}{( 3 - x ) ( x + 3 ) ^{2}} \leq 0

\frac{( 2 m ^{3} n ^{2} ) ^{3}}{- 18 m ^{5} n ^{4}}

p^{2} q^{2} + pq - q^{3} - p^{3}

4 (2 - 4 x) - 3 x = 65

- 6 \leq 4 x - 2 < - 2