S.chand books class 8 maths solution chapter 15 Linear Equations exercise 15 A

EXERCISE 15 A

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Solve the following equations:

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Q1 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 1

3(x+1) + 4x = 24
Sol :

⇒ 3x+3+4x=24

⇒ 7x=24-3

⇒ 7x=21

⇒ $x=\dfrac{21}{7}$

⇒ $x=3$

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Q2 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 2

16-2(3y+5) = 4(y-2)
Sol:

⇒16-6y-10=4y-8

⇒16-10+8=4y+6y

⇒24-10=10y

⇒14=10y

⇒$y=\dfrac{14}{10}$

⇒y=1.4

 

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Q3 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 3

3(z+1)+4(z+0.3) = 20z + 0.95
Sol:

⇒3z+3+4z+1.2=20z+0.95

⇒7z+4.2=20z+0.95

⇒4.2-0.95=20z-7z

⇒3.25=13z

⇒$z=\dfrac{3.25}{13}$

⇒z=0.25

 

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Q4 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 4

7m-4(m+6) = 7(m-8)+4

Sol :

⇒7m-4m-24=7m-56+4

⇒3m-24=7m-52

⇒52-24=7m-3m

⇒28=4m

⇒$m=\dfrac{28}{4}$

⇒m=7

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Q5 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 5

$\dfrac{x}{2}+\dfrac{x}{3} = 15$

Sol :

Take L.C.M of 2 and 3 = 6

⇒$\dfrac{x\times 3 + x\times 2}{6}=15$

⇒$\dfrac{3x+2x}{6}=15$

⇒$\dfrac{5x}{6}=15$

⇒$x=\dfrac{15\times 6}{5}$

⇒x=18

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Q6 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 6

$\dfrac{4(x+2)}{5}=7+\dfrac{5x}{13}$

Sol :
⇒$\dfrac{4x+8)}{5}=7+\dfrac{5x}{13}$

⇒$\dfrac{4x+8}{5}=\dfrac{7\times 13+5x}{13}$

⇒$\dfrac{4x+8}{5}=\dfrac{91+5x}{13}$

⇒13(4x+8)=5(91+5x)

⇒52x+104=455+25x

⇒52x-25x=455-104

⇒27x=351

⇒$x=\dfrac{351}{27}$

⇒x=13

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Q7 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 7

$\dfrac{x+3}{2}-\dfrac{4}{3}=\dfrac{3x-7}{3}$

Sol :
L.C.M of 2 and 3 is 6

⇒$\dfrac{3(x+3)-2(4)}{6}=\dfrac{3x-7}{3}$

⇒$\dfrac{3x+9-8}{6}=\dfrac{3x-7}{3}$

⇒$\dfrac{3x+1}{6}=\dfrac{3x-7}{3}$

⇒3(3x+1)=6(3x-7)

⇒9x+3=18x-42

⇒42+3=18x-9x

⇒9x=45

⇒$x=\dfrac{45}{9}$

⇒x=5

 

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Q8 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 8

$\dfrac{5x+1}{2}-\dfrac{x-2}{6}=\dfrac{2x+4}{3}$
Sol :
L.C.M of 2 and 6 is 6

⇒$\dfrac{3(5x+1)-(x-2)}{6}=\dfrac{2x+4}{3}$

⇒$\dfrac{15x+3-x+2}{6}=\dfrac{2x+4}{3}$

⇒$\dfrac{14x+5}{6}=\dfrac{2x+4}{3}$

⇒3(14x+5)=6(2x+4)

⇒42x+15=12x+24

⇒42x-12x=24-15

⇒30x=9

⇒$x=\dfrac{9}{30}$

⇒$x=\dfrac{3}{10}$

 

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Q9 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 9

$\dfrac{9x-5}{7}=\dfrac{6x+2}{5}$
Sol :

⇒5(9x-5)=7(6x+2)

⇒45x-25=42x+14

⇒45x-42x=14+25

⇒3x=39

⇒$x=\dfrac{39}{3}$

⇒x=13

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Q10 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 10

$\dfrac{9x}{7-6x}=2$
Sol :

⇒2(7-6x)=9x

⇒14-12x=9x

⇒14=9x+12x

⇒14=21x

⇒$x=\dfrac{14}{21}$

⇒$x=\dfrac{2}{3}$

 

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Question 11

$\dfrac{2x-1}{3x+1}=\dfrac{6x-1}{9x-3}$
Sol:

⇒(2x-1)(9x-3)=(6x-1)(3x+1)

⇒$2 x(9 x-3)-1(9 x-3)=6 x(3 x+1)-(3 x+1)$

$\Rightarrow 18 x^{2}-6 x-9 x+3=18 x^{2}+6 x-3 x-1$

$\Rightarrow 18 x^{2}-15 x+3=18 x^{2}+3 x-1$

$\Rightarrow 18 x^{2}-18 x^{2}-15 x-3 x=-1-3$

$\Rightarrow-18 x=-4$

$\Rightarrow x=\frac{-4}{-18}=\frac{2}{9}$

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Q12 | Ex-15A | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 12

$\dfrac{(1-3y)(4-y)}{(2-3y)(1-y)}=1$
Sol :

⇒(1-3y)(4-y)=(2-3y)(1-y)

⇒1(4-y)-3y(4-y)=2(1-y)-3y(1-y)

⇒$4-y-12 y+3 y^{2}=2-2 y-3 y+3 y^{2}$

⇒y-13y=2-5y

⇒2=8y

⇒$y=\frac{2}{8}$

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Question 13

$\dfrac{(4x+4)-(7x-8)}{4x+5}=-\dfrac{4}{3}$
Sol :

⇒$\dfrac{4x+4-7x+8}{4x+5}=-\dfrac{4}{3}$

⇒$\dfrac{-3x+12}{4x+5}=-\dfrac{4}{3}$

⇒3(-3x+12)=-4(4x+5)

⇒-9x+36=-16x-20

⇒16x-9x=-20-36

⇒7x=-56

⇒x=-8

 

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