SVN CLASS 8 MATHS CHAPTER 2 EXERCISE 2(B) SANAKRDEV BIDYA NIKETAN শংকৰদেৱ শিশু নিকেটন অনুশীলনী – 2(B)
শংকৰদেৱ বিদ্যা নিকেতনৰ অষ্টম শ্ৰেণীৰ গণিত পাঠ্যক্ৰমৰ অনুশীলনী 2(B) ৰ সম্পূৰ্ণ সমাধান এই পৃষ্ঠাত আগবঢ়োৱা হৈছে। এই অধ্যায়ত বীজগণিতীয় ৰাশিৰ যোগ-বিয়োগ, সমীকৰণ সমাধান, আৰু প্ৰায়োগিক সমস্যাৰ সমাধানৰ ওপৰত গুৰুত্ব দিয়া হৈছে।
EXERCISE 2(B)
1.তলৰ প্ৰতিটো সমীকৰণৰ সমাধান নিৰ্ণয় কৰা———
i)\( \frac{5x-7}{3x} = 2 \)
Solution:
\( \Rightarrow 5x – 7 = 2 \times 3x \)
\( \Rightarrow 5x – 7 = 6x \)
\( \Rightarrow 5x – 6x = 7 \)
\( \Rightarrow -x = 7 \)
\( \Rightarrow x = -7 \)
ii)\( \frac{4x+5}{3+2x} = \frac{3}{2} \)
Solution:
\( \Rightarrow 2(4x + 5) = 3(3 + 2x) \)
\( \Rightarrow 8x + 10 = 9 + 6x \)
\( \Rightarrow 8x – 6x = 9 – 10 \)
\( \Rightarrow 2x = -1 \)
\( \Rightarrow x = -\frac{1}{2} \)
iii)\( \frac{4x-3}{2x-5} = \frac{3}{4} \)
Solution:-
\( \Rightarrow 4(4x – 3) = 3(2x – 5) \)
\( \Rightarrow 16x – 12 = 6x – 15 \)
\( \Rightarrow 16x – 6x = -15 + 12 \)
\( \Rightarrow 10x = -3 \)
\( \Rightarrow x = \frac{-3}{10} \)
vi)\(\frac{3(4x-5)-5(x+2)}{2(3-5x)-4(5x-2)} = \frac{2}{3} \)
Solution:-
\( \Rightarrow 3(12x – 15 – 5x – 10) = 2(6 – 10x – 20x + 8) \)
\( \Rightarrow 3(7x – 25) = 2(-30x + 14) \)
\( \Rightarrow 21x – 75 = -60x + 28 \)
\( \Rightarrow 21x + 60x = 28 + 75 \)
\( \Rightarrow 81x = 103 \)
\( \Rightarrow x = \frac{103}{81} \)
vii)\(\frac{17(2-x)-5(x+12)}{1-7x} = 8 \)
Solution:
\( \Rightarrow 17(2-x) – 5(x+12) = 8(1-7x) \)
\( \Rightarrow 34 – 17x – 5x – 60 = 8 – 56x \)
\( \Rightarrow -22x – 26 = 8 – 56x \)
\( \Rightarrow -22x + 56x = 8 + 26 \)
\( \Rightarrow 34x = 34 \)
\( \Rightarrow x = \frac{34}{34} \)
\( \Rightarrow x = 1 \)
viii)\( \frac{3p-6}{4} – \frac{2p+1}{3} = \frac{10}{4} \)
Solution:
\( \Rightarrow \frac{3(3p-6)-4(2p+1)}{12} = \frac{10}{4} \)
\( \Rightarrow \frac{9p-18-8p-4}{12} = \frac{5}{2} \)
\( \Rightarrow \frac{p-22}{12} = \frac{5}{2} \)
\( \Rightarrow 2(p-22) = 5 \times 12 \)
\( \Rightarrow 2p – 44 = 60 \)
\( \Rightarrow 2p = 60 + 44 \)
\( \Rightarrow 2p = 104 \)
\( \Rightarrow p = \frac{104}{2} \)
\( \Rightarrow p = 52 \)
ix) \( \frac{2(u-3)}{4} – \frac{2(u+4)}{3} = 1 + \frac{3u-1}{2} \)
Solution:
\( \Rightarrow \frac{6(u-3)-8(u+4)}{12} = \frac{2+3u-1}{2} \)
\( \Rightarrow \frac{6u-18-8u-32}{12} = \frac{3u+1}{2} \)
\( \Rightarrow \frac{-2u-50}{12} = \frac{3u+1}{2} \)
\( \Rightarrow 2(-2u-50) = 12(3u+1) \)
\( \Rightarrow -4u – 100 = 36u + 12 \)
\( \Rightarrow -4u – 36u = 12 + 100 \)
\( \Rightarrow -40u = 112 \)
\( \Rightarrow u = \frac{112}{-40} \)
\( \Rightarrow u = -\frac{14}{5} \)
x)\(\frac{x-4}{2} + \frac{3x-22}{5} = \frac{2x-5}{3} – \frac{x-3}{2} \)
Solution:-
\( \Rightarrow \frac{5(x-4)+2(3x-22)}{10} = \frac{2(2x-5)-3(x-3)}{6} \)
\( \Rightarrow \frac{5x-20+6x-44}{10} = \frac{4x-10-3x+9}{6} \)
\( \Rightarrow \frac{11x-64}{10} = \frac{x-1}{6} \)
\( \Rightarrow 6(11x-64) = 10(x-1) \)
\( \Rightarrow 66x-384 = 10x-10 \)
\( \Rightarrow 66x-10x = -10+384 \)
\( \Rightarrow 56x = 374 \)
\( \Rightarrow x = \frac{374}{56} \)
\( \Rightarrow x = \frac{187}{28} \)
xi)\(\frac{2x-3}{5} + \frac{x+3}{4} = x + \frac{1-3x}{7} \)
Solution:-
\( \Rightarrow \frac{4(2x-3)+5(x+3)}{20} = \frac{7x+(1-3x)}{7} \)
\( \Rightarrow \frac{8x-12+5x+15}{20} = \frac{7x+1-3x}{7} \)
\( \Rightarrow \frac{13x+3}{20} = \frac{4x+1}{7} \)
\( \Rightarrow 7(13x+3) = 20(4x+1) \)
\( \Rightarrow 91x+21 = 80x+20 \)
\( \Rightarrow 91x-80x = 20-21 \)
\( \Rightarrow 11x = -1 \)
\( \Rightarrow x = -\frac{1}{11} \)
xii)\( \frac{3x+1}{2} = \frac{3x-1}{4} + \frac{3(x-1)}{2} \)
Solution:
\( \Rightarrow \frac{3x+1}{2} = \frac{2(3x-1)+12(x-1)}{8} \)
\( \Rightarrow \frac{3x+1}{2} = \frac{6x-2+12x-12}{8} \)
\( \Rightarrow \frac{3x+1}{2} = \frac{18x-14}{8} \)
\( \Rightarrow 8(3x+1) = 2(18x-14) \)
\( \Rightarrow 24x+8 = 36x-28 \)
\( \Rightarrow 24x-36x = -28-8 \)
\( \Rightarrow -12x = -36 \)
\( \Rightarrow x = \frac{-36}{-12} \)
\( \Rightarrow x = 3 \)
2. তলত দিয়া সমীকৰণবিলাকৰ মূল নিৰ্ণয় কৰা আৰু ইয়াৰ ফলশুদ্ধি(সত্যৰাসত্য) পৰীক্ষা কৰা—
(i)\(\frac{x-7}{x+3} – 1 = \frac{1}{3} + \frac{2(x-2)}{x+3} \)
Solution:
\( \Rightarrow \frac{(x-7)-(x+3)}{x+3} = \frac{(x+3)+6(x-2)}{3(x+3)} \)
\( \Rightarrow \frac{x-7-x-3}{x+3} = \frac{x+3+6x-12}{3(x+3)} \)
\( \Rightarrow \frac{-10}{x+3} = \frac{7x-9}{3(x+3)} \)
\( \Rightarrow -10 \times 3 = 7x – 9 \)
\( \Rightarrow -30 = 7x – 9 \)
\( \Rightarrow -30 + 9 = 7x \)
\( \Rightarrow -21 = 7x \)
\( \Rightarrow x = -\frac{21}{7} \)
\( \Rightarrow x = -3 \)
সত্যাসত্য,
LHS,\( \frac{x-7}{x+3} – 1 = \frac{-3-7}{-3+3} – 1 \)
\( = \frac{-10}{0} – 1 = -1\)
RHS:
\( \frac{1}{3} + \frac{2(x-2)}{x+3} = \frac{1}{3} + \frac{2(-3-2)}{-3+3} \)
\( = \frac{1}{3} + \frac{-10}{0} =\frac{1}{3}\)
\(\therefore LHS\neq RHS\)
b)\( \frac{x+1}{8} + \frac{x-2}{5} = \frac{x+3}{10} + \frac{3x-1}{20} \)
Solution:
\( \Rightarrow \frac{5(x+1)+8(x-2)}{40} = \frac{2(x+3)+(3x-1)}{20} \)
\( \Rightarrow \frac{5x+5+8x-16}{40} = \frac{2x+6+3x-1}{20} \)
\( \Rightarrow \frac{13x-11}{40} = \frac{5x+5}{20} \)
\( \Rightarrow 20(13x – 11) = 40(5x + 5) \)
\( \Rightarrow 260x – 220 = 200x + 200 \)
\( \Rightarrow 260x – 200x = 200 + 220 \)
\( \Rightarrow 60x = 420 \)
\( \Rightarrow x = \frac{420}{60} = 7 \)
সত্যাসত্য
LHS:
\( \frac{7+1}{8} + \frac{7-2}{5} \)
\( = \frac{8}{8} + \frac{5}{5} \)
\( = 1 + 1 \)
\( = 2 \)
RHS:
\( \frac{7+3}{10} + \frac{3(7)-1}{20} \)
\( = \frac{10}{10} + \frac{20}{20} \)
\( = 1 + 1 \)
\( = 2 \)
\( \text{LHS} = \text{RHS} = 2 \)
\(\therefore x = 7 \)
C)\(\frac{3}{5}(Y-4) + 2 = \frac{1}{4}(Y-1) + \frac{1}{3}(2Y-9) \)
Solution:
\( \Rightarrow \frac{3Y-12}{5} + 2 = \frac{Y-1}{4} + \frac{2Y-9}{3} \)
\( \Rightarrow \frac{3Y-12}{5} + \frac{10}{5} = \frac{3(Y-1)+4(2Y-9)}{12} \)
\( \Rightarrow \frac{3Y-2}{5} = \frac{3Y-3+8Y-36}{12} \)
\( \Rightarrow \frac{3Y-2}{5} = \frac{11Y-39}{12} \)
\( \Rightarrow 12(3Y-2) = 5(11Y-39) \)
\( \Rightarrow 36Y-24 = 55Y-195 \)
\( \Rightarrow -19Y = -171 \)
\( \Rightarrow Y = \frac{171}{19} \)
\( \Rightarrow Y = 9 \)
সত্যাসত্য,
LHS,
\(\frac{3}{5}(9-4) + 2 \)
\( = \frac{3}{5} \times 5 + 2 \)
\( = 3 + 2 \)
\( = 5 \)
RHS,
\(\frac{1}{4}(9-1) + \frac{1}{3}(2 \times 9 – 9) \)
\( = \frac{8}{4} + \frac{9}{3} \)
\( = 2 + 3 \)
\( = 5 \)
\(\text{LHS} = \text{RHS} = 5\)
\(\therefore Y = 9 \) is correct.
d)\(\frac{3x+5}{18} + \frac{6x-2}{9} = \frac{1}{3} \)
Solution:
\( \Rightarrow \frac{3x+5}{18} + \frac{2(6x-2)}{18} = \frac{1}{3} \)
\( \Rightarrow \frac{3x+5+12x-4}{18} = \frac{1}{3} \)
\( \Rightarrow \frac{15x+1}{18} = \frac{1}{3} \)
\( \Rightarrow 3(15x+1) = 18 \)
\( \Rightarrow 45x+3 = 18 \)
\( \Rightarrow 45x = 15 \)
\( \Rightarrow x = \frac{15}{45} \)
\( \Rightarrow x = \frac{1}{3} \)
সত্যাসত্য,
LHS:
\( \frac{3(\frac{1}{3})+5}{18} + \frac{6(\frac{1}{3})-2}{9} \)
\( = \frac{1+5}{18} + \frac{2-2}{9} \)
\( = \frac{6}{18} + 0 \)
\( = \frac{1}{3} \)
RHS:
\( \frac{1}{3} \)
\(\text{LHS} = \text{RHS} = \frac{1}{3}\)
\(\therefore x = \frac{1}{3} \) is correct.
e)\( (x-1) + \frac{x-12}{6} = \frac{5x}{6} – \frac{2x-3}{9} \)
Solution:
\( \Rightarrow \frac{6(x-1)+(x-12)}{6} = \frac{15x-2(2x-3)}{18} \)
\( \Rightarrow \frac{6x-6+x-12}{6} = \frac{15x-4x+6}{18} \)
\( \Rightarrow \frac{7x-18}{6} = \frac{11x+6}{18} \)
\( \Rightarrow 18(7x-18) = 6(11x+6) \)
\( \Rightarrow 126x – 324 = 66x + 36 \)
\( \Rightarrow 126x – 66x = 36 + 324 \)
\( \Rightarrow 60x = 360 \)
\( \Rightarrow x = \frac{360}{60} \)
\( \Rightarrow x = 6 \)
সত্যাসত্য
LHS:
\( (6-1) + \frac{6-12}{6} \)
\( = 5 + \frac{-6}{6} \)
\( = 5 – 1 \)
\( = 4 \)
RHS:
\( \frac{5 \times 6}{6} – \frac{2 \times 6 – 3}{9} \)
\( = \frac{30}{6} – \frac{9}{9} \)
\( = 5 – 1 \)
\( = 4 \)
\(\text{LHS} = \text{RHS} = 4\)
\(\therefore x = 6 \) is correct.
f)\( 3(7x-1) – \left(2x + \frac{x-1}{2}\right) = \frac{19}{3} \)
Solution:
\( \Rightarrow \frac{21x-3}{4} – \left(\frac{4x + x – 1}{2}\right) = \frac{19}{3} \)
\( \Rightarrow \frac{21x-3}{4} – \frac{5x-1}{2} = \frac{19}{3} \)
\( \Rightarrow \frac{21x-3 – 2(5x-1)}{4} = \frac{19}{3} \)
\( \Rightarrow \frac{21x-3-10x+2}{4} = \frac{19}{3} \)
\( \Rightarrow \frac{11x-1}{4} = \frac{19}{3} \)
\( \Rightarrow 3(11x-1) = 4 \times 19 \)
\( \Rightarrow 33x – 3 = 76 \)
\( \Rightarrow 33x = 79 \)
\( \Rightarrow x = \frac{79}{33} \)
সত্যাসত্য,
\( \frac{3(7 \times \frac{79}{33} – 1)}{4} – \left(2 \times \frac{79}{33} + \frac{\frac{79}{33}-1}{2}\right) \)
\( = \frac{3(\frac{553}{33}-1)}{4} – \left(\frac{158}{33} + \frac{\frac{46}{33}}{2}\right) \)
\( = \frac{\frac{520}{33}}{4} – \left(\frac{158}{33} + \frac{23}{33}\right) \)
\( = \frac{520}{132} – \frac{181}{33} \)
\( = \frac{130}{33} – \frac{181}{33} \)
\( = \frac{-51}{33} \)
\( = -\frac{17}{11} \neq \frac{19}{3} \)
g)\((x – 3) + \frac{5}{6} = 3x + 2 \left( \frac{3x}{4} – \frac{2}{3} \right) \)
Solution:
\( \Rightarrow \frac{6(x-3)+5}{6} = 3x + 2 \left( \frac{9x-8}{12} \right) \)
\( \Rightarrow \frac{6x-18+5}{6} = 3x + \frac{18x-16}{12} \)
\( \Rightarrow \frac{6x-13}{6} = \frac{36x+18x-16}{12} \)
\( \Rightarrow \frac{6x-13}{6} = \frac{54x-16}{12} \)
\( \Rightarrow 12(6x – 13) = 6(54x – 16) \)
\( \Rightarrow 72x – 156 = 324x – 96 \)
\( \Rightarrow -156 + 96 = 324x – 72x \)
\( \Rightarrow -60 = 252x \)
\( \Rightarrow x = -\frac{60}{252} = -\frac{5}{21} \)
সত্যাসত্য,
LHS,
\( \left(-\frac{5}{21} – 3\right) + \frac{5}{6} \)
\( = \left(-\frac{5}{21} – \frac{63}{21}\right) + \frac{5}{6} \)
\( = -\frac{68}{21} + \frac{5}{6} \)
\( = -\frac{136}{42} + \frac{35}{42} \)
\( = -\frac{101}{42} \)
RHS,
\( 3\left(-\frac{5}{21}\right) + 2\left(\frac{3(-\frac{5}{21})}{4} – \frac{2}{3}\right) \)
\( = -\frac{5}{7} + 2\left(-\frac{15}{84} – \frac{2}{3}\right) \)
\( = -\frac{5}{7} + 2\left(-\frac{15}{84} – \frac{56}{84}\right) \)
\( = -\frac{5}{7} + 2\left(-\frac{71}{84}\right) \)
\( = -\frac{5}{7} – \frac{71}{42} \)
\( = -\frac{30}{42} – \frac{71}{42} \)
\( = -\frac{101}{42} \)
\(\text{LHS} = \text{RHS} = -\frac{101}{42}\)
\(\therefore x = -\frac{5}{21} \) is correct.
h)\( \frac{7m+3}{2} + \frac{m-2}{3} = m + 15 \)
Solution:
\( \Rightarrow \frac{3(7m+3)+2(m-2)}{6} = m + 15 \)
\( \Rightarrow \frac{21m+9+2m-4}{6} = m + 15 \)
\( \Rightarrow \frac{23m+5}{6} = m + 15 \)
\( \Rightarrow 23m + 5 = 6(m + 15) \)
\( \Rightarrow 23m + 5 = 6m + 90 \)
\( \Rightarrow 23m – 6m = 90 – 5 \)
\( \Rightarrow 17m = 85 \)
\( \Rightarrow m = \frac{85}{17} \)
\( \Rightarrow m = 5 \)
সসত্যাসত্য,
LHS:
\( \frac{7(5)+3}{2} + \frac{5-2}{3} \)
\( = \frac{35+3}{2} + \frac{3}{3} \)
\( = \frac{38}{2} + 1 \)
\( = 19 + 1 \)
\( = 20 \)
RHS:
\( 5 + 15 \)
\( = 20 \)
\(\text{LHS} = \text{RHS} = 20\)
\(\therefore m = 5 \) is correct.