S.chand publication New Learning Composite mathematics solution of class 7 Chapter 6 Linear Equations in One Variable Exercise 6B

 Exercise 6B

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Q1 | Ex-6B |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 1

$\frac{x}{2}+ 5 = 8$

Sol :

$\frac{x}{2}=3$

x=6

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Q2 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 2

$\frac{y}{3} – 1 = 7$

Sol :

$\frac{y}{3}=8$

y=24

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Q3 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 3

$2 – \frac{m}{8} = 5$

Sol :

$\frac{-m}{8}=5-2$

$\frac{-m}{8}=3$

-m=24

m=24

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Q4 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 4

$\frac{m}{40} – 7 = 0$

Sol :

$\frac{m}{40}=7$

m=280

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Q5 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 5

$\frac{2x}{5} – 18 = -24$

Sol :

$\frac{2x}{5}=-24+18$

$\frac{2x}{5}=-6$

2x=-30

x=-15

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Q6 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 6

$\frac{n}{2} + \frac{n}{3} = 5$

Sol :

L.C.M of 2 and 3 is 6, multiplying both sides by 6

$\frac{n}{2}\times 6 +\frac{n}{3} \times 6=5 \times 6$

3n+2n=30

$n=\frac{30}{5}=6$

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Q7 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 7

$6\frac{1}{2} + 1\frac{1}{2} (1 – 2x) = -4$

Sol :

$\frac{13}{2}+\frac{3}{2} (1-2x)=-4$

$\frac{31}{2}+\frac{3}{2}-3x=-4$

$-3x=-4-\frac{13}{2}-\frac{3}{2}$

$-3x=-\left(4+\frac{31}{2}+\frac{3}{2}\right)$

$3x=\frac{8+13+3}{2}$

$3x=\frac{24}{2}$

3x=12

$x=\frac{12}{3}=4$

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Q8 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 8

$\frac{4x-3}{5}-\frac{5x-3}{8}=1$

Sol :

L.C.M of 5 and 8 is 40, multiplying both sides by 40

$\left(\frac{4x-3}{5}\times 40\right)-\left(\frac{5x-3}{8}\times 40\right)=1\times 40$

(32x-24)-(25x-15)=40

32x-24-25x+15=40

32x-25x=40+24-15

7x=49

$x=\frac{49}{7}=7$

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Q9 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 9

$\dfrac{4 z+3}{3}+\dfrac{1}{3}=\dfrac{3 z-1}{2}$

Sol :

$\dfrac{(4z+3)+(1)}{3}=\dfrac{3z-1}{2}$

$\dfrac{4 z+4}{3}=\dfrac{3 z-1}{2}$

2(4z+4)=3(3z-1)

8z+8=9z-3

8+3=+9z-8z 

11=z or z=11

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Q10 | Ex-6B |Class 7 |S.Chand |New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 10

$\frac{1}{2}(8-6n) = n$

Sol 

4-3n=n

-3n-n=-4

-4n=-4

n=1