Class 10 NCERT Solutions
Chapter 3: Pair of Linear Equations in Two Variables
Master the graphical method, the substitution and elimination techniques, and the consistency of linear equations with our step-by-step logic.
Exercise 3.1
1(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.
Let number of boys $= x$ and number of girls $= y$
Setting up the equations based on the given conditions:
$x + y = 10$ …(i)
$y = x + 4$ …(ii)
To find the solution, we plot both equations on a graph.
Some points on line (i) are:
$\begin{array}{|c|c|c|c|} \hline x & 4 & 3 & 5 \\ \hline y & 6 & 7 & 5 \\ \hline \end{array}$
Some points on line (ii) are:
$\begin{array}{|c|c|c|c|} \hline x & 0 & 3 & 2 \\ \hline y & 4 & 7 & 6 \\ \hline \end{array}$
The two lines meet at $\mathrm{A}(3,7)$, i.e., $x = 3,\ y = 7$.
Hence, number of boys $= 3$, number of girls $= 7$.
Setting up the equations based on the given conditions:
$x + y = 10$ …(i)
$y = x + 4$ …(ii)
To find the solution, we plot both equations on a graph.
Some points on line (i) are:
$\begin{array}{|c|c|c|c|} \hline x & 4 & 3 & 5 \\ \hline y & 6 & 7 & 5 \\ \hline \end{array}$
Some points on line (ii) are:
$\begin{array}{|c|c|c|c|} \hline x & 0 & 3 & 2 \\ \hline y & 4 & 7 & 6 \\ \hline \end{array}$
The two lines meet at $\mathrm{A}(3,7)$, i.e., $x = 3,\ y = 7$.
Hence, number of boys $= 3$, number of girls $= 7$.
1(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.
Let cost of one pencil $= ₹\, x$ and cost of one pen $= ₹\, y$
Translating the given information into a pair of linear equations:
$5x + 7y = 50$ …(i)
$7x + 5y = 46$ …(ii)
Some points on line (i) are:
$\begin{array}{|c|c|c|} \hline x & 10 & 3 \\ \hline y & 0 & 5 \\ \hline \end{array}$
Some points on line (ii) are:
$\begin{array}{|c|c|c|} \hline x & 0 & 5 \\ \hline y & 9.2 & 2.2 \\ \hline \end{array}$
Plotting the two equations on the graph, the two lines intersect at the point $\mathrm{A}(3,5)$, i.e., $x = 3,\ y = 5$.
Hence, cost of one pencil $= ₹\,3$ and cost of one pen $= ₹\,5$.
Translating the given information into a pair of linear equations:
$5x + 7y = 50$ …(i)
$7x + 5y = 46$ …(ii)
Some points on line (i) are:
$\begin{array}{|c|c|c|} \hline x & 10 & 3 \\ \hline y & 0 & 5 \\ \hline \end{array}$
Some points on line (ii) are:
$\begin{array}{|c|c|c|} \hline x & 0 & 5 \\ \hline y & 9.2 & 2.2 \\ \hline \end{array}$
Plotting the two equations on the graph, the two lines intersect at the point $\mathrm{A}(3,5)$, i.e., $x = 3,\ y = 5$.
Hence, cost of one pencil $= ₹\,3$ and cost of one pen $= ₹\,5$.
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) $6x – 3y + 10 = 0;\ 2x – y + 9 = 0$
(i) $5x – 4y + 8 = 0;\ 7x + 6y – 9 = 0$
(ii) $9x + 3y + 12 = 0;\ 18x + 6y + 24 = 0$
(iii) $6x – 3y + 10 = 0;\ 2x – y + 9 = 0$
(i) From $5x – 4y + 8 = 0$: $a_1 = 5,\ b_1 = -4,\ c_1 = 8$
From $7x + 6y – 9 = 0$: $a_2 = 7,\ b_2 = 6,\ c_2 = -9$
Comparing: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e., $\frac{5}{7} \neq \frac{-4}{6}$
Hence, the lines intersect at a point.
(ii) From $9x + 3y + 12 = 0$: $a_1 = 9,\ b_1 = 3,\ c_1 = 12$
From $18x + 6y + 24 = 0$: $a_2 = 18,\ b_2 = 6,\ c_2 = 24$
Comparing: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$, i.e., $\frac{9}{18} = \frac{3}{6} = \frac{12}{24} = \frac{1}{2}$
Hence, the lines are coincident.
(iii) From $6x – 3y + 10 = 0$: $a_1 = 6,\ b_1 = -3,\ c_1 = 10$
From $2x – y + 9 = 0$: $a_2 = 2,\ b_2 = -1,\ c_2 = 9$
Comparing: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, i.e., $\frac{6}{2} = \frac{-3}{-1} \neq \frac{10}{9}$
Hence, the lines are parallel.
From $7x + 6y – 9 = 0$: $a_2 = 7,\ b_2 = 6,\ c_2 = -9$
Comparing: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e., $\frac{5}{7} \neq \frac{-4}{6}$
Hence, the lines intersect at a point.
(ii) From $9x + 3y + 12 = 0$: $a_1 = 9,\ b_1 = 3,\ c_1 = 12$
From $18x + 6y + 24 = 0$: $a_2 = 18,\ b_2 = 6,\ c_2 = 24$
Comparing: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$, i.e., $\frac{9}{18} = \frac{3}{6} = \frac{12}{24} = \frac{1}{2}$
Hence, the lines are coincident.
(iii) From $6x – 3y + 10 = 0$: $a_1 = 6,\ b_1 = -3,\ c_1 = 10$
From $2x – y + 9 = 0$: $a_2 = 2,\ b_2 = -1,\ c_2 = 9$
Comparing: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, i.e., $\frac{6}{2} = \frac{-3}{-1} \neq \frac{10}{9}$
Hence, the lines are parallel.
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 pairs 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) $\frac{4}{3}x + 2y = 8;\ 2x + 3y = 12$
(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) $\frac{4}{3}x + 2y = 8;\ 2x + 3y = 12$
(i) $a_1 = 3,\ b_1 = 2,\ c_1 = -5$; $a_2 = 2,\ b_2 = -3,\ c_2 = -7$
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e., $\frac{3}{2} \neq \frac{2}{-3}$
Hence, the pair is consistent.
(ii) $a_1 = 2,\ b_1 = -3,\ c_1 = -8$; $a_2 = 4,\ b_2 = -6,\ c_2 = -9$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, i.e., $\frac{2}{4} = \frac{-3}{-6} \neq \frac{-8}{-9}$
Hence, the pair is inconsistent.
(iii) Rewriting $\frac{3}{2}x + \frac{5}{3}y = 7$ gives $9x + 10y – 42 = 0$.
$a_1 = 9,\ b_1 = 10,\ c_1 = -42$; $a_2 = 9,\ b_2 = -10,\ c_2 = -14$
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e., $\frac{9}{9} \neq \frac{10}{-10}$
Hence, the pair is consistent.
(iv) $a_1 = 5,\ b_1 = -3,\ c_1 = -11$; $a_2 = -10,\ b_2 = 6,\ c_2 = 22$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, i.e., $\frac{5}{-10} = \frac{-3}{6} = \frac{-11}{22}$
Hence, the pair is dependent (consistent).
(v) Rewriting $\frac{4}{3}x + 2y = 8$ gives $4x + 6y – 24 = 0$.
$a_1 = 4,\ b_1 = 6,\ c_1 = -24$; $a_2 = 2,\ b_2 = 3,\ c_2 = -12$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, i.e., $\frac{4}{2} = \frac{6}{3} = \frac{-24}{-12}$
Hence, the pair is dependent (consistent).
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e., $\frac{3}{2} \neq \frac{2}{-3}$
Hence, the pair is consistent.
(ii) $a_1 = 2,\ b_1 = -3,\ c_1 = -8$; $a_2 = 4,\ b_2 = -6,\ c_2 = -9$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, i.e., $\frac{2}{4} = \frac{-3}{-6} \neq \frac{-8}{-9}$
Hence, the pair is inconsistent.
(iii) Rewriting $\frac{3}{2}x + \frac{5}{3}y = 7$ gives $9x + 10y – 42 = 0$.
$a_1 = 9,\ b_1 = 10,\ c_1 = -42$; $a_2 = 9,\ b_2 = -10,\ c_2 = -14$
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e., $\frac{9}{9} \neq \frac{10}{-10}$
Hence, the pair is consistent.
(iv) $a_1 = 5,\ b_1 = -3,\ c_1 = -11$; $a_2 = -10,\ b_2 = 6,\ c_2 = 22$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, i.e., $\frac{5}{-10} = \frac{-3}{6} = \frac{-11}{22}$
Hence, the pair is dependent (consistent).
(v) Rewriting $\frac{4}{3}x + 2y = 8$ gives $4x + 6y – 24 = 0$.
$a_1 = 4,\ b_1 = 6,\ c_1 = -24$; $a_2 = 2,\ b_2 = 3,\ c_2 = -12$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, i.e., $\frac{4}{2} = \frac{6}{3} = \frac{-24}{-12}$
Hence, the pair is dependent (consistent).
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) $2x – 2y – 2 = 0,\ 4x – 4y – 5 = 0$
(i) $x + y = 5,\ 2x + 2y = 10$
(ii) $x – y = 8,\ 3x – 3y = 16$
(iii) $2x + y – 6 = 0,\ 4x – 2y – 4 = 0$
(iv) $2x – 2y – 2 = 0,\ 4x – 4y – 5 = 0$
(i) $a_1 = 1,\ b_1 = 1,\ c_1 = -5$; $a_2 = 2,\ b_2 = 2,\ c_2 = -10$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$
The pair is consistent (infinitely many solutions).
Some points for $x + y = 5$: $\begin{array}{|c|c|c|c|} \hline x & 0 & 5 & 2 \\ \hline y & 5 & 0 & 3 \\ \hline \end{array}$
From the graph, the lines are overlapping. The pair has infinitely many solutions, e.g., $x = 2,\ y = 3$ and $x = 0,\ y = 5$.
(ii) $a_1 = 1,\ b_1 = -1,\ c_1 = -8$; $a_2 = 3,\ b_2 = -3,\ c_2 = -16$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Hence, the pair is inconsistent (no solution).
(iii) $a_1 = 2,\ b_1 = 1,\ c_1 = -6$; $a_2 = 4,\ b_2 = -2,\ c_2 = -4$
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e., $\frac{2}{4} \neq \frac{1}{-2}$
The pair is consistent (unique solution).
Some points for $2x + y – 6 = 0$: $\begin{array}{|c|c|c|c|} \hline x & 0 & 2 & 3 \\ \hline y & 6 & 2 & 0 \\ \hline \end{array}$
Some points for $4x – 2y – 4 = 0$: $\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 3 \\ \hline y & -2 & 0 & 4 \\ \hline \end{array}$
The two lines meet at $\mathrm{A}(2,2)$. Hence, solution is $x = 2,\ y = 2$.
(iv) $a_1 = 2,\ b_1 = -2,\ c_1 = -2$; $a_2 = 4,\ b_2 = -4,\ c_2 = -5$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, i.e., $\frac{2}{4} = \frac{-2}{-4} \neq \frac{-2}{-5}$
Hence, the pair is inconsistent (no solution).
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$
The pair is consistent (infinitely many solutions).
Some points for $x + y = 5$: $\begin{array}{|c|c|c|c|} \hline x & 0 & 5 & 2 \\ \hline y & 5 & 0 & 3 \\ \hline \end{array}$
From the graph, the lines are overlapping. The pair has infinitely many solutions, e.g., $x = 2,\ y = 3$ and $x = 0,\ y = 5$.
(ii) $a_1 = 1,\ b_1 = -1,\ c_1 = -8$; $a_2 = 3,\ b_2 = -3,\ c_2 = -16$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Hence, the pair is inconsistent (no solution).
(iii) $a_1 = 2,\ b_1 = 1,\ c_1 = -6$; $a_2 = 4,\ b_2 = -2,\ c_2 = -4$
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e., $\frac{2}{4} \neq \frac{1}{-2}$
The pair is consistent (unique solution).
Some points for $2x + y – 6 = 0$: $\begin{array}{|c|c|c|c|} \hline x & 0 & 2 & 3 \\ \hline y & 6 & 2 & 0 \\ \hline \end{array}$
Some points for $4x – 2y – 4 = 0$: $\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 3 \\ \hline y & -2 & 0 & 4 \\ \hline \end{array}$
The two lines meet at $\mathrm{A}(2,2)$. Hence, solution is $x = 2,\ y = 2$.
(iv) $a_1 = 2,\ b_1 = -2,\ c_1 = -2$; $a_2 = 4,\ b_2 = -4,\ c_2 = -5$
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, i.e., $\frac{2}{4} = \frac{-2}{-4} \neq \frac{-2}{-5}$
Hence, the pair is inconsistent (no solution).
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.
Let length of the garden $= x$ m and breadth $= y$ m.
Translating both conditions into equations:
$x + y = 36$ …(i)
$x = y + 4$ …(ii)
Solving (i) and (ii), we get $x = 20,\ y = 16$.
Therefore, length $= 20$ m and breadth $= 16$ m.
Translating both conditions into equations:
$x + y = 36$ …(i)
$x = y + 4$ …(ii)
Solving (i) and (ii), we get $x = 20,\ y = 16$.
Therefore, length $= 20$ m and breadth $= 16$ m.
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
(i) intersecting lines (ii) parallel lines (iii) coincident lines
Given linear equation: $2x + 3y – 8 = 0$
(i) Intersecting lines: We need $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Consider the equation $5x + 3y – 1 = 0$.
Check: $\frac{2}{5} \neq \frac{3}{3}$ ✓
(ii) Parallel lines: We need $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Consider the equation $4x + 6y + 1 = 0$.
Check: $\frac{2}{4} = \frac{3}{6} \neq \frac{-8}{1}$ ✓
(iii) Coincident lines: We need $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Consider the equation $4x + 6y – 16 = 0$.
Check: $\frac{2}{4} = \frac{3}{6} = \frac{-8}{-16} = \frac{1}{2}$ ✓
(i) Intersecting lines: We need $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Consider the equation $5x + 3y – 1 = 0$.
Check: $\frac{2}{5} \neq \frac{3}{3}$ ✓
(ii) Parallel lines: We need $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Consider the equation $4x + 6y + 1 = 0$.
Check: $\frac{2}{4} = \frac{3}{6} \neq \frac{-8}{1}$ ✓
(iii) Coincident lines: We need $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Consider the equation $4x + 6y – 16 = 0$.
Check: $\frac{2}{4} = \frac{3}{6} = \frac{-8}{-16} = \frac{1}{2}$ ✓
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.
First equation: $x – y + 1 = 0 \Rightarrow y = x + 1$
Some points on graph:
$\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 2 \\ \hline y & 1 & 2 & 3 \\ \hline \end{array}$
Second equation: $3x + 2y – 12 = 0 \Rightarrow y = \frac{12 – 3x}{2}$
Some points on graph:
$\begin{array}{|c|c|c|c|} \hline x & 0 & 2 & 4 \\ \hline y & 6 & 3 & 0 \\ \hline \end{array}$
Plotting these points on a graph, the coordinates of the vertices of triangle ABC are:
$\mathrm{A}(2,3),\ \mathrm{B}(-1,0),\ \mathrm{C}(4,0)$.
Some points on graph:
$\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 2 \\ \hline y & 1 & 2 & 3 \\ \hline \end{array}$
Second equation: $3x + 2y – 12 = 0 \Rightarrow y = \frac{12 – 3x}{2}$
Some points on graph:
$\begin{array}{|c|c|c|c|} \hline x & 0 & 2 & 4 \\ \hline y & 6 & 3 & 0 \\ \hline \end{array}$
Plotting these points on a graph, the coordinates of the vertices of triangle ABC are:
$\mathrm{A}(2,3),\ \mathrm{B}(-1,0),\ \mathrm{C}(4,0)$.
Exercise 3.2
1. Solve the following pairs 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) $\frac{3x}{2} – \frac{5y}{3} = -2;\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
(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) $\frac{3x}{2} – \frac{5y}{3} = -2;\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
(i) From $x – y = 4$: $x = y + 4$
Substituting in $x + y = 14$: $y + 4 + y = 14 \Rightarrow 2y = 10 \Rightarrow y = 5$
Then $x = 9$.
$\therefore$ Solution: $x = 9,\ y = 5$.
(ii) From $s – t = 3$: $s = t + 3$
Substituting in $\frac{s}{3} + \frac{t}{2} = 6$:
$\frac{t+3}{3} + \frac{t}{2} = 6 \Rightarrow 2t + 6 + 3t = 36 \Rightarrow 5t = 30 \Rightarrow t = 6$
Then $s = 9$.
$\therefore$ Solution: $s = 9,\ t = 6$.
(iii) From $3x – y = 3$: $y = 3x – 3$
Substituting in $9x – 3y = 9$: $9x – 3(3x – 3) = 9 \Rightarrow 9 = 9$ (always true).
Hence, infinitely many solutions. One such solution: $x = 1,\ y = 0$. General solution: $y = 3x – 3$.
(iv) From $0.2x + 0.3y = 1.3$: $y = \frac{1.3 – 0.2x}{0.3}$
Substituting in $0.4x + 0.5y = 2.3$:
$0.4x + 0.5\!\left(\frac{1.3 – 0.2x}{0.3}\right) = 2.3 \Rightarrow 0.12x + 0.65 – 0.1x = 0.69 \Rightarrow 0.02x = 0.04 \Rightarrow x = 2$
Then $y = \frac{1.3 – 0.4}{0.3} = \frac{0.9}{0.3} = 3$.
$\therefore$ Solution: $x = 2,\ y = 3$.
(v) From $\sqrt{2}\,x + \sqrt{3}\,y = 0$: $y = -\dfrac{\sqrt{2}}{\sqrt{3}}\,x$
Substituting in $\sqrt{3}\,x – \sqrt{8}\,y = 0$:
$\sqrt{3}\,x – \sqrt{8}\!\left(\dfrac{-\sqrt{2}}{\sqrt{3}}\,x\right) = 0 \Rightarrow 3x + 4x = 0 \Rightarrow 7x = 0 \Rightarrow x = 0$
Then $y = 0$.
$\therefore$ Solution: $x = 0,\ y = 0$.
(vi) From $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$: $y = 2\!\left(\frac{13}{6} – \frac{x}{3}\right)$
Substituting in $\frac{3x}{2} – \frac{5y}{3} = -2$:
$\frac{3x}{2} – \frac{5}{3} \times 2\!\left(\frac{13}{6} – \frac{x}{3}\right) = -2$
$\Rightarrow \frac{3x}{2} – \frac{65}{9} + \frac{10x}{9} = -2$
$\Rightarrow 27x – 130 + 20x = -36 \Rightarrow 47x = 94 \Rightarrow x = 2$
Then $y = 2\!\left(\frac{13}{6} – \frac{2}{3}\right) = 2 \times \frac{9}{6} = 3$.
$\therefore$ Solution: $x = 2,\ y = 3$.
Substituting in $x + y = 14$: $y + 4 + y = 14 \Rightarrow 2y = 10 \Rightarrow y = 5$
Then $x = 9$.
$\therefore$ Solution: $x = 9,\ y = 5$.
(ii) From $s – t = 3$: $s = t + 3$
Substituting in $\frac{s}{3} + \frac{t}{2} = 6$:
$\frac{t+3}{3} + \frac{t}{2} = 6 \Rightarrow 2t + 6 + 3t = 36 \Rightarrow 5t = 30 \Rightarrow t = 6$
Then $s = 9$.
$\therefore$ Solution: $s = 9,\ t = 6$.
(iii) From $3x – y = 3$: $y = 3x – 3$
Substituting in $9x – 3y = 9$: $9x – 3(3x – 3) = 9 \Rightarrow 9 = 9$ (always true).
Hence, infinitely many solutions. One such solution: $x = 1,\ y = 0$. General solution: $y = 3x – 3$.
(iv) From $0.2x + 0.3y = 1.3$: $y = \frac{1.3 – 0.2x}{0.3}$
Substituting in $0.4x + 0.5y = 2.3$:
$0.4x + 0.5\!\left(\frac{1.3 – 0.2x}{0.3}\right) = 2.3 \Rightarrow 0.12x + 0.65 – 0.1x = 0.69 \Rightarrow 0.02x = 0.04 \Rightarrow x = 2$
Then $y = \frac{1.3 – 0.4}{0.3} = \frac{0.9}{0.3} = 3$.
$\therefore$ Solution: $x = 2,\ y = 3$.
(v) From $\sqrt{2}\,x + \sqrt{3}\,y = 0$: $y = -\dfrac{\sqrt{2}}{\sqrt{3}}\,x$
Substituting in $\sqrt{3}\,x – \sqrt{8}\,y = 0$:
$\sqrt{3}\,x – \sqrt{8}\!\left(\dfrac{-\sqrt{2}}{\sqrt{3}}\,x\right) = 0 \Rightarrow 3x + 4x = 0 \Rightarrow 7x = 0 \Rightarrow x = 0$
Then $y = 0$.
$\therefore$ Solution: $x = 0,\ y = 0$.
(vi) From $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$: $y = 2\!\left(\frac{13}{6} – \frac{x}{3}\right)$
Substituting in $\frac{3x}{2} – \frac{5y}{3} = -2$:
$\frac{3x}{2} – \frac{5}{3} \times 2\!\left(\frac{13}{6} – \frac{x}{3}\right) = -2$
$\Rightarrow \frac{3x}{2} – \frac{65}{9} + \frac{10x}{9} = -2$
$\Rightarrow 27x – 130 + 20x = -36 \Rightarrow 47x = 94 \Rightarrow x = 2$
Then $y = 2\!\left(\frac{13}{6} – \frac{2}{3}\right) = 2 \times \frac{9}{6} = 3$.
$\therefore$ Solution: $x = 2,\ y = 3$.
2. Solve $2x + 3y = 11$ and $2x – 4y = -24$ and hence find the value of $m$ for which $y = mx + 3$.
From $2x + 3y = 11$: $y = \frac{11 – 2x}{3}$
Substituting in $2x – 4y = -24$:
$2x – 4\!\left(\frac{11 – 2x}{3}\right) = -24$
$\Rightarrow 6x – 44 + 8x = -72 \Rightarrow 14x = -28 \Rightarrow x = -2$
Then $y = \frac{11 + 4}{3} = \frac{15}{3} = 5$.
Hence solution is $x = -2,\ y = 5$.
Substituting in $y = mx + 3$: $5 = m(-2) + 3 \Rightarrow -2m = 2 \Rightarrow m = -1$.
Substituting in $2x – 4y = -24$:
$2x – 4\!\left(\frac{11 – 2x}{3}\right) = -24$
$\Rightarrow 6x – 44 + 8x = -72 \Rightarrow 14x = -28 \Rightarrow x = -2$
Then $y = \frac{11 + 4}{3} = \frac{15}{3} = 5$.
Hence solution is $x = -2,\ y = 5$.
Substituting in $y = mx + 3$: $5 = m(-2) + 3 \Rightarrow -2m = 2 \Rightarrow m = -1$.
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 10 km the charge is ₹105 and for 15 km the charge is ₹155. What are the fixed charges and the charge per km? How much for 25 km?
(v) A fraction becomes $\frac{9}{11}$ if 2 is added to both numerator and denominator. If 3 is added to both, 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?
(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 10 km the charge is ₹105 and for 15 km the charge is ₹155. What are the fixed charges and the charge per km? How much for 25 km?
(v) A fraction becomes $\frac{9}{11}$ if 2 is added to both numerator and denominator. If 3 is added to both, 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?
(i) Let numbers be $x$ and $y$ $(x > y)$.
$x – y = 26$ and $x = 3y$
$\Rightarrow 3y – y = 26 \Rightarrow y = 13,\ x = 39$.
The numbers are 39 and 13.
(ii) Let larger angle $= x$, smaller $= y$.
$x + y = 180°$ and $x – y = 18°$
Solving: $x = 99°,\ y = 81°$.
Larger angle $= 99°$, smaller angle $= 81°$.
(iii) Let cost of each bat $= ₹\,x$, each ball $= ₹\,y$.
$7x + 6y = 3800$ and $3x + 5y = 1750$
From second equation: $y = \frac{1750 – 3x}{5}$
Substituting: $35x + 10500 – 18x = 19000 \Rightarrow 17x = 8500 \Rightarrow x = 500$
Then $y = \frac{1750 – 1500}{5} = 50$.
Cost of each bat $= ₹\,500$, each ball $= ₹\,50$.
(iv) Let fixed charge $= ₹\,x$, charge per km $= ₹\,y$.
$x + 10y = 105$ and $x + 15y = 155$
Solving: $x = 5,\ y = 10$.
Fixed charge $= ₹\,5$, charge per km $= ₹\,10$.
Charge for 25 km $= ₹(5 + 25 \times 10) = ₹\,255$.
(v) Let the fraction be $\frac{x}{y}$.
$\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x – 9y = -4$ …(i)
$\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x – 5y = -3$ …(ii)
Solving: $x = 7,\ y = 9$.
Hence, the required fraction is $\dfrac{7}{9}$.
(vi) Let Jacob’s present age $= x$ years, son’s $= y$ years.
$x + 5 = 3(y + 5) \Rightarrow x – 3y = 10$ …(i)
$x – 5 = 7(y – 5) \Rightarrow x – 7y = -30$ …(ii)
Solving: $x = 40,\ y = 10$.
Jacob’s present age $= 40$ years, son’s present age $= 10$ years.
$x – y = 26$ and $x = 3y$
$\Rightarrow 3y – y = 26 \Rightarrow y = 13,\ x = 39$.
The numbers are 39 and 13.
(ii) Let larger angle $= x$, smaller $= y$.
$x + y = 180°$ and $x – y = 18°$
Solving: $x = 99°,\ y = 81°$.
Larger angle $= 99°$, smaller angle $= 81°$.
(iii) Let cost of each bat $= ₹\,x$, each ball $= ₹\,y$.
$7x + 6y = 3800$ and $3x + 5y = 1750$
From second equation: $y = \frac{1750 – 3x}{5}$
Substituting: $35x + 10500 – 18x = 19000 \Rightarrow 17x = 8500 \Rightarrow x = 500$
Then $y = \frac{1750 – 1500}{5} = 50$.
Cost of each bat $= ₹\,500$, each ball $= ₹\,50$.
(iv) Let fixed charge $= ₹\,x$, charge per km $= ₹\,y$.
$x + 10y = 105$ and $x + 15y = 155$
Solving: $x = 5,\ y = 10$.
Fixed charge $= ₹\,5$, charge per km $= ₹\,10$.
Charge for 25 km $= ₹(5 + 25 \times 10) = ₹\,255$.
(v) Let the fraction be $\frac{x}{y}$.
$\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x – 9y = -4$ …(i)
$\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x – 5y = -3$ …(ii)
Solving: $x = 7,\ y = 9$.
Hence, the required fraction is $\dfrac{7}{9}$.
(vi) Let Jacob’s present age $= x$ years, son’s $= y$ years.
$x + 5 = 3(y + 5) \Rightarrow x – 3y = 10$ …(i)
$x – 5 = 7(y – 5) \Rightarrow x – 7y = -30$ …(ii)
Solving: $x = 40,\ y = 10$.
Jacob’s present age $= 40$ years, son’s present age $= 10$ years.
Exercise 3.3
1. Solve the following pairs 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 $x – \frac{y}{3} = 3$
(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 $x – \frac{y}{3} = 3$
(i) Elimination Method:
Multiplying $x + y = 5$ by 3 and adding $2x – 3y = 4$:
$3x + 3y = 15$
$2x – 3y = 4$
Adding: $5x = 19 \Rightarrow x = \frac{19}{5}$
Substituting: $y = 5 – \frac{19}{5} = \frac{6}{5}$
Substitution Method:
From (i): $y = 5 – x$. Substituting in (ii): $2x – 3(5 – x) = 4 \Rightarrow 5x = 19 \Rightarrow x = \frac{19}{5}$
Then $y = \frac{6}{5}$.
$\therefore$ Solution: $x = \dfrac{19}{5},\ y = \dfrac{6}{5}$.
(ii) Elimination Method:
Multiplying $2x – 2y = 2$ by 2 and adding to $3x + 4y = 10$:
$3x + 4y = 10$
$4x – 4y = 4$
Adding: $7x = 14 \Rightarrow x = 2$
Then $6 + 4y = 10 \Rightarrow y = 1$.
$\therefore$ Solution: $x = 2,\ y = 1$.
(iii) Elimination Method:
Multiplying $3x – 5y – 4 = 0$ by 3 and subtracting $9x – 2y – 7 = 0$:
$9x – 15y – 12 = 0$
$9x – 2y – 7 = 0$
Subtracting: $-13y – 5 = 0 \Rightarrow y = -\dfrac{5}{13}$
Substituting: $3x + \frac{25}{13} – 4 = 0 \Rightarrow 3x = \frac{27}{13} \Rightarrow x = \frac{9}{13}$
$\therefore$ Solution: $x = \dfrac{9}{13},\ y = -\dfrac{5}{13}$.
(iv) Elimination Method:
$\frac{x}{2} + \frac{2y}{3} = -1 \Rightarrow 3x + 4y = -6$ …(i)
$x – \frac{y}{3} = 3 \Rightarrow 3x – y = 9$ …(ii)
Subtracting (ii) from (i): $5y = -15 \Rightarrow y = -3$
Substituting: $3x + 3 = 9 \Rightarrow x = 2$
Substitution Method:
From (ii): $y = 3x – 9$. Substituting in (i):
$3x + 4(3x – 9) = -6 \Rightarrow 15x = 30 \Rightarrow x = 2$
Then $y = 6 – 9 = -3$.
$\therefore$ Solution: $x = 2,\ y = -3$.
Multiplying $x + y = 5$ by 3 and adding $2x – 3y = 4$:
$3x + 3y = 15$
$2x – 3y = 4$
Adding: $5x = 19 \Rightarrow x = \frac{19}{5}$
Substituting: $y = 5 – \frac{19}{5} = \frac{6}{5}$
Substitution Method:
From (i): $y = 5 – x$. Substituting in (ii): $2x – 3(5 – x) = 4 \Rightarrow 5x = 19 \Rightarrow x = \frac{19}{5}$
Then $y = \frac{6}{5}$.
$\therefore$ Solution: $x = \dfrac{19}{5},\ y = \dfrac{6}{5}$.
(ii) Elimination Method:
Multiplying $2x – 2y = 2$ by 2 and adding to $3x + 4y = 10$:
$3x + 4y = 10$
$4x – 4y = 4$
Adding: $7x = 14 \Rightarrow x = 2$
Then $6 + 4y = 10 \Rightarrow y = 1$.
$\therefore$ Solution: $x = 2,\ y = 1$.
(iii) Elimination Method:
Multiplying $3x – 5y – 4 = 0$ by 3 and subtracting $9x – 2y – 7 = 0$:
$9x – 15y – 12 = 0$
$9x – 2y – 7 = 0$
Subtracting: $-13y – 5 = 0 \Rightarrow y = -\dfrac{5}{13}$
Substituting: $3x + \frac{25}{13} – 4 = 0 \Rightarrow 3x = \frac{27}{13} \Rightarrow x = \frac{9}{13}$
$\therefore$ Solution: $x = \dfrac{9}{13},\ y = -\dfrac{5}{13}$.
(iv) Elimination Method:
$\frac{x}{2} + \frac{2y}{3} = -1 \Rightarrow 3x + 4y = -6$ …(i)
$x – \frac{y}{3} = 3 \Rightarrow 3x – y = 9$ …(ii)
Subtracting (ii) from (i): $5y = -15 \Rightarrow y = -3$
Substituting: $3x + 3 = 9 \Rightarrow x = 2$
Substitution Method:
From (ii): $y = 3x – 9$. Substituting in (i):
$3x + 4(3x – 9) = -6 \Rightarrow 15x = 30 \Rightarrow x = 2$
Then $y = 6 – 9 = -3$.
$\therefore$ Solution: $x = 2,\ y = -3$.
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.
(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.
(i) Let the fraction be $\frac{x}{y}$.
$\frac{x+1}{y-1} = 1 \Rightarrow x – y = -2$ …(i)
$\frac{x}{y+1} = \frac{1}{2} \Rightarrow 2x – y = 1$ …(ii)
Solving: $x = 3,\ y = 5$.
Hence, the required fraction is $\dfrac{3}{5}$.
(ii) Let Nuri’s present age $= x$ years, Sonu’s $= y$ years.
$x – 5 = 3(y – 5) \Rightarrow x – 3y = -10$ …(i)
$x + 10 = 2(y + 10) \Rightarrow x – 2y = 10$ …(ii)
Solving: $x = 50,\ y = 20$.
Nuri’s age $= 50$ years, Sonu’s age $= 20$ years.
(iii) Let unit’s digit $= y$, ten’s digit $= x$. Number $= 10x + y$.
$x + y = 9$ …(i)
$9(10x + y) = 2(10y + x) \Rightarrow 88x – 11y = 0 \Rightarrow 8x – y = 0$ …(ii)
Adding (i) and (ii): $9x = 9 \Rightarrow x = 1$, then $y = 8$.
Hence, the number is 18.
(iv) Let number of ₹50 notes $= x$, ₹100 notes $= y$.
$x + y = 25$ …(i)
$50x + 100y = 2000 \Rightarrow x + 2y = 40$ …(ii)
Subtracting (i) from (ii): $y = 15$, then $x = 10$.
Number of ₹50 notes $= 10$, ₹100 notes $= 15$.
(v) Let fixed charge $= ₹\,x$, additional charge per day $= ₹\,y$.
$x + 4y = 27$ …(i)
$x + 2y = 21$ …(ii)
Subtracting (ii) from (i): $2y = 6 \Rightarrow y = 3$, then $x = 15$.
Fixed charge for first three days $= ₹\,15$, additional charge per day $= ₹\,3$.
$\frac{x+1}{y-1} = 1 \Rightarrow x – y = -2$ …(i)
$\frac{x}{y+1} = \frac{1}{2} \Rightarrow 2x – y = 1$ …(ii)
Solving: $x = 3,\ y = 5$.
Hence, the required fraction is $\dfrac{3}{5}$.
(ii) Let Nuri’s present age $= x$ years, Sonu’s $= y$ years.
$x – 5 = 3(y – 5) \Rightarrow x – 3y = -10$ …(i)
$x + 10 = 2(y + 10) \Rightarrow x – 2y = 10$ …(ii)
Solving: $x = 50,\ y = 20$.
Nuri’s age $= 50$ years, Sonu’s age $= 20$ years.
(iii) Let unit’s digit $= y$, ten’s digit $= x$. Number $= 10x + y$.
$x + y = 9$ …(i)
$9(10x + y) = 2(10y + x) \Rightarrow 88x – 11y = 0 \Rightarrow 8x – y = 0$ …(ii)
Adding (i) and (ii): $9x = 9 \Rightarrow x = 1$, then $y = 8$.
Hence, the number is 18.
(iv) Let number of ₹50 notes $= x$, ₹100 notes $= y$.
$x + y = 25$ …(i)
$50x + 100y = 2000 \Rightarrow x + 2y = 40$ …(ii)
Subtracting (i) from (ii): $y = 15$, then $x = 10$.
Number of ₹50 notes $= 10$, ₹100 notes $= 15$.
(v) Let fixed charge $= ₹\,x$, additional charge per day $= ₹\,y$.
$x + 4y = 27$ …(i)
$x + 2y = 21$ …(ii)
Subtracting (ii) from (i): $2y = 6 \Rightarrow y = 3$, then $x = 15$.
Fixed charge for first three days $= ₹\,15$, additional charge per day $= ₹\,3$.
