NCERT Solutions: Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables

(i) Let the number of boys be $x$ and the number of girls be $y$. Then $x + y = 10$ and $y = x + 4$, i.e., $-x + y = 4$. Solving, $x + (x + 4) = 10$, so $2x = 6$ and $x = 3$; hence $y = 7$. The two lines meet at $(3, 7)$.
(ii) Let the cost of one pencil be $x$ and one pen be $y$. Then $5x + 7y = 50$ and $7x + 5y = 46$. Subtracting the second equation from the first gives $-2x + 2y = 4$, so $y = x + 2$. Substituting in $5x + 7y = 50$ gives $5x + 7x + 14 = 50$, so $x = 3$ and $y = 5$. The graph of the two lines intersects at $(3, 5)$.

(i) $\dfrac{a_1}{a_2} = \dfrac{5}{7}$ and $\dfrac{b_1}{b_2} = \dfrac{-4}{6}$ are not equal, so the lines intersect at one point.
(ii) $\dfrac{9}{18} = \dfrac{3}{6} = \dfrac{12}{24} = \dfrac{1}{2}$, so the lines are coincident.
(iii) $\dfrac{6}{2} = \dfrac{-3}{-1} = 3$, but $\dfrac{10}{9} \neq 3$, so the lines are parallel.

A pair is consistent if the lines intersect or coincide, and inconsistent if they are parallel.
(i) $\dfrac{3}{2} \neq \dfrac{2}{-3}$, so the pair has a unique solution and is consistent.
(ii) $\dfrac{2}{4} = \dfrac{-3}{-6} \neq \dfrac{-8}{-9}$ in standard form, so the lines are parallel and the pair is inconsistent.
(iii) $\dfrac{3/2}{9} \neq \dfrac{5/3}{-10}$, so the lines intersect and the pair is consistent.
(iv) $\dfrac{5}{-10} = \dfrac{-3}{6} = \dfrac{-11}{22}$ in standard form, so the lines are coincident and the pair is consistent.
(v) $\dfrac{4/3}{2} = \dfrac{2}{3} = \dfrac{-8}{-12}$ in standard form, so the lines are coincident and the pair is consistent.

(i) The second equation is $2(x + y) = 10$, i.e., the same as $x + y = 5$, so the lines are coincident.
(ii) In standard form, $x - y - 8 = 0$ and $3x - 3y - 16 = 0$. Here $\dfrac{1}{3} = \dfrac{-1}{-3}$ but $\dfrac{-8}{-16} \neq \dfrac{1}{3}$, so the lines are parallel.
(iii) From $2x + y - 6 = 0$, $y = 6 - 2x$. From $4x - 2y - 4 = 0$, $y = 2x - 2$. Equating gives $6 - 2x = 2x - 2$, so $x = 2$ and $y = 2$.
(iv) In standard form, $2x - 2y - 2 = 0$ and $4x - 4y - 5 = 0$. The coefficient ratios of $x$ and $y$ are equal but the constant ratio is different, so the lines are parallel.

Let the length be $l$ m and the width be $w$ m. Half the perimeter is $l + w = 36$, and $l = w + 4$. Therefore $w + 4 + w = 36$, so $2w = 32$ and $w = 16$. Hence $l = 20$.

(i) For intersecting lines, choose a line whose $x$ and $y$ coefficient ratios are not both equal to those of $2x + 3y - 8 = 0$.
(ii) For parallel lines, keep the same $x$ and $y$ coefficient ratio but change the constant ratio.
(iii) For coincident lines, take a non-zero multiple of the given equation.

For $x - y + 1 = 0$, we have $y = x + 1$. It meets the $x$-axis when $y = 0$, so $x = -1$ and the point is $(-1, 0)$.
For $3x + 2y - 12 = 0$, it meets the $x$-axis when $y = 0$, so $3x = 12$, $x = 4$, and the point is $(4, 0)$.
The intersection of the two lines is found from $y = x + 1$ and $3x + 2y - 12 = 0$: $3x + 2(x + 1) - 12 = 0$, so $5x - 10 = 0$, $x = 2$, and $y = 3$. Hence the third vertex is $(2, 3)$.

(i) Adding the equations gives $2x = 18$, so $x = 9$ and $y = 5$.
(ii) From $s - t = 3$, $t = s - 3$. Substitute in $\dfrac{s}{3} + \dfrac{t}{2} = 6$: $\dfrac{s}{3} + \dfrac{s - 3}{2} = 6$. Multiplying by 6 gives $2s + 3s - 9 = 36$, so $s = 9$ and $t = 6$.
(iii) The second equation is exactly 3 times the first, so the two equations represent the same line. Hence there are infinitely many solutions, given by $y = 3x - 3$.
(iv) Multiplying by 10 gives $2x + 3y = 13$ and $4x + 5y = 23$. From the first, $x = \dfrac{13 - 3y}{2}$. Substitution gives $y = 3$ and $x = 2$.
(v) From $2x + 3y = 0$, $x = -\dfrac{3y}{2}$. Substituting in $3x - 8y = 0$ gives $-\dfrac{9y}{2} - 8y = 0$, so $y = 0$ and $x = 0$.
(vi) Multiplying the equations by 6 gives $9x - 10y = -12$ and $2x + 3y = 13$. From $2x + 3y = 13$, $x = \dfrac{13 - 3y}{2}$. Substitution gives $y = \dfrac{11}{3}$ and then $x = 1$.

Subtracting $2x - 4y = -24$ from $2x + 3y = 11$ gives $7y = 35$, so $y = 5$. Substituting in $2x + 3y = 11$ gives $2x + 15 = 11$, so $x = -2$. Now $y = mx + 3$ gives $5 = m(-2) + 3$, so $-2m = 2$ and $m = -1$.

(i) Let the larger number be $x$ and the smaller be $y$. Then $x - y = 26$ and $x = 3y$. So $3y - y = 26$, $y = 13$, $x = 39$.
(ii) Let the larger angle be $x^\circ$ and the smaller be $y^\circ$. Then $x + y = 180$ and $x - y = 18$. Solving gives $x = 99$, $y = 81$.
(iii) Let a bat cost ₹$x$ and a ball cost ₹$y$. Then $7x + 6y = 3800$ and $3x + 5y = 1750$. From the second, $x = \dfrac{1750 - 5y}{3}$. Substitution gives $y = 50$ and $x = 500$.
(iv) Let fixed charge be ₹$x$ and charge per km be ₹$y$. Then $x + 10y = 105$ and $x + 15y = 155$. Subtracting gives $5y = 50$, so $y = 10$ and $x = 5$. For 25 km, fare $= 5 + 25 \times 10 = 255$.
(v) Let the fraction be $\dfrac{x}{y}$. Then $\dfrac{x+2}{y+2} = \dfrac{9}{11}$ and $\dfrac{x+3}{y+3} = \dfrac{5}{6}$. These give $11x - 9y = -4$ and $6x - 5y = -3$. Solving gives $x = 7$, $y = 9$.
(vi) Let Jacob’s present age be $x$ years and his son’s present age be $y$ years. Then $x + 5 = 3(y + 5)$ and $x - 5 = 7(y - 5)$. These simplify to $x - 3y = 10$ and $x - 7y = -30$. Subtracting gives $4y = 40$, so $y = 10$ and $x = 40$.

(i) From $x + y = 5$, $x = 5 - y$. Substituting in $2x - 3y = 4$ gives $2(5 - y) - 3y = 4$, so $10 - 5y = 4$, $y = \dfrac{6}{5}$, and $x = \dfrac{19}{5}$.
(ii) From $2x - 2y = 2$, $x - y = 1$, so $x = y + 1$. Substituting in $3x + 4y = 10$ gives $3(y + 1) + 4y = 10$, so $7y = 7$, $y = 1$, $x = 2$.
(iii) The equations are $3x - 5y = 4$ and $9x - 2y = 7$. Multiplying the first by 3 gives $9x - 15y = 12$. Subtracting $9x - 2y = 7$ gives $-13y = 5$, so $y = -\dfrac{5}{13}$. Then $9x - 2\left(-\dfrac{5}{13}\right) = 7$, giving $x = \dfrac{9}{13}$.
(iv) Multiplying $\dfrac{x}{2} + \dfrac{2y}{3} = -1$ by 6 gives $3x + 4y = -6$. Multiplying $x - \dfrac{y}{3} = 3$ by 3 gives $3x - y = 9$. Subtracting gives $5y = -15$, so $y = -3$ and $x = 2$.

(i) Let the fraction be $\dfrac{x}{y}$. Then $\dfrac{x+1}{y-1} = 1$, so $y = x + 2$. Also, $\dfrac{x}{y+1} = \dfrac{1}{2}$, so $2x = y + 1$. Solving gives $x = 3$, $y = 5$.
(ii) Let Nuri’s present age be $x$ and Sonu’s present age be $y$. Five years ago: $x - 5 = 3(y - 5)$, so $x - 3y = -10$. Ten years later: $x + 10 = 2(y + 10)$, so $x - 2y = 10$. Subtracting gives $y = 20$, hence $x = 50$.
(iii) Let the tens digit be $x$ and the units digit be $y$. Then $x + y = 9$. The number is $10x + y$ and the reversed number is $10y + x$. Given $9(10x + y) = 2(10y + x)$, so $88x = 11y$, i.e., $y = 8x$. Hence $9x = 9$, $x = 1$, $y = 8$, and the number is 18.
(iv) Let the number of ₹50 notes be $x$ and ₹100 notes be $y$. Then $x + y = 25$ and $50x + 100y = 2000$, i.e., $x + 2y = 40$. Subtracting gives $y = 15$ and $x = 10$.
(v) Let the fixed charge be ₹$x$ and the charge for each extra day be ₹$y$. For seven days, $x + 4y = 27$; for five days, $x + 2y = 21$. Subtracting gives $2y = 6$, so $y = 3$ and $x = 15$.