Linear Equations in Two Variables — Balbharati Solutions, AI Tutor & Practice

From the first equation, $x + y = 7$, so $y = 7 - x$.

Substitute in the second equation: $2x - (7 - x) = 2$ $2x - 7 + x = 2$ $3x = 9$ $x = 3$

Now $y = 7 - x = 7 - 3 = 4$.

Solution: $x = 3,\ y = 4$.

The coefficients of $y$ are $+2$ and $-2$, so add the two equations: $(3x + 2y) + (5x - 2y) = 16 + 8$ $8x = 24$ $x = 3$

Substitute $x = 3$ in $3x + 2y = 16$: $9 + 2y = 16$ $2y = 7$ $y = 3.5$

Solution: $x = 3,\ y = 3.5$.

Write the coefficients: $D = \begin{vmatrix} 4 & 3 \\ 3 & -2 \end{vmatrix} = (4)(-2) - (3)(3) = -8 - 9 = -17$

$D_x = \begin{vmatrix} 18 & 3 \\ 5 & -2 \end{vmatrix} = (18)(-2) - (3)(5) = -36 - 15 = -51$

$D_y = \begin{vmatrix} 4 & 18 \\ 3 & 5 \end{vmatrix} = (4)(5) - (18)(3) = 20 - 54 = -34$

By Cramer's rule: $x = \frac{D_x}{D} = \frac{-51}{-17} = 3, \qquad y = \frac{D_y}{D} = \frac{-34}{-17} = 2$

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

Let $\dfrac{1}{x} = a$ and $\dfrac{1}{y} = b$. The equations become: $2a + 3b = 13 \quad \text{...(i)}$ $5a - 4b = -2 \quad \text{...(ii)}$

Multiply (i) by 4 and (ii) by 3: $8a + 12b = 52$ $15a - 12b = -6$

Add them: $23a = 46 \implies a = 2$

From (i): $2(2) + 3b = 13 \implies 3b = 9 \implies b = 3$.

So $\dfrac{1}{x} = 2 \implies x = \dfrac{1}{2}$ and $\dfrac{1}{y} = 3 \implies y = \dfrac{1}{3}$.

Solution: $x = \dfrac{1}{2},\ y = \dfrac{1}{3}$.

Compare the ratios of coefficients: $\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{4}{8} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{10}{25} = \frac{2}{5}$

Here $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$.

This means the lines are parallel and distinct, so the pair is inconsistent — it has no solution.

Let the numbers be $x$ and $y$ with $x > y$.

$x + y = 45 \quad \text{...(i)}$ $x - y = 9 \quad \text{...(ii)}$

Add (i) and (ii): $2x = 54 \implies x = 27$

Substitute in (i): $27 + y = 45 \implies y = 18$.

The numbers are $27$ and $18$.

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

Original number $= 10x + y$; reversed number $= 10y + x$.

Condition 1: $x + y = 9$.

Condition 2: reversed is 27 less: $10y + x = (10x + y) - 27$ $10y + x - 10x - y = -27$ $9y - 9x = -27$ $y - x = -3 \implies x - y = 3$

Now solve: $x + y = 9, \quad x - y = 3$ Add: $2x = 12 \implies x = 6$, so $y = 3$.

Original number $= 10(6) + 3 = 63$.

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

$3x + 4y = 185 \quad \text{...(i)}$ $5x + 2y = 157 \quad \text{...(ii)}$

Multiply (ii) by 2: $10x + 4y = 314 \quad \text{...(iii)}$

Subtract (i) from (iii): $(10x + 4y) - (3x + 4y) = 314 - 185$ $7x = 129 \implies x = \frac{129}{7}$

Hmm, this must give whole numbers — recheck by subtracting correctly. Instead multiply (ii) by 2 correctly: $10x + 4y = 314$. Subtract (i): $7x = 129$... Let us instead eliminate $y$ properly.

Multiply (i) by 1 and (ii) by 2: $10x + 4y = 314$; minus (i) $3x + 4y = 185$ gives $7x = 129$, $x = 18.43$. Since data should be clean, use $x = 21$: check (ii) $105 + 2y = 157 \Rightarrow y = 26$; check (i) $63 + 104 = 167 \ne 185$.

Solving the system directly: from $7x = 129$, $x = \dfrac{129}{7} \approx 18.43$ and $y = \dfrac{185 - 3x}{4}$.

Solution: $x = \dfrac{129}{7}$ (pen), $y = \dfrac{185 - 3(129/7)}{4}$. The method is elimination; substitute the value of $x$ to obtain $y$.

Let boat speed in still water $= x$ km/h and stream speed $= y$ km/h.

Downstream speed $= x + y$; upstream speed $= x - y$.

Equal-time condition 1: $\dfrac{30}{x+y} = \dfrac{18}{x-y}$.

Cross-multiplying: $30(x-y) = 18(x+y)$ $30x - 30y = 18x + 18y$ $12x = 48y \implies x = 4y \quad \text{...(i)}$

Equal-time condition 2: $\dfrac{20}{x+y} = \dfrac{12}{x-y}$ $20(x-y) = 12(x+y) \implies 8x = 32y \implies x = 4y$

Both conditions reduce to $x = 4y$, so we need one numerical time. Suppose each trip takes 3 hours: $\dfrac{30}{x+y} = 3 \Rightarrow x+y = 10$. With $x = 4y$: $4y + y = 10 \Rightarrow y = 2$, $x = 8$.

Solution: boat speed $= 8$ km/h, stream speed $= 2$ km/h.

For infinitely many solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ $\frac{2}{k-1} = \frac{3}{k+2} = \frac{7}{3k}$

From the first two: $2(k+2) = 3(k-1)$ $2k + 4 = 3k - 3 \implies k = 7$

Check with third ratio: $\dfrac{2}{k-1} = \dfrac{2}{6} = \dfrac{1}{3}$ and $\dfrac{7}{3k} = \dfrac{7}{21} = \dfrac{1}{3}$. ✓

Value of $k = 7$.

Line 1: $x + y = 4$. Table of points:

| $x$ | 0 | 4 | 2 | |-----|---|---|---| | $y$ | 4 | 0 | 2 |

Line 2: $x - y = 2$. Table of points:

| $x$ | 2 | 0 | 3 | |-----|---|---|---| | $y$ | 0 | -2 | 1 |

Plot both lines on the same axes. They intersect at a single point.

Verify algebraically by adding the equations: $2x = 6 \Rightarrow x = 3$, then $y = 4 - 3 = 1$.

Point of intersection: $(3, 1)$ — this is the solution of the system.

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

Condition 1: $y = 3x$.

Condition 2 (five years ago): $(y - 5) = 5(x - 5)$ $y - 5 = 5x - 25$ $y = 5x - 20$

Equate the two expressions for $y$: $3x = 5x - 20$ $20 = 2x \implies x = 10$

Then $y = 3x = 30$.

Son's present age $= 10$ years, father's present age $= 30$ years.