Pair of Linear Equations in Two Variables

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): $x + y = 10$

| $x$ | 0 | 10 | 5 |
|---|---|---|---|
| $y$ | 10 | 0 | 5 |

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

| $x$ | 0 | 2 | 4 |
|---|---|---|---|
| $y$ | 4 | 6 | 8 |

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): $y = \dfrac{50 - 5x}{7}$

| $x$ | 3 | 10 |
|---|---|---|
| $y$ | 5 | 0 |

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

| $x$ | 3 | 8 |
|---|---|---|
| $y$ | 5 | -2 |

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 $2x + y = 6$:
| $x$ | 0 | 3 |
|---|---|---|
| $y$ | 6 | 0 |

For $4x - 2y = 4$, i.e., $2x - y = 2$:
| $x$ | 0 | 1 |
|---|---|---|
| $y$ | -2 | 0 |

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

| $x$ | 0 | 2 | 4 |
|---|---|---|---|
| $y$ | 1 | 3 | 5 |

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

| $x$ | 0 | 2 | 4 |
|---|---|---|---|
| $y$ | 6 | 3 | 0 |

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.

Given:
$$0.2x + 0.3y = 1.3 \quad \text{...(1)}$$
$$0.4x + 0.5y = 2.3 \quad \text{...(2)}$$

Multiply both equations by 10:
$$2x + 3y = 13 \quad \text{...(3)}$$
$$4x + 5y = 23 \quad \text{...(4)}$$

From equation (3):
$$x = \frac{13 - 3y}{2} \quad \text{...(5)}$$

Substituting (5) in (4):
$$4 \cdot \frac{13 - 3y}{2} + 5y = 23$$
$$2(13 - 3y) + 5y = 23$$
$$26 - 6y + 5y = 23$$
$$-y = -3 \Rightarrow y = 3$$

Substituting $y = 3$ in (5):
$$x = \frac{13 - 9}{2} = \frac{4}{2} = 2$$

Verification: $0.2(2) + 0.3(3) = 0.4 + 0.9 = 1.3$ ✓ and $0.4(2) + 0.5(3) = 0.8 + 1.5 = 2.3$ ✓

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

Given:
$$\sqrt{2}\,x + \sqrt{3}\,y = 0 \quad \text{...(1)}$$
$$\sqrt{3}\,x - \sqrt{8}\,y = 0 \quad \text{...(2)}$$

From equation (1):
$$x = -\frac{\sqrt{3}}{\sqrt{2}}\,y \quad \text{...(3)}$$

Substituting (3) in (2):
$$\sqrt{3} \cdot \left(-\frac{\sqrt{3}}{\sqrt{2}}\,y\right) - \sqrt{8}\,y = 0$$
$$-\frac{3}{\sqrt{2}}\,y - 2\sqrt{2}\,y = 0$$
$$y\left(-\frac{3}{\sqrt{2}} - 2\sqrt{2}\right) = 0$$
$$y\left(\frac{-3 - 4}{\sqrt{2}}\right) = 0$$
$$y \cdot \frac{-7}{\sqrt{2}} = 0 \Rightarrow y = 0$$

Substituting $y = 0$ in (3):
$$x = 0$$

Verification: $\sqrt{2}(0) + \sqrt{3}(0) = 0$ ✓ and $\sqrt{3}(0) - \sqrt{8}(0) = 0$ ✓

Answer: $x = 0,\ y = 0$.

Given:
$$\frac{3x}{2} - \frac{5y}{3} = -2 \quad \text{...(1)}$$
$$\frac{x}{3} + \frac{y}{2} = \frac{13}{6} \quad \text{...(2)}$$

Multiply (1) by 6: $9x - 10y = -12 \quad \text{...(3)}$

Multiply (2) by 6: $2x + 3y = 13 \quad \text{...(4)}$

From equation (4):
$$x = \frac{13 - 3y}{2} \quad \text{...(5)}$$

Substituting (5) in (3):
$$9 \cdot \frac{13 - 3y}{2} - 10y = -12$$
$$\frac{117 - 27y}{2} - 10y = -12$$

Multiply throughout by 2:
$$117 - 27y - 20y = -24$$
$$117 - 47y = -24$$
$$47y = 141 \Rightarrow y = 3$$

Substituting $y = 3$ in (5):
$$x = \frac{13 - 9}{2} = \frac{4}{2} = 2$$

Verification: $\dfrac{3(2)}{2} - \dfrac{5(3)}{3} = 3 - 5 = -2$ ✓ and $\dfrac{2}{3} + \dfrac{3}{2} = \dfrac{4+9}{6} = \dfrac{13}{6}$ ✓

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

Given:
$$2x + 3y = 11 \quad \text{...(1)}$$
$$2x - 4y = -24 \quad \text{...(2)}$$

Subtracting (2) from (1):
$$(2x + 3y) - (2x - 4y) = 11 - (-24)$$
$$7y = 35 \Rightarrow y = 5$$

Substituting $y = 5$ in (1):
$$2x + 15 = 11 \Rightarrow 2x = -4 \Rightarrow x = -2$$

Verification: $2(-2) + 3(5) = -4 + 15 = 11$ ✓ and $2(-2) - 4(5) = -4 - 20 = -24$ ✓

Finding $m$: Substituting $x = -2$ and $y = 5$ in $y = mx + 3$:
$$5 = m(-2) + 3$$
$$5 - 3 = -2m$$
$$2 = -2m \Rightarrow m = -1$$

Answer: $x = -2,\ y = 5$ and $m = -1$.

Let the two numbers be $x$ and $y$ where x > y.

Equations formed:
$$x - y = 26 \quad \text{...(1)}$$
$$x = 3y \quad \text{...(2)}$$

Substituting (2) in (1):
$$3y - y = 26$$
$$2y = 26 \Rightarrow y = 13$$

From (2): $x = 3 \times 13 = 39$

Verification: $39 - 13 = 26$ ✓ and $39 = 3 \times 13$ ✓

Answer: The two numbers are 39 and 13.

Let the larger angle = $x°$ and the smaller angle = $y°$.

Equations formed:

Supplementary angles: $x + y = 180 \quad \text{...(1)}$

Larger exceeds smaller by 18°: $x - y = 18 \quad \text{...(2)}$

From (2): $x = y + 18 \quad \text{...(3)}$

Substituting (3) in (1):
$$(y + 18) + y = 180$$
$$2y = 162 \Rightarrow y = 81°$$

From (3): $x = 81 + 18 = 99°$

Verification: $99 + 81 = 180°$ ✓ and $99 - 81 = 18°$ ✓

Answer: The two supplementary angles are 99° and 81°.

Let cost of one bat = ₹ $x$ and cost of one ball = ₹ $y$.

Equations formed:
$$7x + 6y = 3800 \quad \text{...(1)}$$
$$3x + 5y = 1750 \quad \text{...(2)}$$

From (2):
$$x = \frac{1750 - 5y}{3} \quad \text{...(3)}$$

Substituting (3) in (1):
$$7 \cdot \frac{1750 - 5y}{3} + 6y = 3800$$
$$\frac{12250 - 35y}{3} + 6y = 3800$$

Multiply throughout by 3:
$$12250 - 35y + 18y = 11400$$
$$-17y = -850 \Rightarrow y = 50$$

Substituting $y = 50$ in (3):
$$x = \frac{1750 - 250}{3} = \frac{1500}{3} = 500$$

Verification: $7(500) + 6(50) = 3500 + 300 = 3800$ ✓ and $3(500) + 5(50) = 1500 + 250 = 1750$ ✓

Answer: Cost of each bat = ₹ 500 and cost of each ball = ₹ 50.

Let fixed charge = ₹ $x$ and charge per km = ₹ $y$.

Equations formed:
$$x + 10y = 105 \quad \text{...(1)}$$
$$x + 15y = 155 \quad \text{...(2)}$$

From (1):
$$x = 105 - 10y \quad \text{...(3)}$$

Substituting (3) in (2):
$$(105 - 10y) + 15y = 155$$
$$105 + 5y = 155$$
$$5y = 50 \Rightarrow y = 10$$

Substituting $y = 10$ in (3):
$$x = 105 - 100 = 5$$

Verification: $5 + 10(10) = 105$ ✓ and $5 + 15(10) = 155$ ✓

Charge for 25 km:
$$= x + 25y = 5 + 25(10) = 5 + 250 = ₹\ 255$$

Answer: Fixed charge = ₹ 5, charge per km = ₹ 10, and charge for 25 km = ₹ 255.

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

Condition 1: Adding 2 to both numerator and denominator:
$$\frac{x+2}{y+2} = \frac{9}{11}$$
$$11(x+2) = 9(y+2)$$
$$11x + 22 = 9y + 18$$
$$11x - 9y = -4 \quad \text{...(1)}$$

Condition 2: Adding 3 to both numerator and denominator:
$$\frac{x+3}{y+3} = \frac{5}{6}$$
$$6(x+3) = 5(y+3)$$
$$6x + 18 = 5y + 15$$
$$6x - 5y = -3 \quad \text{...(2)}$$

From (2):
$$x = \frac{5y - 3}{6} \quad \text{...(3)}$$

Substituting (3) in (1):
$$11 \cdot \frac{5y-3}{6} - 9y = -4$$
$$\frac{55y - 33}{6} - 9y = -4$$

Multiply throughout by 6:
$$55y - 33 - 54y = -24$$
$$y = 9$$

Substituting $y = 9$ in (3):
$$x = \frac{45 - 3}{6} = \frac{42}{6} = 7$$

Verification: $\dfrac{7+2}{9+2} = \dfrac{9}{11}$ ✓ and $\dfrac{7+3}{9+3} = \dfrac{10}{12} = \dfrac{5}{6}$ ✓

Answer: The fraction is $\dfrac{7}{9}$.

Let Jacob's present age = $x$ years and his son's present age = $y$ years.

Condition 1: Five years hence:
$$x + 5 = 3(y + 5)$$
$$x + 5 = 3y + 15$$
$$x - 3y = 10 \quad \text{...(1)}$$

Condition 2: Five years ago:
$$x - 5 = 7(y - 5)$$
$$x - 5 = 7y - 35$$
$$x - 7y = -30 \quad \text{...(2)}$$

From (1): $x = 3y + 10 \quad \text{...(3)}$

Substituting (3) in (2):
$$(3y + 10) - 7y = -30$$
$$-4y = -40 \Rightarrow y = 10$$

From (3): $x = 3(10) + 10 = 40$

Verification: $40 + 5 = 45 = 3(15) = 3(10+5)$ ✓ and $40 - 5 = 35 = 7(5) = 7(10-5)$ ✓

Answer: Jacob's present age = 40 years and his son's present age = 10 years.

Given:
$$x + y = 5 \quad \text{...(1)}$$
$$2x - 3y = 4 \quad \text{...(2)}$$

Elimination Method:

Multiply (1) by 3:
$$3x + 3y = 15 \quad \text{...(3)}$$

Adding (2) and (3):
$$5x = 19 \Rightarrow x = \frac{19}{5}$$

Substituting in (1):
$$\frac{19}{5} + y = 5 \Rightarrow y = 5 - \frac{19}{5} = \frac{6}{5}$$

Substitution Method:

From (1): $x = 5 - y \quad \text{...(4)}$

Substituting in (2):
$$2(5 - y) - 3y = 4$$
$$10 - 2y - 3y = 4$$
$$-5y = -6 \Rightarrow y = \frac{6}{5}$$

From (4): $x = 5 - \dfrac{6}{5} = \dfrac{19}{5}$

Verification: $\dfrac{19}{5} + \dfrac{6}{5} = \dfrac{25}{5} = 5$ ✓ and $2\left(\dfrac{19}{5}\right) - 3\left(\dfrac{6}{5}\right) = \dfrac{38-18}{5} = \dfrac{20}{5} = 4$ ✓

Answer: $x = \dfrac{19}{5},\ y = \dfrac{6}{5}$.

Given:
$$3x + 4y = 10 \quad \text{...(1)}$$
$$2x - 2y = 2 \quad \text{...(2)}$$

Elimination Method:

Multiply (2) by 2:
$$4x - 4y = 4 \quad \text{...(3)}$$

Adding (1) and (3):
$$7x = 14 \Rightarrow x = 2$$

Substituting in (2):
$$4 - 2y = 2 \Rightarrow 2y = 2 \Rightarrow y = 1$$

Substitution Method:

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

Substituting in (1):
$$3(y+1) + 4y = 10$$
$$3y + 3 + 4y = 10$$
$$7y = 7 \Rightarrow y = 1$$

From (4): $x = 1 + 1 = 2$

Verification: $3(2) + 4(1) = 10$ ✓ and $2(2) - 2(1) = 2$ ✓

Answer: $x = 2,\ y = 1$.