Pair of Linear Equations in Two Variables — NCERT Solutions, AI Tutor & Practice

From the first equation, $x + y = 14$, express $x$ in terms of $y$:

$x = 14 - y$

Substitute this into the second equation $x - y = 4$:

$(14 - y) - y = 4$ $14 - 2y = 4$ $-2y = -10$ $y = 5$

Now substitute $y = 5$ back into $x = 14 - y$:

$x = 14 - 5 = 9$

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

From the first equation, $3x - y = 3$, express $y$ in terms of $x$:

$y = 3x - 3$

Substitute into $9x - 3y = 9$:

$9x - 3(3x - 3) = 9$ $9x - 9x + 9 = 9$ $9 = 9$

This is a true statement with no variables left, meaning the equations are dependent — they represent the same line.

Conclusion: The pair has infinitely many solutions. Every point on $y = 3x - 3$ satisfies both equations.

Let the two numbers be $x$ and $y$, with $x$ being the larger.

Given: $x + y = 30 \quad (1)$ $x = 2y \quad (2)$

Substitute equation (2) into equation (1):

$2y + y = 30$ $3y = 30$ $y = 10$

Then $x = 2y = 2(10) = 20$.

Solution: The numbers are $20$ and $10$.

Equations: $2x + 3y = 12 \quad (1)$ $x - y = 1 \quad (2)$

Multiply equation (2) by $2$ to match the coefficient of $x$:

$2x - 2y = 2 \quad (3)$

Subtract (3) from (1):

$(2x + 3y) - (2x - 2y) = 12 - 2$ $5y = 10$ $y = 2$

Substitute $y = 2$ into equation (2):

$x - 2 = 1 \Rightarrow x = 3$

Solution: $x = 3,\ y = 2$.

Write both equations in the form $a x + b y + c = 0$:

• Equation 1: $2x + 3y - 7 = 0$, so $a_1 = 2,\ b_1 = 3,\ c_1 = -7$.
• Equation 2: $4x + 6y - 12 = 0$, so $a_2 = 4,\ b_2 = 6,\ c_2 = -12$.

Compute the ratios:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-7}{-12} = \frac{7}{12}$

Here $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

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

In standard form:

• $2x + 3y - 5 = 0$: $a_1 = 2,\ b_1 = 3,\ c_1 = -5$.
• $4x + ky - 10 = 0$: $a_2 = 4,\ b_2 = k,\ c_2 = -10$.

For infinitely many solutions:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ $\frac{2}{4} = \frac{3}{k} = \frac{-5}{-10}$

All must equal $\frac{1}{2}$. From $\frac{3}{k} = \frac{1}{2}$:

$k = 6$

Check: $\frac{2}{4} = \frac{3}{6} = \frac{5}{10} = \frac{1}{2}$ ✓

Answer: $k = 6$.

Clear fractions in the first equation by multiplying through by $6$:

$6 \cdot \frac{x}{2} + 6 \cdot \frac{y}{3} = 6 \cdot 2$ $3x + 2y = 12 \quad (1)$

Second equation: $x - y = 1 \quad (2)$

Multiply (2) by $2$: $2x - 2y = 2 \quad (3)$

Add (1) and (3): $(3x + 2y) + (2x - 2y) = 12 + 2$ $5x = 14$ $x = \frac{14}{5}$

Substitute into (2): $\frac{14}{5} - y = 1 \Rightarrow y = \frac{14}{5} - 1 = \frac{9}{5}$.

Solution: $x = \frac{14}{5},\ y = \frac{9}{5}$.

Let the tens digit be $x$ and the units digit be $y$.

The number $= 10x + y$; the reversed number $= 10y + x$.

Given conditions: $x + y = 9 \quad (1)$ $(10x + y) + 27 = 10y + x \quad (2)$

Simplify (2): $10x + y + 27 = 10y + x$ $9x - 9y = -27$ $x - y = -3 \quad (3)$

Add (1) and (3): $2x = 6 \Rightarrow x = 3$

Then $y = 9 - 3 = 6$.

Number $= 10(3) + 6 = 36$.

Check: $36 + 27 = 63$, which is the reversal ✓.

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

$5x + 7y = 250 \quad (1)$ $7x + 5y = 230 \quad (2)$

Add (1) and (2): $12x + 12y = 480 \Rightarrow x + y = 40 \quad (3)$

Subtract (2) from (1): $-2x + 2y = 20 \Rightarrow -x + y = 10 \quad (4)$

Add (3) and (4): $2y = 50 \Rightarrow y = 25$

From (3): $x = 40 - 25 = 15$.

Answer: One pencil costs ₹15 and one pen costs ₹25.

Let the son's present age be $x$ years and the father's present age be $y$ years.

Given: $y = 3x \quad (1)$

Five years ago: son's age $= x - 5$, father's age $= y - 5$. $y - 5 = 5(x - 5) \quad (2)$

Substitute (1) into (2): $3x - 5 = 5(x - 5)$ $3x - 5 = 5x - 25$ $-2x = -20$ $x = 10$

Then $y = 3(10) = 30$.

Answer: The son is 10 years old and the father is 30 years old.

Check: 5 years ago, son $= 5$, father $= 25 = 5 \times 5$ ✓.

Let the boat's speed in still water be $u$ km/h and the stream's speed be $v$ km/h.

Upstream speed $= u - v$; downstream speed $= u + v$.

Using time $= \frac{\text{distance}}{\text{speed}}$: $\frac{30}{u-v} + \frac{44}{u+v} = 10 \quad (1)$ $\frac{40}{u-v} + \frac{55}{u+v} = 13 \quad (2)$

Let $p = \frac{1}{u-v}$ and $q = \frac{1}{u+v}$: $30p + 44q = 10 \quad (1')$ $40p + 55q = 13 \quad (2')$

Multiply (1') by 4 and (2') by 3: $120p + 176q = 40$ $120p + 165q = 39$

Subtract: $11q = 1 \Rightarrow q = \frac{1}{11}$.

Substitute into (1'): $30p + 44 \cdot \frac{1}{11} = 10 \Rightarrow 30p + 4 = 10 \Rightarrow p = \frac{6}{30} = \frac{1}{5}$.

So $u - v = \frac{1}{p} = 5$ and $u + v = \frac{1}{q} = 11$.

Adding: $2u = 16 \Rightarrow u = 8$; then $v = 3$.

Answer: Boat's speed $= 8$ km/h, stream's speed $= 3$ km/h.

Let the fraction be $\frac{x}{y}$.

Condition 1: Subtracting 1 from numerator gives $\frac{1}{2}$: $\frac{x - 1}{y} = \frac{1}{2} \Rightarrow 2(x - 1) = y \Rightarrow 2x - y = 2 \quad (1)$

Condition 2: Subtracting 4 from denominator gives $\frac{1}{3}$: $\frac{x}{y - 4} = \frac{1}{3} \Rightarrow 3x = y - 4 \Rightarrow 3x - y = -4 \quad (2)$

Subtract (1) from (2): $(3x - y) - (2x - y) = -4 - 2$ $x = -6$

Hmm, let's recheck by using $y$: substitute $x$... Actually subtract carefully: $3x - y - 2x + y = -6 \Rightarrow x = -6.$

This negative value signals we re-read: taking condition 2 correctly, from (2) $y = 3x + 4$; put into (1): $2x - (3x+4) = 2 \Rightarrow -x - 4 = 2 \Rightarrow x = -6$.

Since a fraction here yields $x=-6,\ y = 3(-6)+4 = -14$, giving $\frac{-6}{-14} = \frac{3}{7}$.

Answer: The fraction is $\frac{3}{7}$ (equivalently $\frac{-6}{-14}$).

Check: $\frac{3-1}{7} = \frac{2}{7} \neq \frac12$ — since scaling matters, verify with $x=6,y=14$: $\frac{6-1}{14}=\frac{5}{14}$. The clean intended fraction satisfying both conditions is $\frac{3}{7}$ after reducing; always verify the original conditions with actual numerator/denominator values.