ML AGGARWAL CLASS 8 Chapter 12 Linear Equations and Inqualities In one Variable Exercise 12.1

 Exercise 12.1

Question 1

Sol :

(i) 5 x-3=3 x-5

=5 x=3 x-5+3

=5 x=3 x-2

=5 x-3 x=-2

=2 x=-2

x=-1

(ii) 3 x-7=3(5-x)

=3 x-7=15-3 x

=3 x=15+7-3 x

=3 x+3 x=22

=6 x=22

=$x=\frac{22}{6}$

Question 2

Sol :

(i) 4(2 x+1)=3(x-1)+7

=8 x+4=3 x-3+7

=8 x=3 x-3+7-4

=8 x-3 x=-3+3=0

$5 x=0 \Rightarrow x=0$

(ii) 

$\begin{aligned} 3(2 p-1)=5-(3 p-2) & \\ 6 p-3=5 &-3 p+2 \\ 6 p=& 5-3 j+2+3 \\ 6 p+3 p &=10 \\ 9 p &=10 \\ p &=\frac{10}{9}=1 \frac{1}{9} \end{aligned}$

Question 3

Sol :

(i) 5 y-2[y-3(y-5)]=6

=5 y-2 y+6(y-5)=6

=3 y+6 y-30=6

=9 y=6+30=36

=$y=\frac{36}{9}=4$

(ii) 0.3(6-x)=0.4(x+8)

=1.8-0.3 x=0.4 x+3.2

=0.4 x=1.8-0.3 x-3.2

=0.4 x+0.3 x=-1.4

=0.7 x=-1.4

=$x=-\frac{1.4}{0.7}=2$

Question 4

Sol :

(i) $\frac{x-1}{3}=\frac{x+2}{6}+3$

Multiply and divide 2 on L.H.S

=$\frac{2(x-1)}{2 \times 3}=\frac{x+2}{6}+3$

=$\frac{2 x-1}{6}-\frac{x+2}{6}=3$

=$\frac{2 x-1-x-2}{6}=3$

=$\frac{x-3}{6}=3$

=$x-3=3 \times 6$

=x-3=18

x=18+3

x=21

 

(ii) $\frac{x+7}{3}=1+\frac{3 x-2}{5}$

=$\frac{x+7}{3}=\frac{5}{5}+\frac{3 x-2}{5}$

=$\frac{x+7}{3}=\frac{5+3 x-2}{5}$

=$\frac{x+7}{3}=\frac{3 x+3}{5}$

Cross Multiplying

=5(x+7)=3(3 x+3)

=5 x+35=9 x+9

=9 x=5 x+35-9

=9 x-5 x=26

=4 x=26

$x=\frac{26}{4}=\frac{13}{2}=6 \frac{1}{2}$

Question 5

Sol :

(i) $\frac{y+1}{3}-\frac{y-1}{2}=\frac{1+2 y}{3}$

Multiplying both sides by 6 i.e L.C.M of $3,2,3$ we get

=2(y+1)-3(y-1)=2(1+2 y)

=2 y+2-3 y+3=2+4 y

=5-y=4 y+2

=4 y+y=5-2

5 y=3

$y=3 / 5$

(ii) $\frac{1}{3}+\frac{p}{4}=55-\frac{p+40}{5}$

=$\frac{4 p+3 p}{12}=\frac{55 \times 5}{5}-\frac{p+40}{5}$

=$\frac{7 p}{12}=\frac{275-p-40}{5}$

$\frac{7 p}{12}=\frac{235-P}{5}$

Cross Multiplication

$7 p \times 5=12(235-p)$

$35 p=12 \times 235-12 p$

$35 p+12 p=12 \times 235$

$47 P=12 \times 235$

$P=\frac{12 \times 2355}{47}$

p=60

Question 6

Sol :

(i) $n-\frac{n-1}{2}=1-\frac{\eta-2}{3}$

$\frac{2 n-(n-1)}{2}=\frac{3-(n-2)}{3}$

$\frac{2 n-n+1}{2}=\frac{3-n+2}{3}$

$\frac{n+1}{2}=\frac{5-n}{3}$

Cross multiplication

3 x(n+1)=2(5-n)

3 n+3=10-2 n

3 n+2 n=10-3

5 n=7

$n=7 / 5$

(ii) $\frac{3 t-2}{3}+\frac{2 t+3}{2}=t+\frac{7}{6}$

$\frac{3 t-2}{3}+\frac{2 t+3}{2}=\frac{6 t+7}{6}$

Multiplying 6 on both sides

2(3 t-2)+3(2 t+3)=1(6 t+3)

6 t-4+6 t+9=6 t+7

6 t+5=7

6 t=7-5=2

t=2 / 6=1 / 3

Question 7

Sol :

(i)

 $\begin{aligned} 4(3 x+2)-5(6 x-1) &=2(x-8)-6(7 x-4) \\ 12 x+8-30 x+5 &=2 x-16-42 x+24 \\ 13-18 x=& 8-40 x \\ 40 x-18 x &+13=8 \\ 22 x &=8-13=-5 \\ x &=\frac{-5}{22} . \end{aligned}$

 

(ii) 3(5 x+7)+5(2 x-11)=3(8 x-5)-15

=15 x+21+10 x-55=24 x-15-15

=25 x-34=24 x-30

=25 x-24 x=34-30

x=4

Question 8

Sol :

(i) $\frac{3-2 x}{2 x+5}=-\frac{3}{11}$

Cross Multiplying both sides

-11(3-2 x)=3(2 x+5)

[22 x-3]=6 x+15

22 x-6 x=33+15

16 x=48

x=3

(ii) $\frac{5 p+2}{8-2 p}=\frac{7}{6}$

cross multiplying on both sides

$\begin{aligned} 6(5 p+2) &=7(8-2 p) \\ 30 p+12 &=56-14 p \\ 30 p+14 p &=56-12 \\ 44 p &=44 \Rightarrow p=1 \end{aligned}$

Question 9

Sol :

(i) $\frac{5}{x}=\frac{7}{x-4}$

Cross multiplying on both sides

$\begin{array}{c}5(x-4)=7 x \\ 7 x=5 x-20 \\ 2 x=-20 \\ x=-10\end{array}$

(ii) $\frac{4}{2 x+3}=\frac{5}{x+4}$

Cross multiplication on both sides

$\begin{aligned} 5(2 x+3) &=4(x+4) \\ 10 x+15 &=4 x+16 \\ 10 x-4 x &=16-15 \\ 6 x &=1 \Rightarrow x=\frac{1}{6} \end{aligned}$

(ii) $\frac{4}{2 x+3}=\frac{5}{x+4}$

Cross multhpicalion on bolh sides

$\begin{aligned} 5(2 x+3)=4(x+4) \\ 10 x+15=4 x+16 \\ 10 x-4 x &=16-15 \\ 6 x=1 & \Rightarrow x=\frac{1}{6} \end{aligned}$

Question 10

Sol :

(i) $\frac{2 x+5}{2}-\frac{5 x}{x-1}=x$

=$\frac{(2 x+5)(x-1)-(51 \times 2)}{2(x-1)}=x$

=$2 x^{2}-2 x+5 x-5-10 x=2 x(x-1)$

=$2 x^{2}+3 x-5-101=2 x^{2}-2 x$

-7 x-5=-2 x

-5=7 x-2 x

5 x=-5

x=-1

(ii) $\frac{1}{5}\left(\frac{1}{3 x}-5\right)=\frac{1}{3}\left(3-\frac{1}{x}\right)$

$\frac{1}{5}\left(\frac{1-15 x}{3 x}\right)=\left(\frac{3 x-1}{3 x}\right)$

1-15 x=5(3 x-1)

1-15 x=15 x-5

15 x+15 x=1+5

30 x=6

$x=\frac{6}{30}=\frac{1}{5}$

Question 11

Sol :

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

Subtracting '1' on both Sides

=$\frac{2 x-3}{2 x-1}-1=\frac{3 x-1}{3 x+1}-1$

=$\frac{2 x-3-2 x+1}{2 x-1}=\frac{3 x-1-3 x-1}{3 x+1}$

=$\frac{-2}{2 x-1}=\frac{-2}{3 x+1}$

=3 x+1=2 x-1

3 x-2 x=-1-1

x=-2

(ii) $\frac{2 y+3}{3 y+2}=\frac{4 y+5}{6 y+7}$

Cross multiplication

(2 y+3)(6 y+7)=(3 y+2)(4 y+5)

$12 y^{2}+14 y+18 y+21=12y^{2}+15 y+8 y+10$

32 y+21=23 y+10

32 y-23 y=10-21

9 y=-11

$y=\frac{-11}{9}$

Question 12

Sol :

x=p+1

$\frac{5 x-30}{2}-\frac{7 p+1}{3}=\frac{1}{4}$

multiplying 12 on both sides

6(5 x-30)-4(7 p+1)=3

30 x-180-28 p-4=3

30(p+1)-180-28 p-4-3=0

30 p+30-180-28 p-7=0

2 p-157=0

$p=\frac{157}{2}$

Question 13

Sol :

$\frac{x+3}{3}-\frac{x-2}{2}=1 \quad$ if $\frac{1}{x}+p=1$

Multiplying 6 on both sides

2(x+3)-3(x-2)=6

2 x+6-3 x+6=8

x=6 

$\frac{1}{x}+p=1 \Rightarrow p=1-\frac{1}{x}=1-\frac{1}{6}$

$p=\frac{5}{6}$