What is (5x-4)+(9x-68)>= x-1 ?

The solution to (5x-4)+(9x-68)>= x-1 is x>= 71/13

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(5x-4)+(9x-68)>= x-1

(5 x - 4) + (9 x - 68) \geq x - 1

x \geq \frac{71}{13}

Interval Notation

[\frac{71}{13}, \infty)

Decimal Notation

x \geq 5.46153 \dots

Solution steps

(5 x - 4) + (9 x - 68) \geq x - 1

Remove parentheses:

(a) = a

5 x - 4 + 9 x - 68 \geq x - 1

Group like terms

5 x + 9 x - 4 - 68 \geq x - 1

Add similar elements:

5 x + 9 x = 14 x

14 x - 4 - 68 \geq x - 1

Subtract the numbers:

- 4 - 68 = - 72

14 x - 72 \geq x - 1

Move

72

to the right side

14 x \geq x + 71

Move

x

to the left side

13 x \geq 71

Divide both sides by

13

x \geq \frac{71}{13}

3 (x - 7) = - 21 + 3 x

y^{2} + 9 y + 14

\frac{( x + 11 ) ( x - 13 )}{x + 9} \leq 0

2 x + 2 > 6

4 - 3 (5 - 2 x) = 5 (2 - 3 x)

What is (5x-4)+(9x-68)>= x-1 ?
The solution to (5x-4)+(9x-68)>= x-1 is x>= 71/13