Question

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Answer 1

Question :

Solution set of inequality

$\sf\dfrac{15-2x}{3}\leqslant\dfrac{x}{6}-5$

Theory :

Solution of linear equations :

Same number may be added to (or subtracted from ) both sides of an inequation without changing the sign of inequality.

The sign of inequality is reversed when both sides of an inequation are multipled or divided by a negative number .

Solution :

We have ,

$\sf\dfrac{15-2x}{3}\leqslant\dfrac{x}{6}-5$

$\sf\dfrac{15-2x}{3}\leqslant\dfrac{x-30}{6}$

Now multiply both sides by 3×6 , then

$\sf\:(3\times6)\dfrac{15-2x}{3}\leqslant \dfrac{x-30}{6}(3\times6)$

$\sf\:6(15-2x)\leqslant\:3(x-30)$

$\sf\:90-12x\:\leqslant\:3x-90$

$\sf\:180\leqslant\:15x$

$\sf\:12\leqslant\:x$

Correct answer D)

It is the required solution !

Answer 2

Answer:

Question :

Solution set of inequality

\sf\dfrac{15-2x}{3}\leqslant\dfrac{x}{6}-5

3

15−2x

⩽

6

x

−5

Theory :

Solution of linear equations :

Same number may be added to (or subtracted from ) both sides of an inequation without changing the sign of inequality.

The sign of inequality is reversed when both sides of an inequation are multipled or divided by a negative number .

Solution :

We have ,

\sf\dfrac{15-2x}{3}\leqslant\dfrac{x}{6}-5

3

15−2x

⩽

6

x

−5

\sf\dfrac{15-2x}{3}\leqslant\dfrac{x-30}{6}

3

15−2x

⩽

6

x−30

Now multiply both sides by 3×6 , then

\sf\:(3\times6)\dfrac{15-2x}{3}\leqslant \dfrac{x-30}{6}(3\times6)(3×6)

3

15−2x

⩽

6

x−30

(3×6)

\sf\:6(15-2x)\leqslant\:3(x-30)6(15−2x)⩽3(x−30)

\sf\:90-12x\:\leqslant\:3x-9090−12x⩽3x−90

\sf\:180\leqslant\:15x180⩽15x

\sf\:12\leqslant\:x12⩽x

Correct answer D)