NCERT Solutions for Exercise 3.3 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.3

Q1(i) Solve the following pair of linear equations by the substitution method.

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

Answer:

Given, two equations,

$$\begin{gathered} x+y=14.......(1) \\ x-y=4........(2) \end{gathered}$$

Now, from (1), we have

$y=14-x........(3)$

Substituting this in (2), we get

$x-(14-x)=4$

$\Rightarrow x-14+x=4$

$\Rightarrow 2x=4+14=18$

$\Rightarrow x=9$

Substituting this value of x in (3)

$\Rightarrow y=14-x=14-9=5$

Hence, Solution of the given equations is x = 9 and y = 5.

Q1(ii) Solve the following pair of linear equations by the substitution method

$$\begin{gathered} s-t=3 \\ \frac{s}{3}+\frac{t}{2}=6 \end{gathered}$$

Answer:

Given, two equations,

$$\begin{gathered} s-t=3.........(1) \\ \frac{s}{3}+\frac{t}{2}=6.......(2) \end{gathered}$$

Now, from (1), we have

$s=t+3........(3)$

Substituting this in (2), we get

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

$\Rightarrow \frac{2t+6+3t}{6}=6$

$\Rightarrow 5t+6=36$

$\Rightarrow 5t=30$

$\Rightarrow t=6$

Substituting this value of t in (3)

$\Rightarrow s=t+3=6+3=9$

Hence, Solution of the given equations is s = 9 and t = 6.

Q1(iii) Solve the following pair of linear equations by the substitution method.

$$\begin{gathered} 3x-y=3 \\ 9x-3y=9 \end{gathered}$$

Answer:

Given, two equations,

$$\begin{gathered} 3x-y=3......(1) \\ 9x-3y=9.....(2) \end{gathered}$$

Now, from (1), we have

$y=3x-3........(3)$

Substituting this in (2), we get

$9x-3(3x-3)=9$

$\Rightarrow 9x-9x+9=9$

$\Rightarrow 9=9$

This is always true, and hence this pair of the equation has infinite solutions.

As we have

$y=3x-3$ ,

One of many possible solutions is $x=1,andy=0$ .

Q1(iv) Solve the following pair of linear equations by the substitution method.

$$\begin{gathered} 0.2x+0.3y=1.3 \\ 0.4x+0.5y=2.3 \end{gathered}$$

Answer:

Given, two equations,

$$\begin{gathered} 0.2x+0.3y=1.3 \\ 0.4x+0.5y=2.3 \end{gathered}$$

Now, from (1), we have

$y=\frac{1.3-0.2x}{0.3}........(3)$

Substituting this in (2), we get

$0.4x+0.5\left(\frac{1.3-0.2x}{0.3}\right)=2.3$

$\Rightarrow 0.12x+0.65-0.1x=0.69$

$\Rightarrow 0.02x=0.69-0.65=0.04$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=\left(\frac{1.3-0.2x}{0.3}\right)=\left(\frac{1.3-0.4}{0.3}\right)=\frac{0.9}{0.3}=3$

Hence, Solution of the given equations is,

$x=2andy=3$ .

Q1(v) Solve the following pair of linear equations by the substitution method.

$$\begin{gathered} \sqrt{2}x+\sqrt{3}y=0 \\ \sqrt{3}x-\sqrt{8}y=0 \end{gathered}$$

Answer:

Given, two equations,

$$\begin{gathered} \sqrt{2}x+\sqrt{3}y=0 \\ \sqrt{3}x-\sqrt{8}y=0 \end{gathered}$$

Now, from (1), we have

$y=-\frac{\sqrt{2}x}{\sqrt{3}}........(3)$

Substituting this in (2), we get

$\sqrt{3}x-\sqrt{8}(-\frac{\sqrt{2}x}{\sqrt{3}})=0$

$\Rightarrow \sqrt{3}x=\sqrt{8}(-\frac{\sqrt{2}x}{\sqrt{3}})$

$\Rightarrow \text{3}x=-4x$

$\Rightarrow 7x=0$

$\Rightarrow x=0$

Substituting this value of x in (3)

$\Rightarrow y=-\frac{\sqrt{2}x}{\sqrt{3}}=0$

Hence, Solution of the given equations is,

$x=0,andy=0$ .

Q1(vi) Solve the following pair of linear equations by the substitution method.

$$\begin{gathered} \frac{3x}{2}-\frac{5y}{3}=-2 \\ \frac{x}{3}+\frac{y}{2}=\frac{13}{6} \end{gathered}$$

Answer:

Given,

$$\begin{gathered} \frac{3x}{2}-\frac{5y}{3}=-2.........(1) \\ \frac{x}{3}+\frac{y}{2}=\frac{13}{6}.............(2) \end{gathered}$$

From (1) we have,

$x=\frac{2}{3}(\frac{5y}{3}-2)........(3)$

Putting this in (2) we get,

$\frac{1}{3}\times \frac{2}{3}(\frac{5y}{3}-2)+\frac{y}{2}=\frac{13}{6}$

$\frac{10y}{27}-\frac{4}{9}+\frac{y}{2}=\frac{13}{6}$

$\frac{20y}{54}-\frac{4}{9}+\frac{27y}{54}=\frac{13}{6}$

$\frac{47y}{54}=\frac{13}{6}+\frac{4}{9}$

$\frac{47y}{54}=\frac{117}{54}+\frac{24}{54}$

$47y=117+24$

$47y=141$

$y=\frac{141}{47}$

$y=3$

putting this value in (3) we get,

$x=\frac{2}{3}(\frac{5y}{3}-2)$

$x=\frac{2}{3}(\frac{5\times 3}{3}-2)$

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

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

$x=2$

Hence $x=2andy=3.$

Q2 Solve $2x+3y=11$and $2x-4y=-24$ and hence find the value of ‘ $m$’ for which $y=mx+3$ .

Answer:

Given, two equations,

$2x+3y=11......(1)$

$2x-4y=-24.......(2)$

Now, from (1), we have

$y=\frac{11-2x}{3}........(3)$

Substituting this in (2), we get

$2x-4\left(\frac{11-2x}{3}\right)=-24$

$\Rightarrow 6x-44+8x=-72$

$\Rightarrow 14x=44-72$

$\Rightarrow 14x=-28$

$\Rightarrow x=-2$

Substituting this value of x in (3)

$\Rightarrow y=\left(\frac{11-2x}{3}\right)=\frac{11-2\times (-2)}{3}=\frac{15}{3}=5$

Hence, Solution of the given equations is,

$x=-2,andy=5.$

Now,

As it satisfies $y=mx+3$ ,

$\Rightarrow 5=m(-2)+3$

$\Rightarrow 2m=3-5$

$\Rightarrow 2m=-2$

$\Rightarrow m=-1$

Hence Value of m is -1.

Q3(i) Form the pair of linear equations for the following problem and find their solution by substitution method.

The difference between the two numbers is 26 and one number is three times the other. Find them.

Answer:

Let two numbers be x and y and let the bigger number is y.

Now, According to the question,

$y-x=26......(1)$

And

$y=3x......(2)$

Now, the substituting value of y from (2) in (1) we get,

$3x-x=26$

$\Rightarrow 2x=26$

$\Rightarrow x=13$

Substituting this in (2)

$\Rightarrow y=3x=3(13)=39$

Hence the two numbers are 13 and 39.

Q3(ii) Form the pair of linear equations for the following problem and find their solution by substitution method

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer:

Let the larger angle be x and smaller angle be y

Now, As we know the sum of supplementary angles is 180. so,

$x+y=180.......(1)$

Also given in the question,

$x-y=18.......(2)$

Now, From (2) we have,

$y=x-18.......(3)$

Substituting this value in (1)

$x+x-18=180$

$\Rightarrow 2x=180+18$

$\Rightarrow 2x=198$

$\Rightarrow x=99$

Now, Substituting this value of x in (3), we get

$\Rightarrow y=x-18=99-18=81$

Hence the two supplementary angles are

${99}^{0}and{81}^0.$

Q4 Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Answer:

Let the cost of 1 bat is x and the cost of 1 ball is y.

Now, According to the question,

$7x+6y=3800......(1)$

$3x+5y=1750......(2)$

Now, From (1) we have

$y=\frac{3800-7x}{6}........(3)$

Substituting this value of y in (2)

$3x+5\left(\frac{3800-7x}{6}\right)=1750$

$\Rightarrow 18x+19000-35x=1750\times 6$

$\Rightarrow -17x=10500-19000$

$\Rightarrow -17x=-8500$

$\Rightarrow x=\frac{8500}{17}$

$\Rightarrow x=500$

Now, Substituting this value of x in (3)

$y=\frac{3800-7x}{6}=\frac{3800-7\times 500}{6}=\frac{3800-3500}{6}=\frac{300}{6}=50$

Hence, The cost of one bat is 500 Rs and the cost of one ball 50 Rs.

Q5 Form the pair of linear equations for the following problems and find their solution by substitution method. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

Answer:

Let the fixed charge is x and the per km charge is y.

Now According to the question

$x+10y=105.......(1)$

And

$x+15y=155.......(2)$

Now, From (1) we have,

$x=105-10y........(3)$

Substituting this value of x in (2), we have

$105-10y+15y=155$

$\Rightarrow 5y=155-105$

$\Rightarrow 5y=50$

$\Rightarrow y=10$

Now, Substituting this value in (3)

$x=105-10y=105-10(10)=105-100=5$

Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.

Now, Fair For 25 km :

$\Rightarrow x+25y=5+25(10)=5+250=255$

Hence fair for 25km is 255 Rs.

Q6 Form the pair of linear equations for the following problems and find their solution by substitution method. (v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$ . Find the fraction.

Answer:

Let the numerator of the fraction be x and denominator of the fraction is y

Now According to the question,

$\frac{x+2}{y+2}=\frac{9}{11}$

$\Rightarrow 11(x+2)=9(y+2)$

$\Rightarrow 11x+22=9y+18$

$\Rightarrow 11x-9y=-4...........(1)$

Also,

$\frac{x+3}{y+3}=\frac{5}{6}$

$\Rightarrow 6(x+3)=5(y+3)$

$\Rightarrow 6x+18=5y+15$

$\Rightarrow 6x-5y=-3...........(2)$

Now, From (1) we have

$y=\frac{11x+4}{9}.............(3)$

Substituting this value of y in (2)

$6x-5\left(\frac{11x+4}{9}\right)=-3$

$\Rightarrow 54x-55x-20=-27$

$\Rightarrow -x=20-27$

$\Rightarrow x=7$

Substituting this value of x in (3)

$y=\frac{11x+4}{9}=\frac{11(7)+4}{9}=\frac{81}{9}=9$

Hence the required fraction is

$\frac{x}{y}=\frac{7}{9}.$

Q7 Form the pair of linear equations for the following problems and find their solution by substitution method. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer:

Let x be the age of Jacob and y be the age of Jacob's son,

Now, According to the question

$x+5=3(y+5)$

$\Rightarrow x+5=3y+15$

$\Rightarrow x-3y=10..........(1)$

Also,

$x-5=7(y-5)$

$\Rightarrow x-5=7y-35$

$\Rightarrow x-7y=-30.........(2)$

Now,

From (1) we have,

$x=10+3y...........(3)$

Substituting this value of x in (2)

$10+3y-7y=-30$

$\Rightarrow -4y=-30-10$

$\Rightarrow 4y=40$

$\Rightarrow y=10$

Substituting this value of y in (3),

$x=10+3y=10+3(10)=10+30=40$

Hence, Present age of Jacob is 40 years and the present age of Jacob's son is 10 years.