NCERT Solutions For Class 10 Maths: Pair of Linear Equations in Two Variable

EXERCISE 3.1

Question 1: Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

Solution-

(i) Let $x$ be the number of boys and $y$ be the number of girls.
Total students: $x + y = 10$.
Girls 4 more than boys: $y = x + 4$, or $y - x = 4$.
The pair of linear equations is:
$x + y = 10$
$y - x = 4$
The solution must be obtained graphically by plotting the two lines and finding the coordinates of their single point of intersection.

(ii) Let $x$ be the cost of one pencil and $y$ be the cost of one pen.
First condition: $5x + 7y = 50$.
Second condition: $7x + 5y = 46$.
The pair of linear equations is:
$5x + 7y = 50$
$7x + 5y = 46$
The solution must be obtained graphically by identifying the point of intersection of the two lines, which represents the cost of one pencil and one pen.

Question 2: On comparing the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) $5x – 4y + 8 = 0$
$7x + 6y – 9 = 0$
(ii) $9x + 3y + 12 = 0$
$18x + 6y + 24 = 0$
(iii) $6x – 3y + 10 = 0$
$2x – y + 9 = 0$

Solution-

(i) $5x – 4y + 8 = 0$ and $7x + 6y – 9 = 0$
Ratios: $\frac{a_1}{a_2} = \frac{5}{7}$, $\frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}$.
Comparison: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Interpretation: The lines intersect at a single point.

(ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$
Ratios: $\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$.
Comparison: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Interpretation: The lines are coincident and have infinitely many solutions.

(iii) $6x – 3y + 10 = 0$ and $2x – y + 9 = 0$
Ratios: $\frac{a_1}{a_2} = \frac{6}{2} = 3$, $\frac{b_1}{b_2} = \frac{-3}{-1} = 3$, $\frac{c_1}{c_2} = \frac{10}{9}$.
Comparison: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Interpretation: The lines are parallel and have no solution.

Question 3: On comparing the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$, find out whether the following pair of linear equations are consistent, or inconsistent.
(i) $3x + 2y = 5 ; 2x – 3y = 7$
(ii) $2x – 3y = 8 ; 4x – 6y = 9$
(iii) $\frac{3}{2}x + \frac{5}{3}y = 7 ; 9x – 10y = 14$
(iv) $5x – 3y = 11 ; – 10x + 6y = –22$
(v) $\frac{4}{3}x + 2y = 8 ; 2x + 3y = 12$

Solution-

(i) $3x + 2y – 5 = 0$ and $2x – 3y – 7 = 0$
Comparison: $\frac{3}{2} \neq \frac{2}{-3}$. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair is consistent (unique solution).

(ii) $2x – 3y – 8 = 0$ and $4x – 6y – 9 = 0$
Comparison: $\frac{2}{4} = \frac{1}{2}$; $\frac{-3}{-6} = \frac{1}{2}$; $\frac{-8}{-9} = \frac{8}{9}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair is inconsistent (no solution).

(iii) $\frac{3}{2}x + \frac{5}{3}y = 7$ and $9x – 10y = 14$
Comparison: $\frac{a_1}{a_2} = \frac{3/2}{9} = \frac{1}{6}$; $\frac{b_1}{b_2} = \frac{5/3}{-10} = -\frac{1}{6}$. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair is consistent (unique solution).

(iv) $5x – 3y – 11 = 0$ and $-10x + 6y + 22 = 0$
Comparison: $\frac{5}{-10} = -\frac{1}{2}$; $\frac{-3}{6} = -\frac{1}{2}$; $\frac{-11}{22} = -\frac{1}{2}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair is consistent (dependent and infinitely many solutions).

(v) $\frac{4}{3}x + 2y = 8$ and $2x + 3y = 12$
Comparison: $\frac{4/3}{2} = \frac{2}{3}$; $\frac{2}{3}$; $\frac{-8}{-12} = \frac{2}{3}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair is consistent (dependent and infinitely many solutions).

Question 4: Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) $x + y = 5, 2x + 2y = 10$
(ii) $x – y = 8, 3x – 3y = 16$
(iii) $2x + y – 6 = 0, 4x – 2y – 4 = 0$
(iv) $2x – 2y – 2 = 0, 4x – 4y – 5 = 0$

Solution-

(i) $x + y – 5 = 0$ and $2x + 2y – 10 = 0$. Ratios: $\frac{1}{2} = \frac{1}{2} = \frac{-5}{-10}$.
The pair is consistent (coincident lines).
Graphical Solution: Since the lines are coincident, they have infinitely many solutions.

(ii) $x – y – 8 = 0$ and $3x – 3y – 16 = 0$. Ratios: $\frac{1}{3} = \frac{-1}{-3} \neq \frac{-8}{-16}$.
The pair is inconsistent (parallel lines).

(iii) $2x + y – 6 = 0$ and $4x – 2y – 4 = 0$. Ratios: $\frac{2}{4} = \frac{1}{2}$; $\frac{1}{-2} = -\frac{1}{2}$.
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair is consistent (intersecting lines).
Graphical Solution: We plot the lines and find the unique point of intersection. (Solution is $x=2, y=2$).

(iv) $2x – 2y – 2 = 0$ and $4x – 4y – 5 = 0$. Ratios: $\frac{2}{4} = \frac{1}{2}$; $\frac{-2}{-4} = \frac{1}{2}$; $\frac{-2}{-5} = \frac{2}{5}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair is inconsistent (parallel lines).

Question 5: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution-

Let the length be $x$ and the width be $y$.
Half perimeter: $x + y = 36$ (1).
Length is 4 m more than width: $x = y + 4$, or $x - y = 4$ (2).

Using Elimination/Substitution Method:
Adding (1) and (2): $(x+y) + (x-y) = 36 + 4$.
$2x = 40$, so $x = 20$.
Substituting $x=20$ into (1): $20 + y = 36$, so $y = 16$.
The dimensions of the garden are Length = 20 m and Width = 16 m.

Question 6: Given the linear equation $2x + 3y – 8 = 0$, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident lines

Solution-

(i) Intersecting lines ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$):
Example: $5x + 7y - 12 = 0$.

(ii) Parallel lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$):
Example: $4x + 6y + 1 = 0$.

(iii) Coincident lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$):
Example: $4x + 6y – 16 = 0$.

Question 7: Draw the graphs of the equations $x – y + 1 = 0$ and $3x + 2y – 12 = 0$. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution-

The coordinates of the vertices are determined by finding the intersection points of the lines.

1. Intersection of the two given lines:
   Solving $x - y = -1$ and $3x + 2y = 12$ yields the intersection point $(2, 3)$.
   (Verification: $2-3=-1$; $3(2) + 2(3) = 6+6=12$).

2. Intersection of $x – y + 1 = 0$ with the x-axis ($y=0$):
   $x - 0 + 1 = 0$, so $x = -1$. Point is $(-1, 0)$.

3. Intersection of $3x + 2y – 12 = 0$ with the x-axis ($y=0$):
   $3x + 0 - 12 = 0$, so $3x = 12$, $x = 4$. Point is $(4, 0)$.

The coordinates of the vertices of the triangle formed by these lines and the x-axis are (2, 3), (–1, 0), and (4, 0).

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EXERCISE 3.2

Question 1: Solve the following pair of linear equations by the substitution method.
(i) $x + y = 14$
$x – y = 4$
(ii) $s – t = 3$
$\frac{s}{3} + \frac{t}{2} = 6$
(iii) $3x – y = 3$
$9x – 3y = 9$
(iv) $0.2x + 0.3y = 1.3$
$0.4x + 0.5y = 2.3$
(v) $\sqrt{2}x + \sqrt{3}y = 0$
$\sqrt{3}x – \sqrt{8}y = 0$
(vi) $\frac{3x}{2} – \frac{5y}{3} = -2$
$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

Solution-

(i) $x + y = 14$ and $x – y = 4$. Solution: $x = 9, y = 5$.

(ii) $s – t = 3$ and $2s + 3t = 36$ (simplified form). Solution: $s = 9, t = 6$.

(iii) $3x – y = 3$ and $9x – 3y = 9$. Since $9x – 3y = 3(3x – y)$, the equations are coincident. $9 = 9$ is a true statement involving no variable. This pair has infinitely many solutions.

(iv) $2x + 3y = 13$ and $4x + 5y = 23$. Solution: $x = 2, y = 3$.

(v) $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x – \sqrt{8}y = 0$. Solution: $x = 0, y = 0$.

(vi) $9x – 10y = -12$ and $2x + 3y = 13$. Solution: $x = 2, y = 3$.

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

Solution-

We solve the pair $2x + 3y = 11$ (1) and $2x – 4y = -24$ (2).
Subtracting (2) from (1): $7y = 35$, so $y = 5$.
Substituting $y=5$ into (1): $2x + 3(5) = 11$, $2x = -4$, so $x = -2$.

The solution is $x = -2, y = 5$.

Substitute $x = -2$ and $y = 5$ into $y = mx + 3$:
$5 = m(-2) + 3$
$2 = -2m$
$m = -1$.

Question 3: Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for 3800. Later, she buys 3 bats and 5 balls for 1750. Find the cost of each bat and each ball.
(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 105 and for a journey of 15 km, the charge paid is 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(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.
(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?

Solution-

(i) Equations: $x - y = 26$ and $x = 3y$.
Solution: The numbers are 39 and 13.

(ii) Equations: $x + y = 180$ and $x = y + 18$.
Solution: The angles are 99 degrees and 81 degrees.

(iii) Let $x$ be bat cost, $y$ be ball cost.
Equations: $7x + 6y = 3800$ and $3x + 5y = 1750$.
Solution: Cost of each bat is 500** and cost of each ball is ** 50.

(iv) Let $x$ be fixed charge, $y$ be charge per km.
Equations: $x + 10y = 105$ and $x + 15y = 155$.
Solution: The fixed charge is 5** and the **charge per km is 10.
Charge for 25 km is $5 + 25(10) = \mathbf{` 255}$.

(v) Let the fraction be $\frac{x}{y}$.
Equations: $11x - 9y = -4$ and $6x - 5y = -3$.
Solution: The fraction is $\frac{7}{9}$.

(vi) Let $x$ be Jacob's age, $y$ be son's age.
Equations: $x - 3y = 10$ and $x - 7y = -30$.
Solution: Jacob's present age is 40 years and his son's present age is 10 years.

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EXERCISE 3.3

Question 1: Solve the following pair of linear equations by the elimination method and the substitution method:
(i) $x + y = 5$ and $2x – 3y = 4$
(ii) $3x + 4y = 10$ and $2x – 2y = 2$
(iii) $3x – 5y – 4 = 0$ and $9x = 2y + 7$
(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x – \frac{y}{3} = 3$

Solution-

(i) $x + y = 5$ (1), $2x – 3y = 4$ (2).
Elimination (Multiplying (1) by 3 and adding): $5x = 19$.
Substitution (From (1), $x = 5-y$): $2(5-y) - 3y = 4$.
Solution: $x = \frac{19}{5}, y = \frac{6}{5}$.

(ii) $3x + 4y = 10$ (1), $2x – 2y = 2$ (2).
Elimination (Multiplying (2) by 2 and adding): $7x = 14$.
Substitution (From (2), $x = 1+y$): $3(1+y) + 4y = 10$.
Solution: $x = 2, y = 1$.

(iii) $3x – 5y = 4$ (1), $9x – 2y = 7$ (2).
Elimination (Multiplying (1) by 3 and subtracting): $-13y = 5$.
Substitution (From (1), $x = \frac{5y+4}{3}$): $9(\frac{5y+4}{3}) - 2y = 7$.
Solution: $x = \frac{9}{13}, y = -\frac{5}{13}$.

(iv) $3x + 4y = -6$ (1), $3x – y = 9$ (2).
Elimination (Subtracting (2) from (1)): $5y = -15$.
Substitution (From (2), $x = \frac{9+y}{3}$): $3(\frac{9+y}{3}) + 4y = -6$.
Solution: $x = 2, y = -3$.

Question 2: Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw 2000. She asked the cashier to give her 50 and 100 notes only. Meena got 25 notes in all. Find how many notes of 50 and 100 she received. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution-

(i) Let the fraction be $\frac{x}{y}$. Equations: $x - y = -2$ and $2x - y = 1$.
Elimination (Subtracting): $x = 3$.
Solution: The fraction is $\frac{3}{5}$.

(ii) Let Nuri's age be $x$ and Sonu's age be $y$. Equations: $x - 3y = -10$ and $x - 2y = 10$.
Elimination (Subtracting): $y = 20$.
Solution: Nuri is 50 years old and Sonu is 20 years old.

(iii) Let the digits be $x$ and $y$. Original number is $10x + y$. Reversed number is $10y + x$.
Equations: $x + y = 9$ and $9(10x + y) = 2(10y + x)$, simplified to $88x - 11y = 0$, or $8x - y = 0$.
Elimination (Adding $x+y=9$ and $8x-y=0$): $9x = 9$.
Solution: The number is 18.

(iv) Let $x$ be the number of 50 notes, $y$ be the number of 100 notes.
Equations: $x + y = 25$ and $50x + 100y = 2000$ (simplified to $x + 2y = 40$).
Elimination (Subtracting): $y = 15$.
Solution: Meena received 10 notes of 50** and **15 notes of 100.

(v) Let $x$ be the fixed charge (3 days), $y$ be the additional daily charge.
Saritha (7 days): $x + 4y = 27$ (1).
Susy (5 days): $x + 2y = 21$ (2).
Elimination (Subtracting (2) from (1)): $2y = 6$.
Solution: The fixed charge is 15** and the **charge for each extra day is 3.