NCERT Solutions for Exercise 3.6 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.6

Q1(i) Solve the following pairs of equations by reducing them to a pair of linear equations:

$$\begin{gathered} \frac{1}{2x}+\frac{1}{3y}=2 \\ \frac{1}{3x}+\frac{1}{2y}=\frac{13}{6} \end{gathered}$$

Answer:

Given Equations,

$$\begin{gathered} \frac{1}{2x}+\frac{1}{3y}=2 \\ \frac{1}{3x}+\frac{1}{2y}=\frac{13}{6} \end{gathered}$$

Let,

$\frac{1}{x}=pand\frac{1}{y}=q$

Now, our equation becomes

$\frac{p}{2}+\frac{q}{3}=2$

$\Rightarrow 3p+2q=12........(1)$

And

$\frac{p}{3}+\frac{q}{2}=\frac{13}{6}$

$\Rightarrow 2p+3q=13..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1{c}_2-b_2{c}_1}=\frac{q}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\frac{p}{(2)(-13)-(3)(-12)}=\frac{q}{(-12)(2)-(-13)(3)}=\frac{1}{(3)(3)-(2)(2)}$

$\frac{p}{-26+36}=\frac{q}{-24+39}=\frac{1}{9-4}$

$\frac{p}{10}=\frac{q}{15}=\frac{1}{5}$

$p=2,andq=3$

And Hence,

$x=\frac{1}{2}andy=\frac{1}{3}.$

Q1(ii) Solve the following pairs of equations by reducing them to a pair of linear equations:

$$\begin{gathered} \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \\ \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \end{gathered}$$

Answer:

Given Equations,

$$\begin{gathered} \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \\ \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \end{gathered}$$

Let,

$\frac{1}{\sqrt{x}}=pand\frac{1}{\sqrt{y}}=q$

Now, our equation becomes

$2p+3q=2........(1)$

And

$4p-9q=-1..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1{c}_2-b_2{c}_1}=\frac{q}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\frac{p}{(3)(1)-(-9)(-2)}=\frac{q}{(-2)(4)-(1)(2)}=\frac{1}{(2)(-9)-(4)(3)}$

$\frac{p}{3-18}=\frac{q}{-8-2}=\frac{1}{-18-12}$

$\frac{p}{-15}=\frac{q}{-10}=\frac{1}{-30}$

$p=\frac{1}{2},andq=\frac{1}{3}$

So,

$p=\frac{1}{2}=\frac{1}{\sqrt{x}}\Rightarrow x=4$

$q=\frac{1}{3}=\frac{1}{\sqrt{y}}\Rightarrow y=9$ .

And hence

$x=4andy=9.$

Q1(iii) Solve the following pairs of equations by reducing them to a pair of linear equations:

$$\begin{gathered} \frac{4}{x}+3y=14 \\ \frac{3}{x}-4y=23 \end{gathered}$$

Answer:

Given Equations,

$$\begin{gathered} \frac{4}{x}+3y=14 \\ \frac{3}{x}-4y=23 \end{gathered}$$

Let,

$\frac{1}{x}=pandy=q$

Now, our equation becomes

$\Rightarrow 4p+3q=14........(1)$

And

$\Rightarrow 3p-4q=23..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1{c}_2-b_2{c}_1}=\frac{q}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\frac{p}{(3)(-23)-(-4)(-14)}=\frac{q}{(-14)(3)-(-23)(4)}=\frac{1}{(4)(-4)-(3)(3)}$

$\frac{p}{-69-56}=\frac{q}{-42+92}=\frac{1}{-16-9}$

$\frac{p}{-125}=\frac{q}{50}=-\frac{1}{25}$

$p=5,andq=-2$

And Hence,

$x=\frac{1}{5}andy=-2.$

Q1(iv) Solve the following pairs of equations by reducing them to a pair of linear equations:

$$\begin{gathered} \frac{5}{x-1}+\frac{1}{y-2}=2 \\ \frac{6}{x-1}-\frac{3}{y-2}=1 \end{gathered}$$

Answer:

Given Equations,

$$\begin{gathered} \frac{5}{x-1}+\frac{1}{y-2}=2 \\ \frac{6}{x-1}-\frac{3}{y-2}=1 \end{gathered}$$

Let,

$\frac{1}{x-1}=pand\frac{1}{y-2}=q$

Now, our equation becomes

$5p+q=2........(1)$

And

$6p-3q=1..........(2)$

Multiplying (1) by 3 we get

$15p+3q=6..........(3)$

Now, adding (2) and (3) we get

$21p=7$

$\Rightarrow p=\frac{1}{3}$

Putting this in (2)

$6\left(\frac{1}{3}\right)-3q=1$

$\Rightarrow 3q=1$

$\Rightarrow q=\frac{1}{3}$

Now,

$p=\frac{1}{3}=\frac{1}{x-1}\Rightarrow x-1=3\Rightarrow x=4$

$q=\frac{1}{3}=\frac{1}{y-2}\Rightarrow y-2=3\Rightarrow x=5$

Hence,

$x=4,andy=5.$

Q1(v) Solve the following pairs of equations by reducing them to a pair of linear equations:

$$\begin{gathered} \frac{7x-2y}{xy}=5 \\ \frac{8x+7y}{xy}=15 \end{gathered}$$

Answer:

Given Equations,

$$\begin{gathered} \frac{7x-2y}{xy}=5 \\ \Rightarrow \frac{7}{y}-\frac{2}{x}=5 \\ \frac{8x+7y}{xy}=15 \\ \Rightarrow \frac{8}{y}+\frac{7}{x}=15 \end{gathered}$$

Let,

$\frac{1}{x}=pand\frac{1}{y}=q$

Now, our equation becomes

$7q-2p=5........(1)$

And

$8q+7p=15..........(2)$

By Cross Multiplication method,

$\frac{q}{b_1{c}_2-b_2{c}_1}=\frac{p}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\frac{q}{(-2)(-15)-(7)(-5)}=\frac{p}{(-5)(8)-(-15)(7)}=\frac{1}{(7)(7)-(8)(-2)}$

$\frac{q}{30+35}=\frac{p}{-40+105}=\frac{1}{49+16}$

$\frac{q}{65}=\frac{p}{65}=\frac{1}{65}$

$p=1,andq=1$

And Hence,

$x=1andy=1.$

Q1(vi) Solve the following pairs of equations by reducing them to a pair of linear equations:

$$\begin{gathered} 6x+3y=6xy \\ 2x+4y=5xy \end{gathered}$$

Answer:

Given Equations,

$$\begin{gathered} 6x+3y=6xy \\ \Rightarrow \frac{6x}{xy}+\frac{3y}{xy}=6 \\ \Rightarrow \frac{6}{y}+\frac{3}{x}=6 \\ and \\ 2x+4y=5xy \\ \Rightarrow \frac{2x}{xy}+\frac{4y}{xy}=5 \\ \Rightarrow \frac{2}{y}+\frac{4}{x}=5 \end{gathered}$$

Let,

$\frac{1}{x}=pand\frac{1}{y}=q$

Now, our equation becomes

$6q+3p=6........(1)$

And

$2q+4p=5..........(2)$

By Cross Multiplication method,

$\frac{q}{b_1{c}_2-b_2{c}_1}=\frac{p}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\frac{q}{(3)(-5)-(-6)(4)}=\frac{p}{(6)(2)-(6)(-5)}=\frac{1}{(6)(4)-(3)(2)}$

$\frac{q}{-15+24}=\frac{p}{-12+30}=\frac{1}{24-6}$

$\frac{q}{9}=\frac{p}{18}=\frac{1}{18}$

$q=\frac{1}{2}andp=1$

And Hence,

$x=1andy=2.$

Q1(vii) Solve the following pairs of equations by reducing them to a pair of linear equations:
$$\begin{gathered} \frac{10}{x+y}+\frac{2}{x-y}=4 \\ \frac{15}{x+y}-\frac{5}{x-y}=-2 \end{gathered}$$

Answer:

Given Equations,

$$\begin{gathered} \frac{10}{x+y}+\frac{2}{x-y}=4 \\ \frac{15}{x+y}-\frac{5}{x-y}=-2 \end{gathered}$$

Let,

$\frac{1}{x+y}=pand\frac{1}{x-y}=q$

Now, our equation becomes

$10p+2q=4........(1)$

And

$15p-5q=-2..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1{c}_2-b_2{c}_1}=\frac{q}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\frac{p}{(2)(2)-(-5)(-4)}=\frac{q}{(-4)(15)-(2)(10)}=\frac{1}{(10)(-5)-(15)(2)}$

$\frac{p}{4-20}=\frac{q}{-60-20}=\frac{1}{-50-30}$

$\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}$

$p=\frac{1}{5},andq=1$

Now,

$p=\frac{1}{5}=\frac{1}{x+y}$

$\Rightarrow x+y=5........(3)$

And,

$q=1=\frac{1}{x-y}$

$\Rightarrow x-y=1...........(4)$

Adding (3) and (4) we get,

$\Rightarrow 2x=6$

$\Rightarrow x=3$

Putting this value in (3) we get,

$3+y=5$

$\Rightarrow y=2$

And Hence,

$x=3andy=2.$

Q1(viii) Solve the following pairs of equations by reducing them to a pair of linear equations:

$$\begin{gathered} \frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4} \\ \frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8} \end{gathered}$$

Answer:

Given Equations,

$$\begin{gathered} \frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4} \\ \frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8} \end{gathered}$$

Let,

$\frac{1}{3x+y}=pand\frac{1}{3x-y}=q$

Now, our equation becomes

$p+q=\frac{3}{4}.........(1)$

And

$$\begin{gathered} \frac{p}{2}-\frac{q}{2}=\frac{-1}{8} \\ p-q=\frac{-1}{4}..........(2) \end{gathered}$$

Now, Adding (1) and (2), we get

$2p=\frac{3}{4}-\frac{1}{4}$

$\Rightarrow 2p=\frac{2}{4}$

$\Rightarrow p=\frac{1}{4}$

Putting this value in (1)

$\frac{1}{4}+q=\frac{3}{4}$

$\Rightarrow q=\frac{3}{4}-\frac{1}{4}$

$\Rightarrow q=\frac{2}{4}$

$\Rightarrow q=\frac{1}{2}$

Now,

$p=\frac{1}{4}=\frac{1}{3x+y}$

$\Rightarrow 3x+y=4...........(3)$

And

$q=\frac{1}{2}=\frac{1}{3x-y}$

$\Rightarrow 3x-y=2............(4)$

Now, Adding (3) and (4), we get

$6x=4+2$

$\Rightarrow 6x=6$

$\Rightarrow x=1$

Putting this value in (3),

$3(1)+y=4$

$\Rightarrow y=4-3$

$\Rightarrow y=1$

Hence,

$x=1,andy=1$

Q2(i) Formulate the following problems as a pair of equations and hence find their solutions: Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Answer:

Let the speed of Ritu in still water be x and speed of current be y,

Let's solve this problem by using relative motion concept,

the relative speed when they are going in the same direction (downstream)= x +y

the relative speed when they are going in the opposite direction (upstream)= x - y

Now, As we know,

Relative distance = Relative speed * time .

So, According to the question,

$x+y=\frac{20}{2}$

$\Rightarrow x+y=10.........(1)$

And,

$x-y=\frac{4}{2}$

$\Rightarrow x-y=2...........(2)$

Now, Adding (1) and (2), we get

$2x=10+2$

$\Rightarrow 2x=12$

$\Rightarrow x=6$

Putting this in (2)

$6-y=2$

$\Rightarrow y=6-2$

$\Rightarrow y=4$

Hence,

$x=6andy=4.$

Hence Speed of Ritu in still water is 6 km/hour and the speed of the current is 4 km/hour

Q2(ii) Formulate the following problems as a pair of equations and hence find their solutions: 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Answer:

Let the number of days taken by woman and man be x and y respectively,

The proportion of Work done by a woman in a single day

$=\frac{1}{x}$

The proportion of Work done by a man in a single day

$=\frac{1}{y}$

Now, According to the question,

$4(\frac{2}{x}+\frac{5}{y})=1$

$\Rightarrow (\frac{2}{x}+\frac{5}{y})=\frac{1}{4}$

Also,

$3(\frac{3}{x}+\frac{6}{y})=1$

$\Rightarrow (\frac{3}{x}+\frac{6}{y})=\frac{1}{3}$

Let,

$\frac{1}{x}=pand\frac{1}{y}=q$

Now, our equation becomes

$2p+5q=\frac{1}{4}$

$8p+20q=1........(1)$

And

$3p+6p=\frac{1}{3}$

$\Rightarrow 9p+18p=1.............(2)$

By Cross Multiplication method,

$\frac{p}{b_1{c}_2-b_2{c}_1}=\frac{q}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\frac{p}{(20)(-1)-(18)(-1)}=\frac{q}{(-1)(9)-(8)(-1)}=\frac{1}{(8)(18)-(20)(9)}$

$\frac{p}{-20+18}=\frac{q}{-9+8}=\frac{1}{146-60}$

$\frac{p}{-2}=\frac{q}{-1}=\frac{1}{-36}$

$p=\frac{1}{18},andq=\frac{1}{36}$

So,

$x=18andy=36.$

Q2(iii) Formulate the following problems as a pair of equations and hence find their solutions: Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer:

Let the speed of the train and bus be u and v respectively

Now According to the question,

$\frac{60}{u}+\frac{240}{v}=4$

And

$\frac{100}{u}+\frac{200}{v}=4+\frac{1}{6}$

$\Rightarrow \frac{100}{u}+\frac{200}{v}=\frac{25}{6}$

Let,

$\frac{1}{u}=pand\frac{1}{v}=q$

Now, our equation becomes

$60p+140q=4$

$\Rightarrow 15p+60q=1.........(1)$

And

$100p+200q=\frac{25}{6}$

$\Rightarrow 4p+8q=\frac{1}{6}$

$\Rightarrow 24p+48q=1..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1{c}_2-b_2{c}_1}=\frac{q}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\frac{q}{(60)(-1)-(48)(-1)}=\frac{p}{(-1)(24)-(-1)(15)}=\frac{1}{(15)(48)-(60)(24)}$

$\frac{p}{-60+48}=\frac{q}{-24+15}=\frac{1}{720-1440}$

$\frac{p}{-12}=\frac{q}{-9}=\frac{1}{-720}$

$p=\frac{12}{720}=\frac{1}{60},andq=\frac{9}{720}=\frac{1}{80}$

And Hence,

$x=60andy=80$

Hence the speed of the train and bus are 60 km/hour and 80 km/hour respectively.