Question

Topic - Inequations
Class - 8
(Refer the attachment for questions)
You have to solve only (iv) and (v)

Good luck!​

Answer 1

$\textsf{\large{\underline{Question 1}:}}$

Given:

$\sf\implies \dfrac{1}{2}x+2 \geq x-1, x \in \mathbb{N}$

Multiplying both sides by 2, we get:

$\sf\implies2\bigg( \dfrac{1}{2}x+2\bigg) \geq 2(x-1)$

$\sf\implies x+4 \geq 2x-2$

$\sf\implies x+4+-x\geq 2x-2-x$

$\sf\implies 4\geq x-2$

$\sf\implies 4+2\geq x-2+2$

$\sf\implies 6\geq x$

$\sf\implies x\leq 6, x\in\mathbb{N}$

Therefore:

$\sf\implies x=\{1,2,3,4,5,6\}\ \ \ (Answer)$

$\textsf{\large{\underline{Question 2}:}}$

Given:

$\sf\implies 4(x-3)<x+3, x\in \mathbb{W}$

$\sf\implies 4x-12<x+3$

$\sf\implies 4x-12-x<x+3-x$

$\sf\implies 3x-12<3$

$\sf\implies 3x-12+12<3+12$

$\sf\implies 3x<15$

$\sf\implies \dfrac{3x}{3}<\dfrac{15}{3}$

$\sf\implies x<5, x\in\mathbb{W}$

Therefore:

$\sf\implies x=\{0,1,2,3,4\}\ \ \ (Answer)$

$\textsf{\large{\underline{Question 3}:}}$

Given:

$\sf\implies 3(2x-1)\geq2(2x+3), x\in\mathbb{I}$

$\sf\implies 6x-3\geq 4x+6$

$\sf\implies 6x-3-4x\geq 4x+6-4x$

$\sf\implies 2x-3\geq 6$

$\sf\implies 2x-3+3\geq 6+3$

$\sf\implies 2x\geq 9$

$\sf\implies \dfrac{2x}{2}\geq \dfrac{9}{2}$

$\sf\implies x\geq 4\dfrac{1}{2}, x\in\mathbb{I}$

Therefore:

$\sf\implies x=\{5,6,7,8,...\infty\}\ \ \ (Answer)$

$\textsf{\large{\underline{Question 4}:}}$

Given:

$\sf\implies 2(4-3x)\leq 4(x-5), x\in\mathbb{W}$

$\sf\implies 8-6x\leq 4x-20$

$\sf\implies 8-6x+6x\leq 4x-20+6x$

$\sf\implies 8\leq 10x-20$

$\sf\implies 8+20\leq 10x-20+20$

$\sf\implies 28\leq 10x$

$\sf\implies 10x\geq 28$

$\sf\implies \dfrac{10x}{10}\geq \dfrac{28}{10}$

$\sf\implies x\geq 2\dfrac{4}{5},x \in\mathbb{W}$

Therefore:

$\sf\implies x=\{3,4,5,6,...\infty\}\ \ \ (Answer)$

$\textsf{\large{\underline{Question 5}:}}$

Given:

$\sf\implies 2(4x-1)<4(3x+1), x\in \{Set\ of\ negative\ integers\}$

$\sf\implies 8x-2<12x+4$

$\sf\implies 8x-2-4<12x+4-4$

$\sf\implies 8x-6<12x$

$\sf\implies 8x-6-8x<12x-8x$

$\sf\implies -6<4x$

$\sf\implies 4x>-6$

$\sf\implies \dfrac{4x}{4}>\dfrac{-6}{4}$

$\sf\implies x >-1\dfrac{1}{2}$

Therefore:

$\sf\implies x=\{-1\}\ \ \ (Answer)$

Answer 2

Answer:

\textsf{\large{\underline{Question 1}:}}

Question 1

:

Given:

\sf\implies \dfrac{1}{2}x+2 \geq x-1, x \in \mathbb{N}⟹

2

1

x+2≥x−1,x∈N

Multiplying both sides by 2, we get:

\sf\implies2\bigg( \dfrac{1}{2}x+2\bigg) \geq 2(x-1)⟹2(

2

1

x+2)≥2(x−1)

\sf\implies x+4 \geq 2x-2⟹x+4≥2x−2

\sf\implies x+4+-x\geq 2x-2-x⟹x+4+−x≥2x−2−x

\sf\implies 4\geq x-2⟹4≥x−2

\sf\implies 4+2\geq x-2+2⟹4+2≥x−2+2

\sf\implies 6\geq x⟹6≥x

\sf\implies x\leq 6, x\in\mathbb{N}⟹x≤6,x∈N

Therefore:

\sf\implies x=\{1,2,3,4,5,6\}\ \ \ (Answer)⟹x={1,2,3,4,5,6} (Answer)

\textsf{\large{\underline{Question 2}:}}

Question 2

:

Given:

\sf\implies 4(x-3) < x+3, x\in \mathbb{W}⟹4(x−3)<x+3,x∈W

\sf\implies 4x-12 < x+3⟹4x−12<x+3

\sf\implies 4x-12-x < x+3-x⟹4x−12−x<x+3−x

\sf\implies 3x-12 < 3⟹3x−12<3

\sf\implies 3x-12+12 < 3+12⟹3x−12+12<3+12

\sf\implies 3x < 15⟹3x<15

\sf\implies \dfrac{3x}{3} < \dfrac{15}{3}⟹

3

3x

<

3

15

\sf\implies x < 5, x\in\mathbb{W}⟹x<5,x∈W

Therefore:

\sf\implies x=\{0,1,2,3,4\}\ \ \ (Answer)⟹x={0,1,2,3,4} (Answer)

\textsf{\large{\underline{Question 3}:}}

Question 3

:

Given:

\sf\implies 3(2x-1)\geq2(2x+3), x\in\mathbb{I}⟹3(2x−1)≥2(2x+3),x∈I

\sf\implies 6x-3\geq 4x+6⟹6x−3≥4x+6

\sf\implies 6x-3-4x\geq 4x+6-4x⟹6x−3−4x≥4x+6−4x

\sf\implies 2x-3\geq 6⟹2x−3≥6

\sf\implies 2x-3+3\geq 6+3⟹2x−3+3≥6+3

\sf\implies 2x\geq 9⟹2x≥9

\sf\implies \dfrac{2x}{2}\geq \dfrac{9}{2}⟹

2

2x

≥

2

9

\sf\implies x\geq 4\dfrac{1}{2}, x\in\mathbb{I}⟹x≥4

2

1

,x∈I

Therefore:

\sf\implies x=\{5,6,7,8,...\infty\}\ \ \ (Answer)⟹x={5,6,7,8,...∞} (Answer)

\textsf{\large{\underline{Question 4}:}}

Question 4

:

Given:

\sf\implies 2(4-3x)\leq 4(x-5), x\in\mathbb{W}⟹2(4−3x)≤4(x−5),x∈W

\sf\implies 8-6x\leq 4x-20⟹8−6x≤4x−20

\sf\implies 8-6x+6x\leq 4x-20+6x⟹8−6x+6x≤4x−20+6x

\sf\implies 8\leq 10x-20⟹8≤10x−20

\sf\implies 8+20\leq 10x-20+20⟹8+20≤10x−20+20

\sf\implies 28\leq 10x⟹28≤10x

\sf\implies 10x\geq 28⟹10x≥28

\sf\implies \dfrac{10x}{10}\geq \dfrac{28}{10}⟹

10

10x

≥

10

28

\sf\implies x\geq 2\dfrac{4}{5},x \in\mathbb{W}⟹x≥2

5

4

,x∈W

Therefore:

\sf\implies x=\{3,4,5,6,...\infty\}\ \ \ (Answer)⟹x={3,4,5,6,...∞} (Answer)

\textsf{\large{\underline{Question 5}:}}

Question 5

:

Given:

\sf\implies 2(4x-1) < 4(3x+1), x\in \{Set\ of\ negative\ integers\}⟹2(4x−1)<4(3x+1),x∈{Set of negative integers}

\sf\implies 8x-2 < 12x+4⟹8x−2<12x+4

\sf\implies 8x-2-4 < 12x+4-4⟹8x−2−4<12x+4−4

\sf\implies 8x-6 < 12x⟹8x−6<12x

\sf\implies 8x-6-8x < 12x-8x⟹8x−6−8x<12x−8x

\sf\implies -6 < 4x⟹−6<4x

\sf\implies 4x > -6⟹4x>−6

\sf\implies \dfrac{4x}{4} > \dfrac{-6}{4}⟹

4

4x

>

4

−6

\sf\implies x > -1\dfrac{1}{2}⟹x>−1

2

1

Therefore:

\sf\implies x=\{-1\}\ \ \ (Answer)⟹x={−1} (Answer)