NCERT Class 10 Mathematics — Pair of Linear Equations in Two Variables, Exercise 3.1. All 7 questions solved.

Every linear equation in two variables is a straight line, so a pair of them is two lines on the same axes, and there are only three possibilities:

• the lines intersect at one point — exactly one solution, and the pair is consistent;
• the lines are parallel — no solution, and the pair is inconsistent;
• the lines coincide — every point on the line is a solution, and the pair is dependent (and consistent).

For the pair $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, the coefficients decide which case you are in without drawing anything:

$$\frac{a_1}{a_2} \ne \frac{b_1}{b_2}: \text{ intersecting}; \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}: \text{ parallel}; \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}: \text{ coincident}.$$

To draw a line, find two (better, three) points on it from a small table of values and join them.

Key insight. The ratio test and the picture are the same fact. Comparing $\dfrac{a_1}{a_2}$ with $\dfrac{b_1}{b_2}$ compares the slopes of the two lines: if the slopes differ the lines must meet, and if they match, $\dfrac{c_1}{c_2}$ decides whether they are the same line or two parallel ones. So questions 2, 3, 4 and 6 can be settled from the coefficients alone; a graph is needed only to read off an actual solution.

Question 1

Form the pair of linear equations in the following problems, and find their solutions graphically.

Question 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.

Solution. Let the number of girls be $x$ and the number of boys $y$. The two sentences give two equations:

$$x + y = 10, \qquad x – y = 4.$$

Table of values for each line:

Plotting the points and joining them gives two lines that meet at $(7, 3)$:

So $x = 7$ and $y = 3$. Check against the words: $7 + 3 = 10$ students, and $7$ girls is $4$ more than $3$ boys.

Question 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.

Solution. Let one pencil cost ₹$x$ and one pen ₹$y$. Then

$$5x + 7y = 50, \qquad 7x + 5y = 46.$$

The coefficients are awkward, so pick values of $x$ that make $y$ a whole number. From $y = \dfrac{50 – 5x}{7}$ and $y = \dfrac{46 – 7x}{5}$:

The point $(3, 5)$ appears in both tables, and the graph confirms that the lines meet there:

Check: $5 \times 3 + 7 \times 5 = 50$ and $7 \times 3 + 5 \times 5 = 46$.

Question 2

On comparing the ratios $\dfrac{a_1}{a_2}$, $\dfrac{b_1}{b_2}$ and $\dfrac{c_1}{c_2}$, 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. Compare $\dfrac{a_1}{a_2}$ with $\dfrac{b_1}{b_2}$ first; only if they are equal is $\dfrac{c_1}{c_2}$ needed. Keep the signs of the coefficients.

(i) $\dfrac{a_1}{a_2} = \dfrac57$ and $\dfrac{b_1}{b_2} = \dfrac{-4}{6} = -\dfrac23$. These differ, so the lines intersect at a point.

(ii) $\dfrac{a_1}{a_2} = \dfrac{9}{18} = \dfrac12$, $\dfrac{b_1}{b_2} = \dfrac36 = \dfrac12$, $\dfrac{c_1}{c_2} = \dfrac{12}{24} = \dfrac12$. All three are equal — the second equation is the first multiplied by $2$ — so the lines are coincident.

(iii) $\dfrac{a_1}{a_2} = \dfrac62 = 3$, $\dfrac{b_1}{b_2} = \dfrac{-3}{-1} = 3$, but $\dfrac{c_1}{c_2} = \dfrac{10}{9}$. Same slope, different line: parallel.

Question 3

On comparing the ratios $\dfrac{a_1}{a_2}$, $\dfrac{b_1}{b_2}$ and $\dfrac{c_1}{c_2}$, 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) $\dfrac32 x + \dfrac53 y = 7$ ; $\;9x – 10y = 14$

(iv) $5x – 3y = 11$ ; $\;-10x + 6y = -22$

(v) $\dfrac43 x + 2y = 8$ ; $\;2x + 3y = 12$

Solution. These equations are written as $ax + by = c$ rather than $ax + by + c = 0$. Moving the constants across changes the sign of both $c_1$ and $c_2$, so the ratio $\dfrac{c_1}{c_2}$ comes out the same either way and the right-hand sides can be used directly. Consistent means at least one solution — intersecting or coincident lines.

(i) $\dfrac{a_1}{a_2} = \dfrac32$, $\dfrac{b_1}{b_2} = \dfrac{2}{-3}$. Different, so the lines intersect: consistent, with a unique solution.

(ii) $\dfrac{a_1}{a_2} = \dfrac24 = \dfrac12$, $\dfrac{b_1}{b_2} = \dfrac{-3}{-6} = \dfrac12$, $\dfrac{c_1}{c_2} = \dfrac89$. Parallel lines: inconsistent.

(iii) $\dfrac{a_1}{a_2} = \dfrac{3/2}{9} = \dfrac16$ and $\dfrac{b_1}{b_2} = \dfrac{5/3}{-10} = -\dfrac16$. Different, so the lines intersect: consistent.

(iv) $\dfrac{a_1}{a_2} = \dfrac{5}{-10} = -\dfrac12$, $\dfrac{b_1}{b_2} = \dfrac{-3}{6} = -\dfrac12$, $\dfrac{c_1}{c_2} = \dfrac{11}{-22} = -\dfrac12$. All equal: the lines coincide, so the pair is consistent (dependent), with infinitely many solutions.

(v) $\dfrac{a_1}{a_2} = \dfrac{4/3}{2} = \dfrac23$, $\dfrac{b_1}{b_2} = \dfrac23$, $\dfrac{c_1}{c_2} = \dfrac{8}{12} = \dfrac23$. All three are equal — multiplying the first equation by $\dfrac32$ gives exactly $2x + 3y = 12$ — so these too are the same line: consistent (dependent). The fractions disguise it, which is why the ratios are worth computing carefully.

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. Use the ratio test to decide, and draw only the consistent pairs.

(i) $\dfrac12 = \dfrac12 = \dfrac{5}{10}$: all three ratios are equal, so the lines coincide and the pair is consistent, with infinitely many solutions. Drawing both lines gives one line through $(0, 5)$, $(2, 3)$ and $(5, 0)$:

Every point on it is a solution: $y = 5 – x$, with $x$ taking any value.

(ii) $\dfrac{a_1}{a_2} = \dfrac13$, $\dfrac{b_1}{b_2} = \dfrac{-1}{-3} = \dfrac13$, $\dfrac{c_1}{c_2} = \dfrac{8}{16} = \dfrac12$. Parallel lines: inconsistent, no solution to find.

(iii) $\dfrac{a_1}{a_2} = \dfrac24 = \dfrac12$ and $\dfrac{b_1}{b_2} = \dfrac{1}{-2} = -\dfrac12$. Different, so consistent with a unique solution. Tables of values:

The lines meet at $(2, 2)$. Check: $2(2) + 2 – 6 = 0$ and $4(2) – 2(2) – 4 = 0$.

(iv) $\dfrac{a_1}{a_2} = \dfrac24 = \dfrac12$, $\dfrac{b_1}{b_2} = \dfrac{-2}{-4} = \dfrac12$, $\dfrac{c_1}{c_2} = \dfrac{-2}{-5} = \dfrac25$. Parallel lines: inconsistent.

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 $x$ m and the width $y$ m. Half the perimeter of a rectangle is length plus width, so

$$x + y = 36, \qquad x – y = 4.$$

Graphically: $x + y = 36$ passes through $(36, 0)$, $(0, 36)$ and $(20, 16)$, and $x – y = 4$ passes through $(4, 0)$, $(20, 16)$ and $(24, 20)$. The point common to both tables, $(20, 16)$, is where the lines cross. (Adding the two equations gives $2x = 40$ at once, which confirms it.)

Check: $20 + 16 = 36$, and $20$ is $4$ more than $16$.

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. Work backwards from the ratio conditions. There are infinitely many correct answers to each part; any equation meeting the condition will do.

(i) We need $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$, so choose coefficients in a different proportion from $2 : 3$. For example $3x + 2y – 7 = 0$: $\dfrac23 \ne \dfrac32$.

(ii) We need $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$. Keep $2x + 3y$ and change only the constant: $2x + 3y – 12 = 0$ gives $\dfrac22 = \dfrac33 \ne \dfrac{-8}{-12}$.

(iii) All three ratios equal: multiply the whole equation by a non-zero number. For example, doubling gives $4x + 6y – 16 = 0$, with every ratio equal to $\dfrac12$.

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. Two of the vertices are where each line meets the $x$-axis, found by putting $y = 0$; the third is where the lines meet each other.

With $y = 0$, the first line gives $x = -1$ and the second $x = 4$. Both tables contain $(2, 3)$, so that is where the lines cross. The triangle, shaded:

Common mistakes

• Question 1(i), losing track of which variable is which. Once $x$ is “girls” and $y$ is “boys”, the intersection $(7, 3)$ means $7$ girls. Reading it the other way round gives an answer that fails “girls are 4 more than boys”.
• Question 2(i), dropping a minus sign in a ratio. $\dfrac{b_1}{b_2} = \dfrac{-4}{6}$ is negative. Here the verdict survives either way, but in other pairs the sign is exactly what separates intersecting from parallel.
• Question 3, worrying about the form of the equations. Writing $ax + by = c$ instead of $ax + by + c = 0$ flips the sign of both constants, so $\dfrac{c_1}{c_2}$ is unchanged. There is no need to rewrite every equation first.
• Question 3(v), missing that the lines coincide. The fraction $\dfrac43$ hides the fact that the first equation is $\dfrac23$ of the second. Computing $\dfrac{4/3}{2} = \dfrac23$ carefully, rather than guessing, shows all three ratios agree.
• Question 4(i), giving a single solution. Coincident lines share every point, so the answer is the whole line $y = 5 – x$, not just $(0, 5)$ or $(5, 0)$.
• Question 6(ii), writing $4x + 6y – 16 = 0$ for parallel lines. That is the given line doubled — coincident, not parallel. For parallel lines the constant must change out of proportion with the other two coefficients.
• Question 7, using the $y$-axis. The triangle is bounded by the $x$-axis, so its base vertices come from $y = 0$. Setting $x = 0$ instead gives $(0, 1)$ and $(0, 6)$, which belong to a different triangle.

Practise next

• Exercise 3.2 — the substitution method, which finds the same solutions algebraically and copes with answers that are not whole numbers.
• Chapter 2, Exercise 2.1 — reading zeroes off a graph, the same skill of reading where a graph meets a line.