Pair of Linear Equations in Two Variables — NCERT Solutions

Given: Total students = 10; Number of girls is 4 more than number of boys.

Let number of boys = $x$ and number of girls = $y$.

Equations formed:
$$x + y = 10 \quad \text{...(1)}$$
$$y - x = 4 \quad \text{...(2)}$$

Solutions for Equation (1): $x + y = 10$

Solutions for Equation (2): $y = x + 4$

Plot these points and draw both lines on the same graph.

The two lines intersect at the point $(3, 7)$.

$$\therefore x = 3, \quad y = 7$$

Verification: $3 + 7 = 10$ ✓ and $7 - 3 = 4$ ✓

Answer: Number of boys = 3 and number of girls = 7.

Let cost of one pencil = ₹ $x$ and cost of one pen = ₹ $y$.

Equations formed:
$$5x + 7y = 50 \quad \text{...(1)}$$
$$7x + 5y = 46 \quad \text{...(2)}$$

Solutions for Equation (1): $y = \dfrac{50 - 5x}{7}$

Solutions for Equation (2): $y = \dfrac{46 - 7x}{5}$

Plot these points and draw both lines on the same graph.

The two lines intersect at the point $(3, 5)$.

$$\therefore x = 3, \quad y = 5$$

Verification: $5(3) + 7(5) = 15 + 35 = 50$ ✓ and $7(3) + 5(5) = 21 + 25 = 46$ ✓

Answer: Cost of one pencil = ₹ 3 and cost of one pen = ₹ 5.

Given equations:
$$5x - 4y + 8 = 0 \quad \Rightarrow a_1=5,\ b_1=-4,\ c_1=8$$
$$7x + 6y - 9 = 0 \quad \Rightarrow a_2=7,\ b_2=6,\ c_2=-9$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{5}{7}, \qquad \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$$

Since $\dfrac{5}{7} \neq \dfrac{-2}{3}$, i.e., $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.

Conclusion: The lines intersect at a point (unique solution; consistent pair).

Given equations:
$$9x + 3y + 12 = 0 \quad \Rightarrow a_1=9,\ b_1=3,\ c_1=12$$
$$18x + 6y + 24 = 0 \quad \Rightarrow a_2=18,\ b_2=6,\ c_2=24$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} = \dfrac{1}{2}$.

Conclusion: The lines are coincident (infinitely many solutions; dependent and consistent pair).

Given equations:
$$6x - 3y + 10 = 0 \quad \Rightarrow a_1=6,\ b_1=-3,\ c_1=10$$
$$2x - y + 9 = 0 \quad \Rightarrow a_2=2,\ b_2=-1,\ c_2=9$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{6}{2} = 3, \qquad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \qquad \frac{c_1}{c_2} = \frac{10}{9}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = 3$ but $\dfrac{c_1}{c_2} = \dfrac{10}{9} \neq 3$.

So $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.

Conclusion: The lines are parallel (no solution; inconsistent pair).

Rewriting in standard form:
$$3x + 2y - 5 = 0 \quad \Rightarrow a_1=3,\ b_1=2,\ c_1=-5$$
$$2x - 3y - 7 = 0 \quad \Rightarrow a_2=2,\ b_2=-3,\ c_2=-7$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{3}{2}, \qquad \frac{b_1}{b_2} = \frac{2}{-3} = -\frac{2}{3}$$

Since $\dfrac{3}{2} \neq -\dfrac{2}{3}$, i.e., $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.

Conclusion: The pair of linear equations is consistent (unique solution).

Rewriting in standard form:
$$2x - 3y - 8 = 0 \quad \Rightarrow a_1=2,\ b_1=-3,\ c_1=-8$$
$$4x - 6y - 9 = 0 \quad \Rightarrow a_2=4,\ b_2=-6,\ c_2=-9$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{1}{2}$ but $\dfrac{c_1}{c_2} = \dfrac{8}{9} \neq \dfrac{1}{2}$.

Conclusion: The pair of linear equations is inconsistent (no solution; parallel lines).

Rewriting in standard form:
$$\frac{3}{2}x + \frac{5}{3}y - 7 = 0 \quad \Rightarrow a_1=\frac{3}{2},\ b_1=\frac{5}{3},\ c_1=-7$$
$$9x - 10y - 14 = 0 \quad \Rightarrow a_2=9,\ b_2=-10,\ c_2=-14$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{3/2}{9} = \frac{3}{18} = \frac{1}{6}$$
$$\frac{b_1}{b_2} = \frac{5/3}{-10} = \frac{5}{-30} = -\frac{1}{6}$$

Since $\dfrac{1}{6} \neq -\dfrac{1}{6}$, i.e., $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.

Conclusion: The pair of linear equations is consistent (unique solution).

Rewriting in standard form:
$$5x - 3y - 11 = 0 \quad \Rightarrow a_1=5,\ b_1=-3,\ c_1=-11$$
$$-10x + 6y + 22 = 0 \quad \Rightarrow a_2=-10,\ b_2=6,\ c_2=22$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} = -\dfrac{1}{2}$.

Conclusion: The pair of linear equations is consistent (infinitely many solutions; coincident lines).

Rewriting in standard form:
$$\frac{4}{3}x + 2y - 8 = 0 \quad \Rightarrow a_1=\frac{4}{3},\ b_1=2,\ c_1=-8$$
$$2x + 3y - 12 = 0 \quad \Rightarrow a_2=2,\ b_2=3,\ c_2=-12$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{4/3}{2} = \frac{4}{6} = \frac{2}{3}, \qquad \frac{b_1}{b_2} = \frac{2}{3}, \qquad \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} = \dfrac{2}{3}$.

Conclusion: The pair of linear equations is consistent (infinitely many solutions; coincident lines).

Rewriting in standard form:
$$x + y - 5 = 0 \quad \Rightarrow a_1=1,\ b_1=1,\ c_1=-5$$
$$2x + 2y - 10 = 0 \quad \Rightarrow a_2=2,\ b_2=2,\ c_2=-10$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the equations are consistent (coincident lines — infinitely many solutions).

Graphical solution: Both equations represent the same line $x + y = 5$.

Some solutions: $(0, 5),\ (5, 0),\ (2, 3)$, etc.

Answer: The pair is consistent with infinitely many solutions — every point on the line $x + y = 5$ is a solution.

Rewriting in standard form:
$$x - y - 8 = 0 \quad \Rightarrow a_1=1,\ b_1=-1,\ c_1=-8$$
$$3x - 3y - 16 = 0 \quad \Rightarrow a_2=3,\ b_2=-3,\ c_2=-16$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{1}{3}, \qquad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}, \qquad \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{1}{3}$ but $\dfrac{c_1}{c_2} = \dfrac{1}{2} \neq \dfrac{1}{3}$.

Conclusion: The pair is inconsistent (parallel lines, no solution).

Given:
$$2x + y - 6 = 0 \quad \Rightarrow a_1=2,\ b_1=1,\ c_1=-6$$
$$4x - 2y - 4 = 0 \quad \Rightarrow a_2=4,\ b_2=-2,\ c_2=-4$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$$

Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the pair is consistent (unique solution).

Graphical solution:

For $2x + y = 6$:

For $4x - 2y = 4$, i.e., $2x - y = 2$:

Plotting and drawing both lines, they intersect at $(2, 2)$.

Verification: $2(2)+2-6=0$ ✓ and $4(2)-2(2)-4=0$ ✓

Answer: The pair is consistent. Solution: $x = 2,\ y = 2$.

Given:
$$2x - 2y - 2 = 0 \quad \Rightarrow a_1=2,\ b_1=-2,\ c_1=-2$$
$$4x - 4y - 5 = 0 \quad \Rightarrow a_2=4,\ b_2=-4,\ c_2=-5$$

Comparing ratios:
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$$

Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{1}{2}$ but $\dfrac{c_1}{c_2} = \dfrac{2}{5} \neq \dfrac{1}{2}$.

Conclusion: The pair is inconsistent (parallel lines, no solution).

Let length = $l$ metres and width = $b$ metres.

Condition 1: Length is 4 m more than width:
$$l = b + 4 \quad \Rightarrow l - b = 4 \quad \text{...(1)}$$

Condition 2: Half the perimeter = 36 m:
$$l + b = 36 \quad \text{...(2)}$$

Adding equations (1) and (2):
$$2l = 40 \Rightarrow l = 20 \text{ m}$$

Substituting in (2):
$$20 + b = 36 \Rightarrow b = 16 \text{ m}$$

Verification: $l - b = 20 - 16 = 4$ ✓ and $l + b = 20 + 16 = 36$ ✓

Answer: Length = 20 m and Width = 16 m.

Given equation: $2x + 3y - 8 = 0$, where $a_1 = 2,\ b_1 = 3,\ c_1 = -8$.

(i) Intersecting lines:
Condition: $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

Choose any equation where the ratio of coefficients of $x$ and $y$ is different.

Example: $3x + 2y - 8 = 0$

Here $\dfrac{2}{3} \neq \dfrac{3}{2}$ ✓

(ii) Parallel lines:
Condition: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Choose $a_2 : b_2 = 2 : 3$ but $c_2 \neq -8k$ for the same $k$.

Example: $2x + 3y - 12 = 0$

Here $\dfrac{2}{2} = \dfrac{3}{3} = 1$ but $\dfrac{-8}{-12} = \dfrac{2}{3} \neq 1$ ✓

(iii) Coincident lines:
Condition: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

Multiply the entire equation by any non-zero constant.

Example: $4x + 6y - 16 = 0$

Here $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{-8}{-16} = \dfrac{1}{2}$ ✓

(Note: Many other valid answers are possible for each part.)

Equation 1: $x - y + 1 = 0 \Rightarrow y = x + 1$

Equation 2: $3x + 2y - 12 = 0 \Rightarrow y = \dfrac{12 - 3x}{2}$

Plot these points and draw both lines.

Finding intersection of the two lines:
From Eq. 1: $y = x + 1$. Substitute in Eq. 2:
$$3x + 2(x+1) = 12 \Rightarrow 5x + 2 = 12 \Rightarrow x = 2,\ y = 3$$
Intersection point: $A(2, 3)$.

Finding where each line meets the $x$-axis ($y = 0$):

• Line 1: $x - 0 + 1 = 0 \Rightarrow x = -1$ → Point $B(-1, 0)$
• Line 2: $3x + 0 - 12 = 0 \Rightarrow x = 4$ → Point $C(4, 0)$

Vertices of the triangle:
$$A(2,\ 3),\quad B(-1,\ 0),\quad C(4,\ 0)$$

Shade the triangular region $ABC$ on the graph.

Answer: The vertices of the triangle are $(-1, 0)$, $(4, 0)$, and $(2, 3)$.

Given:
$$x + y = 14 \quad \text{...(1)}$$
$$x - y = 4 \quad \text{...(2)}$$

From equation (2):
$$x = 4 + y \quad \text{...(3)}$$

Substituting (3) in (1):
$$(4 + y) + y = 14$$
$$4 + 2y = 14$$
$$2y = 10 \Rightarrow y = 5$$

Substituting $y = 5$ in (3):
$$x = 4 + 5 = 9$$

Verification: $9 + 5 = 14$ ✓ and $9 - 5 = 4$ ✓

Answer: $x = 9,\ y = 5$.

Given:
$$s - t = 3 \quad \text{...(1)}$$
$$\frac{s}{3} + \frac{t}{2} = 6 \quad \text{...(2)}$$

From equation (1):
$$s = t + 3 \quad \text{...(3)}$$

Substituting (3) in (2):
$$\frac{t+3}{3} + \frac{t}{2} = 6$$

Multiplying throughout by 6:
$$2(t+3) + 3t = 36$$
$$2t + 6 + 3t = 36$$
$$5t = 30 \Rightarrow t = 6$$

Substituting $t = 6$ in (3):
$$s = 6 + 3 = 9$$

Verification: $9 - 6 = 3$ ✓ and $\dfrac{9}{3} + \dfrac{6}{2} = 3 + 3 = 6$ ✓

Answer: $s = 9,\ t = 6$.

Given:
$$3x - y = 3 \quad \text{...(1)}$$
$$9x - 3y = 9 \quad \text{...(2)}$$

Observe: Equation (2) = 3 × Equation (1), so both equations are the same.

From equation (1):
$$y = 3x - 3 \quad \text{...(3)}$$

Substituting in (2): $9x - 3(3x-3) = 9 \Rightarrow 9x - 9x + 9 = 9 \Rightarrow 9 = 9$ (always true).

Conclusion: The equations are dependent and have infinitely many solutions.

The solution is $y = 3x - 3$, i.e., every point on the line $3x - y = 3$ is a solution.