Class 7 Maths Sankardev Sishu Niketan Chapter – 4 Solution সৰল সমীকৰণ অনুশীলনী – 4(B)
অনুশীলনী – 4(B)
1) পৰ্যবেক্ষণৰ সহায়ত চলকৰ কি মানে তলৰ সমীকৰণ একোটা সিদ্ধ কৰে ঠাৱৰ কৰা_
(i) x-3=0
Answer:- x = 3
(ii) y + 5 = 0
Answer:- y= -5
(iii) x-3=8
Answer:- x = 11
(iv) u + 5 = 10
Answer:- u = 5
(v) x-4=-8
Answer:- x = -4
(vi) y + 3 = 3
Answer:- y = 0
(vii) 5x = 40
Answer:- x = 8
(viii) \(\frac{x}{9}\)= 3
Answer:- x = 27
(ix) 4x+1=9
Answer:- x =2
(x)\(\frac{x}{3}\)-1= 10
Answer:- x = 33
2. তলৰ সমীকৰণবোৰৰ সমাধান কৰা আৰু প্ৰতিটোতে ফলাফল পৰীক্ষা কৰা (অর্থাৎ সত্যাপন কৰা)
(i) \(\ 6x – 5 = 13 \)
সমাধানঃ (i) \(\ 6x – 5 = 13 \)
\(\Rightarrow 6x = 13 + 5 \)
\(\Rightarrow 6x = 18 \)
\(\Rightarrow x = \frac{18}{6} \)
\(\Rightarrow x = 3 \)
\(\therefore \) সমীকৰণৰ মূল = \( 3 \)
সত্যাপনঃ
বাওঁফলঃ \(\ 6x – 5 \)
\(\Rightarrow 6 \times 3 – 5 = 18 – 5 = 13 \)
সোঁফলঃ\(13\)
\(\therefore\) বাওঁফল = সোঁফল ✅
(ii)\(\ 5x – 6 = 2x + 9 \)
সমাধানঃ (ii) \(\ 5x – 6 = 2x + 9 \)
\(\Rightarrow 5x – 2x = 9 + 6 \)
\(\Rightarrow 3x = 15 \)
\(\Rightarrow x = \frac{15}{3} \)
\(\Rightarrow x = 5 \)
\(\therefore \) সমীকৰণৰ মূল = \( 5 \)
সত্যাপনঃ
বাওঁফলঃ \(\ 5x – 6 \)
\(\Rightarrow 5 \times 5 – 6 = 25 – 6 = 19 \)
সোঁফলঃ \(\ 2x + 9 \)
\(\Rightarrow 2 \times 5 + 9 = 10 + 9 = 19 \)
\(\therefore\) বাওঁফল = সোঁফল ✅
(iii)\(\ \frac{1}{2}x + 3 = \frac{3x}{4} \)
সমাধানঃ (iii) \(\ \frac{1}{2}x + 3 = \frac{3x}{4} \)
\(\Rightarrow \frac{1}{2}x – \frac{3x}{4} = -3 \)
\(\Rightarrow \left( \frac{2x – 3x}{4} \right) = -3 \)
\(\Rightarrow \frac{-x}{4} = -3 \)
\(\Rightarrow x = (-3) \times (-4) \)
\(\Rightarrow x = 12 \)
\(\therefore\) সমীকৰণৰ মূল = \( 12 \)
সত্যাপনঃ
বাওঁফলঃ \(\ \frac{1}{2}x + 3 \)
\(\Rightarrow \frac{1}{2} \times 12 + 3 = 6 + 3 = 9 \)
সোঁফলঃ \(\ \frac{3x}{4} \)
\(\Rightarrow \frac{3 \times 12}{4} = \frac{36}{4} = 9 \)
\(\therefore\) বাওঁফল = সোঁফল
(iv)\(\ 2(x + 5) – 3(x – 6) = 40 \)
সমাধানঃ (iv) \(\ 2(x + 5) – 3(x – 6) = 40 \)
\(\Rightarrow 2x + 10 – 3x + 18 = 40 \)
\(\Rightarrow (2x – 3x) + (10 + 18) = 40 \)
\(\Rightarrow -x + 28 = 40 \)
\(\Rightarrow -x = 40 – 28 \)
\(\Rightarrow -x = 12 \)
\(\Rightarrow x = -12 \)
\(\therefore\) সমীকৰণৰ মূল = \( -12 \)
সত্যাপনঃ
বাওঁফলঃ \(\ 2(x + 5) – 3(x – 6) \)
\(\Rightarrow 2(-12 + 5) – 3(-12 – 6) \)
\(\Rightarrow 2(-7) – 3(-18) = -14 + 54 = 40 \)
সোঁফলঃ\(40\)
\(\therefore\) বাওঁফল = সোঁফল
(v)\(\ 2(x + 7) – 5 = 3(x – 2) \)
সমাধানঃ (v) \(\ 2(x + 7) – 5 = 3(x – 2) \)
\(\Rightarrow 2x + 14 – 5 = 3x – 6 \)
\(\Rightarrow 2x + 9 = 3x – 6 \)
\(\Rightarrow 9 + 6 = 3x – 2x \)
\(\Rightarrow 15 = x \)
\(\therefore\) সমীকৰণৰ মূল = \( 15 \)
সত্যাপনঃ
বাওঁফলঃ \(\ 2(x + 7) – 5 \)
\(\Rightarrow 2(15 + 7) – 5 = 2 \times 22 – 5 = 44 – 5 = 39 \)
সোঁফলঃ \(\ 3(x – 2) \)
\(\Rightarrow 3(15 – 2) = 3 \times 13 = 39 \)
\(\therefore\) বাওঁফল = সোঁফল
(vi)\(\ x – 22 = \frac{x}{12} \)
সমাধানঃ (vi) \(\ x – 22 = \frac{x}{12} \)
\(\Rightarrow x – \frac{x}{12} = 22 \)
\(\Rightarrow \frac{12x – x}{12} = 22 \)
\(\Rightarrow \frac{11x}{12} = 22 \)
\(\Rightarrow 11x = 22 \times 12 = 264 \)
\(\Rightarrow x = \frac{264}{11} = 24 \)
\(\therefore\) সমীকৰণৰ মূল = \( 24 \)
সত্যাপনঃ
বাওঁফলঃ\(\ x – 22 = 24 – 22 = 2 \)
সোঁফলঃ \(\ \frac{x}{12} = \frac{24}{12} = 2 \)
\(\therefore\) বাওঁফল = সোঁফল
(vii) \(\ 3x – 2(2x – 5) = 2(x + 3) – 8 \)
সমাধানঃ (vii) \(\ 3x – 2(2x – 5) = 2(x + 3) – 8 \)
\(\Rightarrow 3x – (4x – 10) = 2x + 6 – 8 \)
\(\Rightarrow 3x – 4x + 10 = 2x – 2 \)
\(\Rightarrow -x + 10 = 2x – 2 \)
\(\Rightarrow 10 + 2 = 2x + x = 3x \)
\(\Rightarrow 12 = 3x \Rightarrow x = \frac{12}{3} = 4 \)
\(\therefore\) সমীকৰণৰ মূল = \( 4 \)
সত্যাপনঃ
বাওঁফলঃ \(\ 3x – 2(2x – 5) = 12 – 2(8 – 5) = 12 – 6 = 6 \)
সোঁফলঃ\(\ 2(x + 3) – 8 = 2(7) – 8 = 14 – 8 = 6 \)
\(\therefore\) বাওঁফল = সোঁফল
(viii)\(\ 18x – 62 = -13x \)
সমাধানঃ (viii) \(\ 18x + 13x = 62 \)
\(\Rightarrow 31x = 62 \Rightarrow x = \frac{62}{31} = 2 \)
\(\therefore\) সমীকৰণৰ মূল = \( 2 \)
সত্যাপনঃ
বাওঁফলঃ \(\ 18x – 62 = 36 – 62 = -26 \)
সোঁফলঃ \(\ -13x = -13 \times 2 = -26 \)
\(\therefore\) বাওঁফল = সোঁফল
(ix)\(\ y^3 = 8 – \frac{y – 2}{5} \)
সমাধানঃ (ix) \(\ y^3 = 8 – \frac{y – 2}{5} \)
Let’s try \(y = 2\):
LHS: \(2^3 = 8\)
RHS: \(8 – \frac{2 – 2}{5} = 8 – 0 = 8\)
\(\therefore\) সমীকৰণৰ মূল = \( 2 \)
সত্যাপনঃ
বাওঁফল = সোঁফল
(x)\(\ \frac{5}{2} = \frac{p}{2} + \frac{1}{2} \)
সমাধানঃ (x) \(\ \frac{5}{2} = \frac{p + 1}{2} \)
\(\Rightarrow 5 = p + 1 \Rightarrow p = 4 \)
\(\therefore\) সমীকৰণৰ মূল = \( 4 \)
সত্যাপনঃ
\(\ \frac{p}{2} + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2} \)
\(\therefore\) বাওঁফল = সোঁফল
(xi)\(\ 7(x – 2) – 8(4 – 3x) = 47 \)
সমাধানঃ (xi) \(\ 7x – 14 – 32 + 24x = 47 \)
\(\Rightarrow (7x + 24x) – (14 + 32) = 47 \)
\(\Rightarrow 31x – 46 = 47 \Rightarrow 31x = 93 \)
\(\Rightarrow x = \frac{93}{31} = 3 \)
\(\therefore\) সমীকৰণৰ মূল = \( 3 \)
সত্যাপনঃ
LHS = \(7(3 – 2) – 8(4 – 9) = 7(1) – 8(-5) = 7 + 40 = 47\)
RHS = \(47\)
\(\therefore\) বাওঁফল = সোঁফল
(xii)\(\ \frac{2u}{5} + \frac{4}{3} = \frac{11}{5} \)
সমাধানঃ (xii) \(\ \frac{2u}{5} = \frac{11}{5} – \frac{4}{3} \)
\(\Rightarrow \frac{2u}{5} = \frac{33 – 20}{15} = \frac{13}{15} \)
\(\Rightarrow 2u = \frac{13 \times 5}{15} = \frac{65}{15} \)
\(\Rightarrow u = \frac{65}{30} = \frac{13}{6} \)
\(\therefore\) সমীকৰণৰ মূল = \( \frac{13}{6} \)
সত্যাপনঃ
LHS = \(\ \frac{2 \times \frac{13}{6}}{5} + \frac{4}{3} = \frac{26}{30} + \frac{4}{3} \)
\(\Rightarrow \frac{13}{15} + \frac{20}{15} = \frac{33}{15} = \frac{11}{5} \)
\(\therefore\) বাওঁফল = সোঁফল
(xiii)\(\ \frac{x + 5}{(2x + 2) + 7} = \frac{1}{4} \)
সমাধানঃ (xiii) \(\ \frac{x + 5}{2x + 9} = \frac{1}{4} \)
Cross-multiplying:
\(\Rightarrow 4(x + 5) = 2x + 9 \)
\(\Rightarrow 4x + 20 = 2x + 9 \Rightarrow 2x = -11 \)
\(\Rightarrow x = \frac{-11}{2} \)
\(\therefore\) সমীকৰণৰ মূল = \( \frac{-11}{2} \)
সত্যাপনঃ
LHS = \(\ \frac{-11/2 + 5}{2(-11/2) + 9} = \frac{-1/2}{-11 + 9} = \frac{-1/2}{-2} = \frac{1}{4} \)
RHS = \( \frac{1}{4} \)
\(\therefore\) বাওঁফল = সোঁফল
(3) কি অৱস্থাত \(\ ax = b \) সমীকৰণৰ মূল \(\ \frac{b}{a} \) হব?
সমাধানঃ\( a \ne 0 \) হ’লে।
(4)এনেধৰণৰ তিনিটা সমীকৰণ লিখা যিবিলাকৰ প্ৰতিটোৰে মূল বা বীজ 2 (অর্থাৎ \( x = 2 \))।
সমাধানঃ-
\(\implies x – 2 = 0 \)
\(\implies 3x = 6 \)
\(\implies 2(x + 1) = 6 \)