Pair of Linear Equations in Two Variables

Q. For what value of $k$ will the pair $kx + 3y – 3 = 0$ and $12x + ky – 6 = 0$ have
no solution?

Solution: No solution means parallel lines, so we need
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$.
$\frac{k}{12} = \frac{3}{k} \Rightarrow k^2 = 36 \Rightarrow k = 6$ or $k = -6$.
Now check the third ratio, $\frac{c_1}{c_2} = \frac{-3}{-6} = \frac{1}{2}$.
For $k = 6$: $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, equal to $\frac{c_1}{c_2}$, so the lines
would be coincident. Reject.
For $k = -6$: $\frac{b_1}{b_2} = \frac{3}{-6} = -\frac{1}{2} \ne \frac{1}{2}$. Parallel.
So $k = -6$.

Q. (Type of NCERT Exercise 3.1, Q6.) Given $2x + 3y – 8 = 0$, write a second linear equation so that the pair
gives (i) intersecting lines, (ii) parallel lines, (iii) coincident lines.

Solution: Here $a_1 = 2$, $b_1 = 3$, $c_1 = -8$. Many answers are possible.
(i) Intersecting: $x – y + 1 = 0$, since $\frac{2}{1} = 2$ and $\frac{3}{-1} = -3$, so $\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$.
(ii) Parallel: $2x + 3y + 5 = 0$, since $\frac{2}{2} = \frac{3}{3} = 1$ but $\frac{c_1}{c_2} = \frac{-8}{5} \ne 1$.
(iii) Coincident: $4x + 6y – 16 = 0$ (the given equation times 2), since $\frac{2}{4} = \frac{3}{6} = \frac{-8}{-16} = \frac{1}{2}$.

Q. Is the pair $x + 3y = 6$ and $2x – 3y = 12$ consistent? If so, solve it graphically.

Solution: First find two points on each line:

Plot A(0, 2), B(6, 0) and P(0, −4), Q(3, −2), and draw lines AB and PQ. Both lines pass
through B(6, 0), so the pair is consistent and the solution is $x = 6$, $y = 0$.

Q. Find whether $5x – 8y + 1 = 0$ and $3x – \frac{24}{5}y + \frac{3}{5} = 0$ have no
solution, a unique solution or infinitely many solutions.

Solution: Multiply the second equation by $\frac{5}{3}$: it becomes
$5x – 8y + 1 = 0$, the first equation exactly. The lines are coincident, so the pair has
infinitely many solutions.

Q. At a sale, Champa’s skirts are two less than twice her pants, and also four less
than four times her pants. How many of each did she buy?

Solution: Let pants $= x$, skirts $= y$. Then $y = 2x – 2$ and $y = 4x – 4$.
Points for the first line: (2, 2), (0, −2). Points for the second: (0, −4), (1, 0).
The lines meet at (1, 0), so $x = 1$, $y = 0$: she bought 1 pant and no skirt.
Check: $0 = 2(1) – 2$ and $0 = 4(1) – 4$.

Q. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were
then. Also, three years from now, I shall be three times as old as you will be.” Find their ages.

Solution: Let Aftab’s age be $s$ years and his daughter’s $t$ years.
Seven years ago: $s – 7 = 7(t – 7)$, i.e. $s – 7t + 42 = 0$ …(1)
Three years from now: $s + 3 = 3(t + 3)$, i.e. $s – 3t = 6$ …(2)
From (2), $s = 3t + 6$. Put in (1): $(3t + 6) – 7t + 42 = 0 \Rightarrow 4t = 48 \Rightarrow t = 12$.
Then $s = 3(12) + 6 = 42$. Aftab is 42 years and his daughter
12 years old.

Q. (NCERT Exercise 3.2, Q2.) Solve $2x + 3y = 11$ and $2x – 4y = -24$ and hence find the value of ‘m’ for which $y = mx + 3$.

Solution: From the first, $2x = 11 – 3y$. Put this in the second: $(11 – 3y) – 4y = -24 \Rightarrow -7y = -35 \Rightarrow y = 5$,
and then $2x = 11 – 15 = -4$, so $x = -2$.
The line $y = mx + 3$ passes through $(-2, 5)$: $5 = -2m + 3 \Rightarrow$ $m = -1$.
Check: $2(-2) + 3(5) = 11$ and $2(-2) – 4(5) = -24$.

Q. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures (money
spent) is 4 : 3. If each saves ₹2000 per month, find their monthly incomes.

Solution: Let the incomes be ₹$9x$ and ₹$7x$, the expenditures ₹$4y$ and ₹$3y$.
Saving = income − expenditure, so

Monthly incomes: $9 \times 2000 =$ ₹18,000 and $7 \times 2000 =$
₹14,000. Check: expenditures are ₹16,000 and ₹12,000, and $16000 : 12000 = 4 : 3$.

Q. Find all solutions of $2x + 3y = 8$ and $4x + 6y = 7$.

Solution: Multiply the first by 2: $4x + 6y = 16$. Subtract $4x + 6y = 7$:
$0 = 9$, which is false. The pair has no solution.

Q. The sum of a two-digit number and the number obtained by reversing its digits is 66.
If the digits differ by 2, find the number. How many such numbers are there?

Solution: Let the tens digit be $x$ and the units digit $y$: the number is $10x + y$
(e.g. $56 = 10(5) + 6$), the reversed number $10y + x$.
$(10x + y) + (10y + x) = 66 \Rightarrow 11(x + y) = 66 \Rightarrow x + y = 6$ …(1)
“Differ by 2” gives two cases: $x – y = 2$ …(2) or $y – x = 2$ …(3).
(1) with (2): $x = 4$, $y = 2$, number 42. (1) with (3): $x = 2$, $y = 4$, number 24.
There are two such numbers, 42 and 24.

Q. The denominator of a fraction is 3 more than its numerator. If 1 is added to both,
the fraction becomes $\frac{2}{3}$. Find the fraction.

Solution: Let the fraction be $\frac{x}{y}$. Then $y = x + 3$ …(1) and
$\frac{x + 1}{y + 1} = \frac{2}{3} \Rightarrow 3x + 3 = 2y + 2 \Rightarrow 3x – 2y = -1$ …(2).
Put (1) in (2): $3x – 2x – 6 = -1 \Rightarrow x = 5$, so $y = 8$. The fraction is
$\frac{5}{8}$. Check: $\frac{6}{9} = \frac{2}{3}$.

Q. (Type of NCERT Exercise 3.3, Q2(iv).) A taxi charge has a fixed part and a part per km. For 10 km the charge is ₹105 and
for 15 km it is ₹155. Find the fixed charge and the charge per km.

Solution: Take ₹$x$ as the fixed charge and ₹$y$ as the charge for each km.
$x + 10y = 105$ …(1) and $x + 15y = 155$ …(2).
(2) − (1): $5y = 50$, so $y = 10$; then $x = 105 – 100 = 5$.
Fixed charge ₹5, charge per km ₹10.
Check: $5 + 10 \times 10 = 105$ and $5 + 15 \times 10 = 155$.

Q. Draw $x + y = 5$ and $2x – y + 2 = 0$. Find the vertices of the triangle these lines
form with the $x$-axis.

Solution: On the $x$-axis, $y = 0$.
$x + y = 5$ cuts it at $(5, 0)$; $2x – y + 2 = 0$ cuts it at $(-1, 0)$.
The lines meet where $x + y = 5$ and $2x – y = -2$: adding, $3x = 3$, so $x = 1$ and $y = 4$.
The vertices are $(-1, 0)$, $(5, 0)$ and $(1, 4)$; shade the region
between them.