Given that $x + 5 > 1$ and $1 - 3 x > - 5$ , $x$ cannot take the value given in one of the following options,which is :

- 6

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- 3

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0

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\frac{1}{2}

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Solution

The correct option is B $- 6$
Given, $x + 5 > 1$
$\Rightarrow x > - 4$
Also given, $1 - 3 x > - 5$
$\Rightarrow 3 x < 6$ or $x < 2$
$∴ x \in (- 4, 2)$
Thus it cannot take the value of $- 6$ .

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Similar questions

Q. Solve the following pair of equations.

\frac{5}{x - 1} + \frac{1}{y - 2} = 2

\frac{6}{x - 1} - \frac{3}{y - 2} = 1

Here, x

\neq

1 and y

\neq

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Choose the correct answer from the given options.

Q. Which among the given options is one of the factors of

x^{2} + \frac{x}{6} - \frac{1}{6}

In each of the following, determine whether the given values are solutions (roots) of the equation or not :
(i) $3 x^{2} - 2 x - 1$ = 0; x = 1
(ii) $x^{2} + 6 x + 5$ = 0; x = – 1, x = – 5
(iii) $x^{2} + x - 4 = 0$ ; x = – 2

Q. Find the value of the expression

3 x - 1

for the given values of

x

x = 1, 5, - 6, \frac{3}{7}

Q. Find the value of the expression

3 x - 1

for the given values of

x :

x = 1, 5, - 6, \frac{3}{7}

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Given that x+5>1 and 1−3x>−5 , x cannot take the value given in one of the following options,which is :

The correct option is B −6Given, x+5>1⇒x>−4Also given, 1−3x>−5⇒3x<6 or x<2∴x∈(−4,2)Thus it cannot take the value of −6.

Given that $x + 5 > 1$ and $1 - 3 x > - 5$ , $x$ cannot take the value given in one of the following options,which is :

The correct option is B $- 6$
Given, $x + 5 > 1$
$\Rightarrow x > - 4$
Also given, $1 - 3 x > - 5$
$\Rightarrow 3 x < 6$ or $x < 2$
$∴ x \in (- 4, 2)$
Thus it cannot take the value of $- 6$ .