SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Linear Inequalities

40 questions · 40 still being checked

EXERCISE 5.1 11–20 (part 2 of 5)

  1. Solve the inequalities in Exercises $\displaystyle 5$ to $\displaystyle 16$ for real \(\displaystyle x\).

    Exercise 11

    3(x−2)5≤5(2−x)3\displaystyle \frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}

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    NCERT’s answer
    $\displaystyle (-\infty, 2]$
    Clear the denominators. Multiply both sides by \(\displaystyle 15\), the LCM of \(\displaystyle 5\) and \(\displaystyle 3\).\[\frac{3(x-2)}{5}\leq\frac{5(2-x)}{3}\] \[9(x-2)\leq 25(2-x)\] \[9x-18\leq 50-25x\] \[9x+25x\leq 50+18\] \[34x\leq 68 \;\Longrightarrow\; x\leq 2\]At \(\displaystyle x=2\) both sides equal \(\displaystyle 0\), so \(\displaystyle 2\) is included.\(\displaystyle x\in(-\infty,\,2]\)

  2. Exercise 12

    12(3x5+4)≥13(x−6)\displaystyle \frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)

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    NCERT’s answer
    $\displaystyle (-\infty, 120]$
    Simplify each side, then clear the denominators.\[\frac{1}{2}\left(\frac{3x}{5}+4\right)=\frac{3x}{10}+2,\qquad \frac{1}{3}(x-6)=\frac{x}{3}-2\]\[\frac{3x}{10}+2\geq\frac{x}{3}-2\]Multiply throughout by \(\displaystyle 30\) (the LCM of \(\displaystyle 10\) and \(\displaystyle 3\), and positive, so the sign stands):\[9x+60\geq 10x-60\] \[60+60\geq 10x-9x\] \[x\leq 120\]\(\displaystyle x\in(-\infty,\,120]\)

  3. Exercise 13

    2(2x+3)−10<6(x−2)\displaystyle 2(2 x+3)-10<6(x-2)

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    NCERT’s answer
    $\displaystyle (4, \infty)$
    Expand, then transpose.\[2(2x+3)-10<6(x-2)\] \[4x+6-10<6x-12\] \[4x-4<6x-12\] \[-4+12<6x-4x\] \[8<2x \;\Longrightarrow\; x>4\]\(\displaystyle x\in(4,\,\infty)\)

  4. Exercise 14

    37−(3x+5)≥9x−8(x−3)\displaystyle 37-(3 x+5) \geq 9 x-8(x-3)

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    NCERT’s answer
    $\displaystyle (-\infty, 2]$
    Expand both sides carefully, watching the signs.\[37-(3x+5)\geq 9x-8(x-3)\] \[37-3x-5\geq 9x-8x+24\] \[32-3x\geq x+24\] \[32-24\geq x+3x\] \[8\geq 4x \;\Longrightarrow\; x\leq 2\]\(\displaystyle x\in(-\infty,\,2]\)

  5. Exercise 15

    x4<(5x−2)3−(7x−3)5\displaystyle \frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}

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    NCERT’s answer
    $\displaystyle (4, \infty)$
    Clear the denominators. The LCM of \(\displaystyle 4,3,5\) is \(\displaystyle 60\); multiplying by a positive number keeps the sign.\[\frac{x}{4}<\frac{5x-2}{3}-\frac{7x-3}{5}\] \[15x<20(5x-2)-12(7x-3)\] \[15x<100x-40-84x+36\] \[15x<16x-4\] \[4<16x-15x \;\Longrightarrow\; x>4\]\(\displaystyle x\in(4,\,\infty)\)

  6. Exercise 16

    (2x−1)3≥(3x−2)4−(2−x)5\displaystyle \frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}

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    NCERT’s answer
    $\displaystyle (-\infty, 2]$
    Clear the denominators. The LCM of \(\displaystyle 3,4,5\) is \(\displaystyle 60\).\[\frac{2x-1}{3}\geq\frac{3x-2}{4}-\frac{2-x}{5}\] \[20(2x-1)\geq 15(3x-2)-12(2-x)\] \[40x-20\geq 45x-30-24+12x\] \[40x-20\geq 57x-54\] \[54-20\geq 57x-40x\] \[34\geq 17x \;\Longrightarrow\; x\leq 2\]\(\displaystyle x\in(-\infty,\,2]\)
  7. Solve the inequalities in Exercises $\displaystyle 17$ to $\displaystyle 20$ and show the graph of the solution in each case on number line

    Exercise 17

    3x−2<2x+1\displaystyle 3 x-2<2 x+1

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    NCERT’s answer
    $\displaystyle (-\infty, 3)$, ![](https://cdn.mathpix.com/cropped/952f948a-c38e-4ca5-a31f-e9bbea10a1ba-07.jpg?height=$\displaystyle 98$&width=$\displaystyle 330$&top_left_y=$\displaystyle 1493$&top_left_x=$\displaystyle 434$)
    Transposition, then graph on the number line.\[3x-2<2x+1\] \[3x-2x<1+2\] \[x<3\]So the solution set is \(\displaystyle (-\infty,\,3)\).On the number line, mark \(\displaystyle 3\) with a hollow (unfilled) circle — the inequality is strict, so \(\displaystyle 3\) itself is not a solution — and shade the entire ray to the left of \(\displaystyle 3\), continuing indefinitely towards \(\displaystyle -\infty\).\(\displaystyle x\in(-\infty,\,3)\)

  8. Exercise 18

    5x−3≥3x−5\displaystyle 5 x-3 \geq 3 x-5

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    NCERT’s answer
    $\displaystyle [-1, \infty)$, ![](https://cdn.mathpix.com/cropped/952f948a-c38e-4ca5-a31f-e9bbea10a1ba-07.jpg?height=$\displaystyle 92$&width=$\displaystyle 336$&top_left_y=$\displaystyle 1495$&top_left_x=$\displaystyle 1036$)
    Transposition, then graph on the number line.\[5x-3\geq 3x-5\] \[5x-3x\geq -5+3\] \[2x\geq -2\] \[x\geq -1\]So the solution set is \(\displaystyle [-1,\,\infty)\).On the number line, mark \(\displaystyle -1\) with a solid (filled) dot — the inequality is \(\displaystyle \geq\), so \(\displaystyle -1\) is a solution (check: \(\displaystyle 5(-1)-3=-8=3(-1)-5\)) — and shade the entire ray to the right of \(\displaystyle -1\), continuing indefinitely towards \(\displaystyle +\infty\).\(\displaystyle x\in[-1,\,\infty)\)

  9. Exercise 19

    3(1−x)<2(x+4)\displaystyle 3(1-x)<2(x+4)

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    NCERT’s answer
    $\displaystyle (-1, \infty)$, ![](https://cdn.mathpix.com/cropped/952f948a-c38e-4ca5-a31f-e9bbea10a1ba-07.jpg?height=$\displaystyle 100$&width=$\displaystyle 334$&top_left_y=$\displaystyle 1637$&top_left_x=$\displaystyle 401$)
    Transposing method. Open the brackets, then collect the \(\displaystyle x\)-terms on one side.\[3(1-x) < 2(x+4)\] \[3 - 3x < 2x + 8\]Add \(\displaystyle 3x\) to both sides and subtract \(\displaystyle 8\) from both sides:\[3 - 8 < 2x + 3x \quad\Longrightarrow\quad -5 < 5x\]Dividing by \(\displaystyle 5\) (a positive number) leaves the sense of the inequality unchanged:\[-1 < x\]Check: \(\displaystyle x=0\) gives \(\displaystyle 3<8\), true; \(\displaystyle x=-1\) gives \(\displaystyle 6<6\), false — so \(\displaystyle -1\) itself is not included.On the number line, mark \(\displaystyle -1\) with a hollow (open) circle and shade the whole ray to its right.Solution set: \(\displaystyle x > -1\), i.e. \(\displaystyle (-1,\ \infty)\).

  10. Exercise 20

    x2≥(5x−2)3−(7x−3)5\displaystyle \frac{x}{2} \geq \frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}

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    NCERT’s answer
    $\displaystyle \left[-\frac{2}{7}, \infty\right)$, ![](https://cdn.mathpix.com/cropped/952f948a-c38e-4ca5-a31f-e9bbea10a1ba-07.jpg?height=$\displaystyle 70$&width=$\displaystyle 334$&top_left_y=$\displaystyle 1657$&top_left_x=$\displaystyle 1038$)
    Clearing the denominators. The LCM of \(\displaystyle 2,\,3,\,5\) is \(\displaystyle 30\); multiplying both sides by \(\displaystyle 30\) (positive) keeps the sense of the inequality.\[\frac{x}{2} \ \geq\ \frac{5x-2}{3} - \frac{7x-3}{5}\] \[15x \ \geq\ 10(5x-2) - 6(7x-3)\] \[15x \ \geq\ 50x - 20 - 42x + 18\] \[15x \ \geq\ 8x - 2\]Subtract \(\displaystyle 8x\) from both sides:\[7x \geq -2 \quad\Longrightarrow\quad x \geq -\frac{2}{7}\]Check the endpoint: at \(\displaystyle x=-\tfrac27\) both sides equal \(\displaystyle -\tfrac17\), so equality holds and \(\displaystyle -\tfrac27\) is included.On the number line, mark \(\displaystyle -\tfrac{2}{7}\) with a solid (filled) dot and shade everything to its right.Solution set: \(\displaystyle x \geq -\dfrac{2}{7}\), i.e. \(\displaystyle \left[-\dfrac{2}{7},\ \infty\right)\).