What are the real values of x that satisfy the inequations $6 x + 9 < 3 x + 5$ and $4 x + 7 > 2 x - 5$ ?

(- 6, \frac{- 4}{3})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(\frac{- 2}{3}, 5)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(\frac{7}{3}, 10)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 5, \frac{2}{3})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(- 6, \frac{- 4}{3})$
Given, $6 x + 9 < 3 x + 5$
$=> 3 x < - 4$
$=> x < - \frac{4}{3}$ --- (1)
And
$4 x + 7 > 2 x - 5$
$=> 2 x > - 12$
$=> x > - 6$ --- (2)
From, solutions $1, 2$ , we get $(- 6, - \frac{4}{3})$

Suggest Corrections

Similar questions

Q. The interval(s) of

x

which satisfy the inequality

\frac{(1 - 3 x)^{7} (1 + 5 x)^{4} (x + 5)^{6}}{(x^{2} - 1) (x - 6)^{3} (1 - 2 x)^{4}} > 0

is/are

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Q. Solve the system of inequalities:

(x + 5) - 7 (x - 2) \geq 4 x + 9, 2 (x - 3) - 7 (x + 5) \leq 3 x - 9

Q. What are the possible values of x, satisfying the equation

| 4 x + 6 | + 2 = 2 x + 3

Q. Find the sum of

6 x^{3} - 7 x^{2} 2 x + 4

and

9 x^{3} - 5 x^{2} - 4 x + 6

. Subtract it from the sum of

3 x^{3} + 2 x^{2} - 9

and

4 x^{3} + 7

Join BYJU'S Learning Program

What are the real values of x that satisfy the inequations 6x+9<3x+5 and 4x+7>2x−5?

The correct option is A (−6,−43)Given, 6x+9<3x+5=>3x<−4=>x<−43 --- (1)And 4x+7>2x−5=>2x>−12=>x>−6 --- (2)From, solutions 1,2, we get (−6,−43)

What are the real values of x that satisfy the inequations $6 x + 9 < 3 x + 5$ and $4 x + 7 > 2 x - 5$ ?

The correct option is A $(- 6, \frac{- 4}{3})$
Given, $6 x + 9 < 3 x + 5$
$=> 3 x < - 4$
$=> x < - \frac{4}{3}$ --- (1)
And
$4 x + 7 > 2 x - 5$
$=> 2 x > - 12$
$=> x > - 6$ --- (2)
From, solutions $1, 2$ , we get $(- 6, - \frac{4}{3})$