Simultaneous Linear Equations

Given,

x + y = 14 ......(i)

x - y = 4 .........(ii)

From eqn. (ii) we get,

x = 4 + y.

Substituting above value of x in eqn. (i) we get,

⟹ (4 + y) + y = 14
⟹ 4 + 2y = 14
⟹ 2y = 14 - 4
⟹ 2y = 10
⟹ y = 5.

∴ x = y + 4 = 5 + 4 = 9.

Hence, x = 9 and y = 5.

Given,

s - t = 3 ........(i)

$\frac{s}{3} + \frac{t}{2} = 6$ ........(ii)

From eqn. (i) we get,

s = 3 + t.

Substituting above value of s in eqn. (ii) we get,

$$\begin{aligned}\Rightarrow \frac{s}{3} + \frac{t}{2} = 6 \\[1em] \Rightarrow \frac{3 + t}{3} + \frac{t}{2} = 6 \\[1em] \Rightarrow \frac{2(3 + t) + 3t}{6} = 6 \\[1em] \Rightarrow 6 + 2t + 3t = 36 \\[1em] \Rightarrow 5t = 36 - 6 \\[1em] \Rightarrow 5t = 30 \\[1em] \Rightarrow t = 6.\end{aligned}$$

∴ s = 3 + t = 3 + 6 = 9.

Hence, t = 6 and s = 9.

Given,

2x + 3y = 9 ......(i)

3x + 4y = 5 .......(ii)

Solving eqn. (i) we get,

⟹ 2x + 3y = 9

⟹ 2x = 9 - 3y

⟹ x = $\frac{9 - 3y}{2}$

Substituting above value of x in eqn. (ii) we get,

$$\begin{aligned}\Rightarrow 3\Big(\frac{9 - 3y}{2}\Big) + 4y = 5 \\[1em] \Rightarrow \frac{27 - 9y}{2} + 4y = 5 \\[1em] \Rightarrow \frac{27 - 9y + 8y}{2} = 5 \\[1em] \Rightarrow 27 - y = 10 \\[1em] \Rightarrow y = 27 - 10 = 17.\end{aligned}$$

Solving for x,

$$\begin{aligned}x = \frac{9 - 3y}{2} = \frac{9 - 3(17)}{2} \\[1em] = \frac{9 - 51}{2} \\[1em] = \frac{-42}{2} \\[1em] = -21.\end{aligned}$$

Hence, x = -21 and y = 17.

Given,

3x - 5y = 4 .......(i)

9x - 2y = 7 .......(ii)

Solving eqn. (i) we get,

⟹ 3x - 5y = 4

⟹ 3x = 4 + 5y

⟹ x = $\frac{4 + 5y}{3}$.

Substituting above value of x in eqn. (ii) we get,

$$\begin{aligned}\Rightarrow 9x - 2y = 7 \\[1em] \Rightarrow 9\Big(\frac{4 + 5y}{3}\Big) - 2y = 7 \\[1em] \Rightarrow 3(4 + 5y) - 2y = 7 \\[1em] \Rightarrow 12 + 15y - 2y = 7 \\[1em] \Rightarrow 12 + 13y = 7 \\[1em] \Rightarrow 13y = 7 - 12 \\[1em] \Rightarrow 13y = -5 \\[1em] \Rightarrow y = -\frac{5}{13}.\end{aligned}$$

Solving for x by substituting value of y,

$$\begin{aligned}\Rightarrow x = \frac{4 + 5y}{3} \\[1em] = \frac{4 + 5 \times \frac{-5}{13}}{3} \\[1em] = \frac{4 - \frac{25}{13}}{3} \\[1em] = \frac{\frac{52 - 25}{13}}{3} \\[1em] = \frac{27}{39} \\[1em] = \frac{9}{13}.\end{aligned}$$

Hence, x = $\frac{9}{13}\text{ and y } = -\frac{5}{13}.$

Given,

3x - 5y = -2 ......................(1)

7x - 3y = -9 ......................(2)

Solving equation (1), we get :

⇒ 3x - 5y = -2

⇒ 3x = -2 + 5y

⇒ x = $\frac{-2 + 5y}{3}$

Substituting above value of x in equation (2), we get :

$$\begin{aligned}\Rightarrow 7\Big(\frac{-2 + 5y}{3}\Big) - 3y = -9\\[1em] \Rightarrow \frac{7(-2 + 5y) - 3y \times 3}{3} = -9\\[1em] \Rightarrow \frac{7(-2 + 5y) - 9y}{3} = -9\\[1em] \Rightarrow 7({-2 + 5y}) - 9y = -27\\[1em] \Rightarrow -14 + 35y - 9y = -27\\[1em] \Rightarrow -14 + 26y = -27\\[1em] \Rightarrow 26y = -27 + 14\\[1em] \Rightarrow 26y = -13\\[1em] \Rightarrow y = -\frac{13}{26}\\[1em] \Rightarrow y = -\frac{1}{2}.\end{aligned}$$

Substituting value of y in x = $\frac{-2 + 5y}{3}$, we get :

$$\begin{aligned}\Rightarrow x = \frac{-2 + 5 \times \frac{-1}{2}}{3}\\[1em] \Rightarrow x = \frac{-2 - \frac{5}{2}}{3} \\[1em] \Rightarrow x = \frac{\frac{-4 - 5}{2}}{3} \\[1em] \Rightarrow x = \frac{-4 - 5}{6}\\[1em] \Rightarrow x = \frac{-9}{6}\\[1em] \Rightarrow x = -\frac{3}{2}.\end{aligned}$$

Hence, x = $-\frac{3}{2}$ and y = $-\frac{1}{2}$.

Given,

5x + 4y - 4 = 0 ......(i)

x - 20 = 12y ........(ii)

From eqn. (ii) we get,

x = 12y + 20 .......(iii)

Substituting value of x from eqn. (iii) in eqn. (i) we get,

⟹ 5(12y + 20) + 4y - 4 = 0

⟹ 60y + 100 + 4y - 4 = 0

⟹ 64y + 96 = 0

⟹ 64y = -96

⟹ y = $-\frac{96}{64} = -\frac{3}{2}$.

Substituting value of y in eqn. (iii) we get,

$$\begin{aligned}x = 12 \times \frac{-3}{2} + 20 \\[1em] = 6 \times -3 + 20 \\[1em] = -18 + 20 \\[1em] = 2.\end{aligned}$$

Hence, x = 2 and y = $-\frac{3}{2}.$

Given,

$2x - \frac{3y}{4} = 3$ .......(i)

5x - 2y - 7 = 0 .......(ii)

Multiplying eqn. (i) by 4 we get,

$$\begin{aligned}\Rightarrow 4\Big(2x - \frac{3y}{4}\Big) = 3 \times 4 \\[1em] \Rightarrow 8x - 3y = 12 \\[1em] \Rightarrow 8x = 12 + 3y \\[1em] \Rightarrow x = \frac{12 + 3y}{8} ......(iii)\end{aligned}$$

Putting value of x from eqn. (iii) in eqn. (ii),

$$\begin{aligned}\Rightarrow 5\Big(\frac{12 + 3y}{8}\Big) - 2y - 7 = 0 \\[1em] \Rightarrow \frac{60 + 15y}{8} - 2y - 7 = 0 \\[1em] \Rightarrow \frac{60 + 15y - 16y - 56}{8} = 0 \\[1em] \Rightarrow 4 - y = 0 \\[1em] \Rightarrow y = 4.\end{aligned}$$

Substituting value of y in eqn. (iii) we get,

$$\begin{aligned}\Rightarrow x = \frac{12 + 3y}{8} \\[1em] = \frac{12 + 3(4)}{8} \\[1em] = \frac{12 + 12}{8} \\[1em] = \frac{24}{8} \\[1em] = 3.\end{aligned}$$

Hence, x = 3 and y = 4.

Given,

2x + 3y = 23 ......(i)

5x - 20 = 8y ......(ii)

Solving (i) we get,

⟹ 2x + 3y = 23

⟹ 2x = 23 - 3y

⟹ x = $\frac{23 - 3y}{2}$ .....(iii)

Substituting value of x from eqn. (iii) in eqn. (ii) we get,

$$\begin{aligned}\Rightarrow 5\Big(\frac{23 - 3y}{2}\Big) - 20 = 8y \\[1em] \Rightarrow \frac{115 - 15y}{2} - 20 = 8y \\[1em] \Rightarrow \frac{115 - 15y - 40}{2} = 8y \\[1em] \Rightarrow 115 - 15y - 40 = 16y \\[1em] \Rightarrow 75 - 15y = 16y \\[1em] \Rightarrow 31y = 75 \\[1em] \Rightarrow y = \frac{75}{31} = 2\frac{13}{31}.\end{aligned}$$

Substituting value of y in eqn.(iii) we get,

$$\begin{aligned}\Rightarrow x = \frac{23 - 3y}{2} \\[1em] = \frac{23 - 3 \times \frac{75}{31}}{2} \\[1em] = \frac{\frac{23 \times 31 - 3 \times 75}{31}}{2} \\[1em] = \frac{713 - 225}{31 \times 2} \\[1em] = \frac{488}{62} \\[1em] = \frac{244}{31} = 7\frac{27}{31}.\end{aligned}$$

Hence, x = $7\frac{27}{31}\text{ and y =} 2\frac{13}{31}$.

Given,

mx - ny = m² + n² ......(i)

x + y = 2m .......(ii)

From eqn. (ii) we get,

x = 2m - y .......(iii)

Substituting value of x from eqn. (iii) in eqn. (i) we get,

⟹ m(2m - y) - ny = m² + n²

⟹ 2m² - my - ny = m² + n²

⟹ 2m² - m² - n² = my + ny

⟹ m² - n² = y(m + n)

⟹ y = $\frac{\text{m}^2 - \text{n}^2}{\text{m} + \text{n}} = \frac{(\text{m} - \text{n})(\text{m} + \text{n})}{\text{m} + \text{n}}$ = (m - n).

Substituting value of y in eqn. (iii) we get,

x = 2m - y = 2m - (m - n) = 2m - m + n = m + n.

Hence, x = m + n and y = m - n.

Given,

$\frac{2x}{a} + \frac{y}{b} = 2$ .......(i)

$\frac{x}{a} - \frac{y}{b} = 4$ ........(ii)

Multiplying both side of eqn (i) by ab we get,

$$\begin{aligned}\Rightarrow ab\Big(\frac{2x}{a} + \frac{y}{b}\Big) = 2ab \\[1em] \Rightarrow 2bx + ay = 2ab ......(iii)\end{aligned}$$

Multiplying both side of eqn (ii) by ab we get,

$$\begin{aligned}\Rightarrow ab\Big(\frac{x}{a} - \frac{y}{b}\Big) = 4ab \\[1em] \Rightarrow bx - ay = 4ab \\[1em] \Rightarrow bx = 4ab + ay ......(iv)\end{aligned}$$

Substituting value of bx in eq. (iii) we get,

⟹ 2(4ab + ay) + ay = 2ab

⟹ 8ab + 2ay + ay = 2ab

⟹ 8ab + 3ay = 2ab

⟹ 3ay = 2ab - 8ab

⟹ 3ay = -6ab

⟹ y = $\frac{-6ab}{3a}$ = -2b.

Substituting value of y in eqn. (iv) we get,

⟹ bx = 4ab - 2ab

⟹ bx = 2ab

⟹ x = 2a.

Hence, x = 2a and y = -2b.

Given,

2x + y = 35 ......(i)

3x + 4y = 65 ......(ii)

From (i) we get,

y = 35 - 2x ......(iii)

Substituting value of y from eqn. (iii) in eqn. (ii) we get,

⟹ 3x + 4(35 - 2x) = 65

⟹ 3x + 140 - 8x = 65

⟹ 140 - 5x = 65

⟹ 5x = 140 - 65

⟹ 5x = 75

⟹ x = 15.

Substituting value of x in eqn. (iii) we get,

⟹ y = 35 - 2x = 35 - 2(15) = 35 - 30 = 5.

⟹ $\frac{x}{y} = \frac{15}{5} = 3.$

Hence, x = 15, y = 5 and $\frac{x}{y} = 3.$

Given,

3x - y = 5 ......(i)

4x - 3y = -1 .....(ii)

Solving (i) we get,

⟹ y = 3x - 5 .....(iii)

Substituting value of y from eqn. (iii) in eqn. (ii) we get,

⟹ 4x - 3(3x - 5) = -1

⟹ 4x - 9x + 15 = -1

⟹ -5x = -1 - 15

⟹ -5x = -16

⟹ x = $\frac{16}{5}$.

Substituting value of x in eqn. (iii) we get,

$$\begin{aligned}\Rightarrow y = 3 \times \frac{16}{5} - 5 \\[1em] = \frac{48}{5} - 5 \\[1em] = \frac{48 - 25}{5} \\[1em] = \frac{23}{5}\end{aligned}$$

Given, y = px - 3. Substituting value of x and y in equation,

$$\begin{aligned}\Rightarrow \frac{23}{5} = \frac{16}{5}p - 3 \\[1em] \Rightarrow \frac{23}{5} = \frac{16p - 15}{5} \\[1em] \Rightarrow 23 = 16p - 15 \\[1em] \Rightarrow 16p = 38 \\[1em] \Rightarrow p = \frac{38}{16} = \frac{19}{8}.\end{aligned}$$

Hence, x =$\frac{16}{5},\text{ y }= \frac{23}{5}\text{ and p} = \frac{19}{8}.$

Given,

3x + 4y = 10 ......(i)

2x - 2y = 2 .......(ii)

Multiplying eq. (ii) by 2 we get,

4x - 4y = 4 ......(iii)

Adding eq. (i) and (iii) we get,

⇒ 3x + 4y + 4x - 4y = 10 + 4

⇒ 7x = 14

⇒ x = 2.

Substituting value of x in eq. (ii) we get,

⇒ 2(2) - 2y = 2

⇒ 4 - 2y = 2

⇒ 2y = 4 - 2

⇒ 2y = 2

⇒ y = 1.

Hence, x = 2 and y = 1.

Given,

2x = 5y + 4 or 2x - 5y - 4 = 0 ........(i)

3x - 2y + 16 = 0 ......(ii)

Multiplying eq. (i) by 3 and eq. (ii) by 2 we get,

6x - 15y - 12 = 0 .......(iii)

6x - 4y + 32 = 0 .......(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ 6x - 4y + 32 - (6x - 15y - 12) = 0

⇒ 6x - 6x - 4y + 15y + 32 + 12 = 0

⇒ 11y + 44 = 0

⇒ 11y = -44

⇒ y = -4.

Substituting value of y in eq. (ii) we get,

⇒ 3x - 2(-4) + 16 = 0

⇒ 3x + 8 + 16 = 0

⇒ 3x + 24 = 0

⇒ 3x = -24

⇒ x = -8.

Hence, x = -8 and y = -4.

Given,

$\frac{3}{4}x - \frac{2}{3}y = 1$ ........(i)

$\frac{3}{8}x - \frac{1}{6}y = 1$ .........(ii)

Multiplying eq. (ii) by 2 we get,

$\frac{3}{4}x - \frac{1}{3}y = 2$ ........(iii)

Subtracting eq. (i) from (iii) we get,

$$\begin{aligned}\Rightarrow \frac{3}{4}x - \frac{1}{3}y - \Big(\frac{3}{4}x - \frac{2}{3}y\Big) = 2 - 1 \\[1em] \Rightarrow \frac{3}{4}x - \frac{3}{4}x - \frac{1}{3}y + \frac{2}{3}y = 1 \\[1em] \Rightarrow \frac{1}{3}y = 1 \\[1em] \Rightarrow y = 3.\end{aligned}$$

Substituting value of y in eq. (i) we get,

$$\begin{aligned}\Rightarrow \frac{3}{4}x - \frac{2}{3} \times 3 = 1 \\[1em] \Rightarrow \frac{3}{4}x - 2 = 1 \\[1em] \Rightarrow \frac{3}{4}x = 3 \\[1em] \Rightarrow x = \frac{3 \times 4}{3} \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, x = 4 and y = 3.

Given,

2x - 3y - 3 = 0 ...........(i)

$\frac{2x}{3} + 4y + \frac{1}{2} = 0$ ........(ii)

Multiplying eq. (ii) by 3 we get,

$2x + 12y + \frac{3}{2} = 0$ .........(iii)

Subtracting eq. (i) from (iii) we get,

$$\begin{aligned}\Rightarrow 2x + 12y + \frac{3}{2} - (2x - 3y - 3) = 0 \\[1em] \Rightarrow 2x - 2x + 12y + 3y + \frac{3}{2} + 3 = 0 \\[1em] \Rightarrow 15y + \frac{9}{2} = 0 \\[1em] \Rightarrow 15y = -\frac{9}{2} \\[1em] \Rightarrow y = \frac{-9}{2 \times 15} \\[1em] \Rightarrow y = -\frac{3}{10}.\end{aligned}$$

Substituting value of y in eq. (i) we get,

$$\begin{aligned}\Rightarrow 2x - 3y - 3 = 0 \\[1em] \Rightarrow 2x - 3 \times -\frac{3}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \frac{9}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \frac{9 - 30}{10} = 0 \\[1em] \Rightarrow 2x - \frac{21}{10} = 0 \\[1em] \Rightarrow x = \frac{21}{20}.\end{aligned}$$

Hence, $x = \frac{21}{20} \text{and y} = -\frac{3}{10}.$

Given,

15x - 14y = 117 .......(i)

14x - 15y = 115 .......(ii)

Multiplying eq. (i) by 14 and eq. (ii) by 15 we get,

210x - 196y = 1638 .......(iii)

210x - 225y = 1725 .......(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ 210x - 225y - (210x - 196y) = 1725 - 1638

⇒ 210x - 210x - 225y + 196y = 87

⇒ -29y = 87

⇒ y = -3.

Substituting value of y in eq. (ii) we get,

⇒ 14x - 15(-3) = 115

⇒ 14x + 45 = 115

⇒ 14x = 115 - 45

⇒ 14x = 70

⇒ x = 5.

Hence, x = 5 and y = -3.

Given,

41x + 53y = 135 ......(i)

53x + 41y = 147 .......(ii)

Multiplying eq. (i) by 53 and eq. (ii) by 41 we get,

2173x + 2809y = 7155 ......(iii)

2173x + 1681y = 6027 ......(iv)

Subtracting eq. (iv) from (iii) we get,

⇒ 2173x + 2809y - (2173x + 1681y) = 7155 - 6027

⇒ 2173x - 2173x + 2809y - 1681y = 1128

⇒ 1128y = 1128

⇒ y = 1.

Substituting value of y in eq. (i) we get,

⇒ 41x + 53(1) = 135

⇒ 41x = 135 - 53

⇒ 41x = 82

⇒ x = 2.

Hence, x = 2 and y = 1.

Given,

$\frac{x}{6} = y - 6$ .......(i)

$\frac{3x}{4} = 1 + y$ ......(ii)

Subtracting eq. (i) from (ii) we get,

$$\begin{aligned}\Rightarrow \frac{3x}{4} - \frac{x}{6} = 1 + y - (y - 6) \\[1em] \Rightarrow \frac{9x - 2x}{12} = 7 \\[1em] \Rightarrow \frac{7x}{12} = 7 \\[1em] \Rightarrow x = \frac{7 \times 12}{7} \\[1em] \Rightarrow x = 12.\end{aligned}$$

Substituting value of x in eq. (i) we get,

$$\begin{aligned}\Rightarrow \frac{12}{6} = y - 6 \\[1em] \Rightarrow 2 = y - 6 \\[1em] \Rightarrow y = 2 + 6 = 8.\end{aligned}$$

Hence, x = 12 and y = 8.

Given,

$x - \frac{2}{3}y = \frac{8}{3}$ ........(i)

$\frac{2x}{5} - y = \frac{7}{5}$ ........(ii)

Multiplying eq. (i) by 6 and eq. (ii) by 15 we get,

6x - 4y = 16 .......(iii)

6x - 15y = 21 ......(iv)

Subtracting eq. (iv) from (iii) we get,

⇒ 6x - 4y - (6x - 15y) = 16 - 21

⇒ 6x - 6x - 4y + 15y = -5

⇒ 11y = -5

⇒ y = $-\frac{5}{11}$.

Substituting value of y in eq. (iv) we get,

$$\begin{aligned}\Rightarrow 6x - 15 \times -\frac{5}{11} = 21 \\[1em] \Rightarrow 6x + \frac{75}{11} = 21 \\[1em] \Rightarrow 6x = 21 - \frac{75}{11} \\[1em] \Rightarrow 6x = \frac{231 - 75}{11} \\[1em] \Rightarrow 6x = \frac{156}{11} \\[1em] \Rightarrow x = \frac{156}{66} \\[1em] \Rightarrow x = \frac{26}{11}.\end{aligned}$$

Hence, $x = \frac{26}{11} \text{ and } y = -\frac{5}{11}.$

Given,

9 - (x - 4) = y + 7

⇒ 9 - x + 4 = y + 7

⇒ x + y = 13 - 7

⇒ x + y = 6 .......(i)

2(x + y) = 4 - 3y

⇒ 2x + 2y = 4 - 3y

⇒ 2x + 5y = 4 .........(ii)

Multiplying eq. (i) by 2 we get,

⇒ 2x + 2y = 12 ........(iii)

Subtracting eq. (iii) from (ii) we get,

⇒ 2x + 5y - (2x + 2y) = 4 - 12

⇒ 5y - 2y = -8

⇒ 3y = -8

⇒ y = $-\frac{8}{3}$.

Substituting value of y in eq. (i) we get,

$$\begin{aligned}\Rightarrow x + \Big(-\frac{8}{3}\Big) = 6 \\[1em] \Rightarrow x = 6 + \frac{8}{3} \\[1em] \Rightarrow x = \frac{26}{3}.\end{aligned}$$

Hence, $x = \frac{26}{3} \text{ and } y = -\frac{8}{3}$.

Solving 1st equation,

$$\begin{aligned}\Rightarrow 2x + \frac{x - y}{6} = 2 \\[1em] \Rightarrow \frac{12x + x - y}{6} = 2 \\[1em] \Rightarrow 13x - y = 12 .......(i)\end{aligned}$$

Solving 2nd equation,

$$\begin{aligned}\Rightarrow x - \frac{2x + y}{3} = 1 \\[1em] \Rightarrow \frac{3x - 2x - y}{3} = 1 \\[1em] \Rightarrow x - y = 3 ........(ii)\end{aligned}$$

Multiplying eq. (ii) by 13 we get,

⇒ 13x - 13y = 39 .........(iii)

Subtracting eq. (iii) from (i) we get,

⇒ 13x - y - (13x - 13y) = 12 - 39

⇒ 13x - 13x - y + 13y = -27

⇒ 12y = -27

⇒ y = $-\frac{9}{4}$.

Substituting value of y in eq (ii) we get,

$$\begin{aligned}\Rightarrow x - \Big(-\frac{9}{4}\Big) = 3 \\[1em] \Rightarrow x + \frac{9}{4} = 3 \\[1em] \Rightarrow x = 3 - \frac{9}{4} \\[1em] \Rightarrow x = \frac{12 - 9}{4} \\[1em] \Rightarrow x = \frac{3}{4}.\end{aligned}$$

Hence, $x = \frac{3}{4} \text{ and } y = -\frac{9}{4}.$

x - 3y = 3x - 1

⇒ 3x - x + 3y = 1

⇒ 2x + 3y = 1 .......(i)

3x - 1 = 2x - y

⇒ 3x - 2x + y = 1

⇒ x + y = 1 .......(ii)

Multiplying eq. (ii) by 2 we get,

2x + 2y = 2 ........(iii)

Subtracting eq. (iii) from (i) we get,

⇒ 2x + 3y - (2x + 2y) = 1 - 2

⇒ 2x - 2x + 3y - 2y = -1

⇒ y = -1.

Substituting value of y in eq. (ii) we get,

⇒ x + (-1) = 1

⇒ x = 1 + 1

⇒ x = 2.

Hence, x = 2 and y = -1.

Solving 1st equation,

$$\begin{aligned}\Rightarrow 4x + \frac{x - y}{8} = 17 \\[1em] \Rightarrow \frac{32x + x - y}{8} = 17 \\[1em] \Rightarrow 33x - y = 136 .......(i)\end{aligned}$$

Solving 2nd equation,

$$\begin{aligned}\Rightarrow 2y + x - \frac{5y + 2}{3} = 2 \\[1em] \Rightarrow \frac{6y + 3x - 5y - 2}{3} = 2 \\[1em] \Rightarrow \frac{y + 3x - 2}{3} = 2 \\[1em] \Rightarrow y + 3x - 2 = 6 \\[1em] \Rightarrow y + 3x = 8 ........(ii)\end{aligned}$$

Adding equation (i) and (ii) we get,

⇒ 33x - y + y + 3x = 136 + 8

⇒ 36x = 144

⇒ x = 4.

Substituting value of x in equation (ii) we get,

⇒ y + 3(4) = 8

⇒ y + 12 = 8

⇒ y = -4.

Hence, x = 4 and y = -4.

Solving 1st equation,

$$\begin{aligned}\Rightarrow \frac{x + 1}{2} + \frac{y - 1}{3} = 8 \\[1em] \Rightarrow \frac{3(x + 1) + 2(y - 1)}{6} = 8 \\[1em] \Rightarrow \frac{3x + 3 + 2y - 2}{6} = 8 \\[1em] \Rightarrow \frac{3x + 2y + 1}{6} = 8 \\[1em] \Rightarrow 3x + 2y + 1 = 48 \\[1em] \Rightarrow 3x + 2y = 47 ......(i)\end{aligned}$$

Solving 2nd equation,

$$\begin{aligned}\Rightarrow \frac{x - 1}{3} + \frac{y + 1}{2} = 9 \\[1em] \Rightarrow \frac{2(x - 1) + 3(y + 1)}{6} = 9 \\[1em] \Rightarrow \frac{2x - 2 + 3y + 3}{6} = 9 \\[1em] \Rightarrow \frac{2x + 3y + 1}{6} = 9 \\[1em] \Rightarrow 2x + 3y + 1 = 54 \\[1em] \Rightarrow 2x + 3y = 53 .......(ii)\end{aligned}$$

Multiplying eq. (i) by 2 and eq. (ii) by 3 we get,

6x + 4y = 94 ......(iii)

6x + 9y = 159 .......(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ 6x + 9y - (6x + 4y) = 159 - 94

⇒ 5y = 65

⇒ y = 13.

Substituting value of y in eq. (iii) we get,

⇒ 6x + 4(13) = 94

⇒ 6x + 52 = 94

⇒ 6x = 42

⇒ x = 7.

Hence, x = 7 and y = 13.

Given,

$\frac{3}{x} + 4y = 7$ .......(i)

$\frac{5}{x} + 6y = 13$ .......(ii)

Multiplying eq. (i) by 3 and eq. (ii) by 2 we get,

$\frac{9}{x} + 12y = 21$ ......(iii)

$\frac{10}{x} + 12y = 26$ .......(iv)

Subtracting eq. (iii) from (iv) we get,

$$\begin{aligned}\Rightarrow \frac{10}{x} - \frac{9}{x} + 12y - 12y = 26 - 21 \\[1em] \Rightarrow \frac{1}{x} = 5 \\[1em] \Rightarrow x = \frac{1}{5}.\end{aligned}$$

Substituting value of x in eq. (i) we get,

$$\begin{aligned}\Rightarrow \frac{3}{x} + 4y = 7 \\[1em] \Rightarrow \frac{3}{\frac{1}{5}} + 4y = 7 \\[1em] \Rightarrow 15 + 4y = 7 \\[1em] \Rightarrow 4y = -8 \\[1em] \Rightarrow y = -2.\end{aligned}$$

Hence, x = $\frac{1}{5}$ and y = -2.

$5x - 9 = \frac{1}{y} \text{ or } 5x - \frac{1}{y} = 9$ ........(i)

$x + \frac{1}{y} = 3$ .......(ii)

Adding eq. (i) and (ii) we get,

$$\begin{aligned}\Rightarrow 5x - \frac{1}{y} + x + \frac{1}{y} = 9 + 3 \\[1em] \Rightarrow 6x = 12 \\[1em] \Rightarrow x = 2.\end{aligned}$$

Substituting value of x in eq. (ii) we get,

$$\begin{aligned}\Rightarrow x + \frac{1}{y} = 3 \\[1em] \Rightarrow 2 + \frac{1}{y} = 3 \\[1em] \Rightarrow \frac{1}{y} = 1 \\[1em] \Rightarrow y = 1.\end{aligned}$$

Hence, x = 2 and y = 1.

Given,

px + qy = p - q .......(i)

qx - py = p + q .......(ii)

Multiplying eq. (i) by q and eq. (ii) by p we get,

pqx + q²y = pq - q² .......(iii)

pqx - p²y = p² + pq .......(iv)

Subtracting eq. (iv) from (iii) we get,

⇒ pqx + q²y - (pqx - p²y) = pq - q² - (p² + pq)

⇒ pqx - pqx + q²y + p²y = pq - pq - q² - p²

⇒ q²y + p²y = -q² - p²

⇒ y(q² + p²) = -(q² + p²)

⇒ y = $\frac{-(\text{q}^2 + \text{p}^2)}{\text{q}^2 + \text{p}^2}$ = -1.

Substituting value of y in eq. (i) we get,

⇒ px + q(-1) = p - q

⇒ px - q = p - q

⇒ px = p - q + q

⇒ px = p

⇒ x = 1.

Hence, x = 1 and y = -1.

$$\begin{aligned}\phantom{\Rightarrow} \frac{x}{a} - \frac{y}{b} = 0 \\[1em] \Rightarrow \frac{bx - ay}{ab} = 0 \\[1em] \Rightarrow bx - ay = 0.\end{aligned}$$

Given equations can be written as,

bx - ay = 0 .......(i)

ax + by = (a² + b²) .......(ii)

Multiplying eq. (i) by a and (ii) by b we get,

abx - a²y = 0 .........(iii)

abx + b²y = b(a² + b²) .......(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ abx + b²y - (abx - a²y) = b(a² + b²)

⇒ abx - abx + b²y + a²y = b(a² + b²)

⇒ y(b² + a²) = b(a² + b²)

⇒ y = b.

Substituting value of y in eq. (i) we get,

⇒ bx - ay = 0

⇒ bx - ab = 0

⇒ bx = ab

⇒ x = a.

Hence, x = a and y = b.

Given,

2x + y = 23 .......(i)

4x - y = 19 .......(ii)

Multiplying eq. (i) by 2 we get,

4x + 2y = 46 ......(iii)

Subtracting eq. (ii) from (iii) we get,

⇒ 4x + 2y - (4x - y) = 46 - 19

⇒ 4x - 4x + 2y + y = 27

⇒ 3y = 27

⇒ y = 9.

Substituting value of y in eq. (ii) we get,

⇒ 4x - 9 = 19

⇒ 4x = 28

⇒ x = 7.

Substituting value of x and y in x - 3y,

⇒ x - 3y = 7 - 3(9) = 7 - 27 = -20.

Substituting value of x and y in 5y - 2x,

⇒ 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31.

Hence, x = 7, y = 9, x - 3y = -20 and 5y - 2x = 31.

Given,

ax + by = 7, when x = 2, y = 1.

⇒ 2a + b = 7 ......(i)

ax + by = 1 when x = -1, y = 1.

⇒ -a + b = 1 .......(ii)

Multiplying eq. (ii) by 2 we get,

⇒ -2a + 2b = 2 .......(iii)

Adding eq. (i) and (iii) we get,

⇒ 2a + b + (-2a + 2b) = 7 + 2

⇒ 2a - 2a + b + 2b = 9

⇒ 3b = 9

⇒ b = 3.

Substituting value of b in eq. (ii) we get,

⇒ -a + 3 = 1

⇒ -a = -2

⇒ a = 2.

Hence, a = 2 and b = 3.

Given,

3x - 7y = 7 ........(i)

11x + 5y = 87 .......(ii)

5x + 4y = 43 .......(iii)

Solving first two equations simultaneously,

Multiplying eq. (i) by 11 and eq. (ii) by 3 we get,

33x - 77y = 77 ......(iv)

33x + 15y = 261 ......(v)

Subtracting eq. (iv) from (v) we get,

⇒ 33x + 15y - (33x - 77y) = 261 - 77

⇒ 33x - 33x + 15y + 77y = 184

⇒ 92y = 184

⇒ y = 2.

Substituting value of y in eq. (i) we get,

⇒ 3x - 7y = 7

⇒ 3x - 7(2) = 7

⇒ 3x - 14 = 7

⇒ 3x = 21

⇒ x = 7.

Substituting x = 7 and y = 2 in L.H.S. of eq. (iii),

5x + 4y = 43 .......(iii)

⇒ 5(7) + 4(2) = 35 + 8 = 43.

Since, L.H.S. = R.H.S. hence following equations can be held simultaneously.

Hence, x = 7 and y = 2.

Given equations can be written as,

3x + 2y - 4 = 0

8x + 5y - 9 = 0

By cross multiplication method,

$$\begin{aligned}\therefore \frac{x}{2 \times (-9) - 5 \times (-4)} = \frac{y}{(-4) \times 8 - (-9) \times 3} = \frac{1}{3 \times 5 - 8 \times 2} \\[1em] \Rightarrow \frac{x}{-18 + 20} = \frac{y}{-32 + 27} = \frac{1}{15 - 16} \\[1em] \Rightarrow \frac{x}{2} = \frac{y}{-5} = \frac{1}{-1} \\[1em] \therefore \frac{x}{2} = \frac{1}{-1} \text{ and } \frac{y}{-5} = \frac{1}{-1} \\[1em] \Rightarrow x = -2 \text{ and } y = 5.\end{aligned}$$

Hence, x = -2 and y = 5.

Given equations can be written as,

3x - 7y + 10 = 0

-2x + y - 3 = 0

By cross multiplication method,

$$\begin{aligned}\therefore \frac{x}{(-7) \times (-3) - 1 \times 10} = \frac{y}{10 \times (-2) - (-3) \times 3} = \frac{1}{3 \times 1 - (-2) \times (-7)} \\[1em] \Rightarrow \frac{x}{21 - 10} = \frac{y}{-20 + 9} = \frac{1}{3 - 14} \\[1em] \Rightarrow \frac{x}{11} = \frac{y}{-11} = \frac{1}{-11} \\[1em] \therefore \frac{x}{11} = \frac{1}{-11} \text{ and } \frac{y}{-11} = \frac{1}{-11} \\[1em] \Rightarrow x = -1 \text{ and } y = 1.\end{aligned}$$

Hence, x = -1 and y = 1.

Given equations can be written as,

2x - 5y + 1 = 0

3x + y - 7 = 0

By cross multiplication method,

$$\begin{aligned}\therefore \frac{x}{-5 \times (-7) - 1 \times 1} = \frac{y}{1 \times 3 - (-7) \times 2} = \frac{1}{2 \times 1 - 3 \times -5}\\[1em] \Rightarrow \frac{x}{35 - 1} = \frac{y}{3 - (-14)} = \frac{1}{2 - (-15)}\\[1em] \Rightarrow \frac{x}{34} = \frac{y}{3 + 14} = \frac{1}{2 + 15}\\[1em] \Rightarrow \frac{x}{34} = \frac{y}{17} = \frac{1}{17}\\[1em] \Rightarrow \frac{x}{34} = \frac{1}{17} \text{ and } \frac{y}{17} = \frac{1}{17}\\[1em] \Rightarrow x = \frac{34}{17} \text{ and }y = \frac{17}{17}\\[1em] \Rightarrow x = 2 \text{ and }y = 1.\end{aligned}$$

Hence, x = 2 and y = 1.

Given equations can be written as,

x + 3y + 4 = 0

3x - y + 2 = 0

By cross multiplication method,

$$\begin{aligned}\therefore \frac{x}{3 \times 2 - (-1) \times 4} = \frac{y}{4 \times 3 - 2 \times 1} = \frac{1}{1 \times (-1) - 3 \times 3}\\[1em] \Rightarrow \frac{x}{6 + 4} = \frac{y}{12 - 2} = \frac{1}{-1 - 9}\\[1em] \Rightarrow \frac{x}{10} = \frac{y}{10} = \frac{1}{-10}\\[1em] \Rightarrow \frac{x}{10} = \frac{1}{-10} \text{ and }\frac{y}{10} = \frac{1}{-10}\\[1em] \Rightarrow x = -\frac{10}{10} \text{ and }y = -\frac{10}{10}\\[1em] \Rightarrow x = -1 \text{ and }y = -1\end{aligned}$$

Hence, x = -1 and y = -1.

Given equations can be written as,

x - y - (a + b) = 0

ax + by - (a² - b²) = 0

By cross multiplication method,

$$\begin{aligned}\therefore \frac{x}{(-1) \times [-(a^2 - b^2)] - b \times [-(a + b)]} = \frac{y}{[-(a + b) \times a] - [-(a^2 - b^2) \times 1] } = \frac{1}{1 \times b - a \times (-1)} \\[1em] \Rightarrow \frac{x}{a^2 - b^2 + ab + b^2} = \frac{y}{-a^2 - ab + a^2 - b^2} = \frac{1}{b + a} \\[1em] \Rightarrow \frac{x}{a^2 + ab} = \frac{y}{-ab - b^2} = \frac{1}{b + a} \\[1em] \therefore \frac{x}{a^2 + ab} = \frac{1}{b + a} \text{ and } \frac{y}{-b(a + b)} = \frac{1}{b + a} \\[1em] \Rightarrow \frac{x}{a(a + b)} = \frac{1}{b + a} \text{ and } \frac{y}{-b(a + b)} = \frac{1}{b + a} \\[1em] \Rightarrow x = \frac{a(a + b)}{b + a} \text{ and } y = \frac{-b(a + b)}{b + a} \\[1em] \Rightarrow x = a \text{ and } y = -b.\end{aligned}$$

Hence, x = a and y = -b.

Given equations can be written as,

2bx + ay - 2ab = 0

bx - ay - 4ab = 0

By cross multiplication method,

$$\begin{aligned}\therefore \frac{x}{a \times -(4ab) - (-a) \times [-2ab]} = \frac{y}{[-(2ab) \times b] - [-(4ab) \times 2b] } = \frac{1}{2b \times (-a) - b \times a} \\[1em] \Rightarrow \frac{x}{-4a^2b - 2a^2b} = \frac{y}{-2ab^2 + 8ab^2} = \frac{1}{-2ab - ab} \\[1em] \Rightarrow -\frac{x}{6a^2b} = \frac{y}{6ab^2} = -\frac{1}{3ab} \\[1em] \therefore -\frac{x}{6a^2b} = -\frac{1}{3ab} \text{ and } \frac{y}{6ab^2} = -\frac{1}{3ab} \\[1em] \Rightarrow x = \frac{6a^2b}{3ab} \text{ and } y = -\frac{6ab^2}{3ab} \\[1em] \Rightarrow x = 2a \text{ and } y = -2b.\end{aligned}$$

Hence, x = 2a and y = -2b.

Substituting $\frac{1}{x} = a \text{ and } \frac{1}{y} = b$ in above equations we get,

$2a + \frac{2}{3}b = \frac{1}{3}$

Multiply by 3:

6a + 2b = 1 .......(i)

2a - b = 2 .......(ii)

⇒ -b = 2 - 2a

⇒ b = 2a - 2

Substituting this value of b in equation (i) we get,

6a + 2(2a - 2) = 1

⇒ 6a + 4a - 4 = 1

⇒ 10a - 4 = 1

⇒ 10a = 1 + 4

⇒ 10a = 5

⇒ a = $\frac{5}{10}$

⇒ a = $\frac{1}{2}$

⇒ $\frac{1}{x} = \frac{1}{2}$

⇒ $x = 2$

From (ii) we have,

b = 2a - 2

Substituting value of a in above equation we get,

b = 2 x $\frac{1}{2}$ - 2

⇒ b = 1 - 2

⇒ b = -1

⇒ $\frac{1}{y}$ = -1

⇒ $y$ = -1

Hence, x = 2 and y = -1.

Substituting $\frac{1}{x} = a \text{ and } \frac{1}{y} = b$ in above equations we get,

$\frac{3}{2}a + \frac{2}{3}b = 5$ ......(i)

5a - 3b = 1 .......(ii)

Solving eq (i) we get,

$$\begin{aligned}\Rightarrow \frac{3}{2}a + \frac{2}{3}b = 5 \\[1em] \Rightarrow \frac{9a + 4b}{6} = 5 \\[1em] \Rightarrow 9a + 4b = 30 .......(iii)\end{aligned}$$

Multiplying eq. (ii) by 4 we get,

20a - 12b = 4 .......(iv)

Multiplying eq. (iii) by 3 we get,

27a + 12b = 90 .......(v)

Adding eq. (iv) and (v) we get,

⇒ 20a - 12b + 27a + 12b = 4 + 90

⇒ 47a = 94

⇒ a = 2.

$$\begin{aligned}\therefore \frac{1}{x} = 2 \\[1em] \Rightarrow x = \frac{1}{2}.\end{aligned}$$

Substituting value of a in eq. (ii) we get,

⇒ 5(2) - 3b = 1

⇒ 10 - 3b = 1

⇒ 3b = 10 - 1

⇒ 3b = 9

⇒ b = 3.

$$\begin{aligned}\therefore \frac{1}{y} = 3 \\[1em] \Rightarrow y = \frac{1}{3}.\end{aligned}$$

Hence, $x = \frac{1}{2} \text{ and } y = \frac{1}{3}.$

Given,

$\frac{7x - 2y}{xy} = 5 \text{ or } \frac{7}{y} - \frac{2}{x} = 5$

$\frac{8x + 7y}{xy} = 15 \text{ or } \frac{8}{y} + \frac{7}{x} = 15$

Substituting $\frac{1}{x} = a \text{ and } \frac{1}{y} = b$ in above equations we get,

7b - 2a = 5 ......(i)

8b + 7a = 15 ......(ii)

Multiplying eq. (i) by 7 and eq. (ii) by 2 we get,

49b - 14a = 35 ......(iii)

16b + 14a = 30 .......(iv)

Adding equations (iii) and (iv) we get,

⇒ 49b - 14a + 16b + 14a = 35 + 30

⇒ 65b = 65

⇒ b = 1.

$$\begin{aligned}\therefore \frac{1}{y} = 1 \\[1em] \Rightarrow y = 1.\end{aligned}$$

Substituting value of b in eq (i) we get,

⇒ 7(1) - 2a = 5

⇒ 7 - 2a = 5

⇒ 2a = 7 - 5

⇒ 2a = 2

⇒ a = 1.

$$\begin{aligned}\therefore \frac{1}{x} = 1 \\[1em] \Rightarrow x = 1.\end{aligned}$$

Hence, x = 1 and y = 1.

Given,

99x + 101y = 499xy

101x + 99y = 501xy

First we note that x = 0, y = 0 is a solution of equations.

Now when x ≠ 0 and y ≠ 0.

Dividing the above equations by xy we get,

$\frac{99}{y} + \frac{101}{x} = 499$ .......(i)

$\frac{101}{y} + \frac{99}{x} = 501$ .......(ii)

Substituting $\frac{1}{x} = p \text{ and } \frac{1}{y} = q$ in both equations and multiplying eq. (i) by 101 and (ii) by 99 we get,

9999q + 10201p = 50399 ......(iii)

9999q + 9801p = 49599 ......(iv)

Subtracting (iv) from (iii) we get,

⇒ 9999q + 10201p - (9999q + 9801p) = 50399 - 49599

⇒ 9999q - 9999q + 10201p - 9801p = 800

⇒ 400p = 800

⇒ p = 2.

$\therefore \frac{1}{x} = 2 \text{ or } x = \frac{1}{2}.$

Substituting value of p in (iii) we get,

⇒ 9999q + 10201(2) = 50399

⇒ 9999q + 20402 = 50399

⇒ 9999q = 29997

⇒ q = 3.

$\therefore \frac{1}{y} = 3 \text{ or } y = \frac{1}{3}$.

Hence, x = 0, y = 0 and x = $\frac{1}{2} \text{ and } y = \frac{1}{3}.$

Given,

3x + 14y = 5xy .......(i)

21y - x = 2xy ........(ii)

First we note that x = 0, y = 0 is a solution of the equations.

Now when x ≠ 0 and y ≠ 0.

Dividing eq. (i) by xy we get,

$$\begin{aligned}\Rightarrow 3x + 14y = 5xy \\[1em] \Rightarrow \frac{3x}{xy} + \frac{14y}{xy} = \frac{5xy}{xy} \\[1em] \Rightarrow \frac{3}{y} + \frac{14}{x} = 5 ......(iii)\end{aligned}$$

Dividing eq. (ii) by xy we get,

$$\begin{aligned}\Rightarrow 21y - x = 2xy \\[1em] \Rightarrow \frac{21y}{xy} - \frac{x}{xy} = 2 \\[1em] \Rightarrow \frac{21}{x} - \frac{1}{y} = 2 .......(iv)\end{aligned}$$

Substituting $\frac{1}{x} = a \text{ and } \frac{1}{y} = b$ in eq. (iii) and (iv) we get,

3b + 14a = 5 ........(v)

21a - b = 2 .........(vi)

Multiplying eq. (vi) by 3 we get,

63a - 3b = 6 .......(vii)

Adding eq. (v) and (vii) we get,

⇒ 3b + 14a + (63a - 3b) = 5 + 6
⇒ 77a = 11
⇒ a = $\frac{11}{77} = \frac{1}{7}$.

$$\begin{aligned}\therefore \frac{1}{x} = \frac{1}{7} \\[1em] \Rightarrow x = 7.\end{aligned}$$

Substituting value of a in eq. (v) we get,

⇒ 3b + $14\times \frac{1}{7}$ = 5

⇒ 3b + 2 = 5

⇒ 3b = 3

⇒ b = 1.

$$\begin{aligned}\therefore \frac{1}{y} = 1 \\[1em] \Rightarrow y = 1.\end{aligned}$$

Hence, x = 0, y = 0 and x = 7, y = 1.

Given,

3x + 5y = 4xy ........(i)

2y - x = xy .......(ii)

First we note that x = 0, y = 0 is a solution of equations.

Now when x ≠ 0 and y ≠ 0.

Dividing above equations by xy we get,

$\frac{3}{y} + \frac{5}{x} = 4$ .......(iii)

$\frac{2}{x} - \frac{1}{y} = 1$ .......(iv)

Substituting $\frac{1}{x} = a \text{ and } \frac{1}{y} = b$ in eq. (iii) and (iv) we get,

3b + 5a = 4 .......(v)

2a - b = 1 .......(vi)

Multiplying eq. (vi) by 3 we get,

6a - 3b = 3 ........(vii)

Adding eq. (v) and (vii) we get,

⇒ 3b + 5a + 6a - 3b = 4 + 3

⇒ 11a = 7

⇒ a = $\frac{7}{11}$.

$$\begin{aligned}\therefore \frac{1}{x} = \frac{7}{11} \\[1em] \Rightarrow x = \frac{11}{7}.\end{aligned}$$

Substituting value of a in eq. (vi) we get,

$$\begin{aligned}\Rightarrow 2 \times \frac{7}{11} - b = 1 \\[1em] \Rightarrow \frac{14}{11} - b = 1 \\[1em] \Rightarrow b = \frac{14}{11} - 1 \\[1em] \Rightarrow b = \frac{3}{11} \\[1em] \therefore \frac{1}{y} = \frac{3}{11} \\[1em] \Rightarrow y = \frac{11}{3}.\end{aligned}$$

Hence, x = 0, y = 0 and x = $\frac{11}{7}, y = \frac{11}{3}.$

Substituting $\frac{1}{x + 1} = a \text{ and } \frac{1}{y - 1} = b$ in above eq. we get,

20a + 4b = 5 .......(i)

10a - 4b = 1 .......(ii)

Multiplying equation (ii) by 2 we get,

20a - 8b = 2 .......(iii)

Subtracting eq. (iii) from (i) we get,

⇒ 20a + 4b - (20a - 8b) = 5 - 2

⇒ 12b = 3

⇒ b = $\frac{3}{12} = \frac{1}{4}$.

$$\begin{aligned}\therefore \frac{1}{y - 1} = \frac{1}{4} \\[1em] \Rightarrow y - 1 = 4 \\[1em] \Rightarrow y = 5.\end{aligned}$$

Substituting value of b in (iii) we get,

⇒ 20a - $8 \times \frac{1}{4}$ = 2

⇒ 20a - 2 = 2

⇒ 20a = 2 + 2

⇒ a = $\frac{4}{20}$

⇒ a = $\frac{1}{5}$

$$\begin{aligned}\therefore \frac{1}{x + 1} = \frac{1}{5} \\[1em] \Rightarrow x + 1 = 5 \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, x = 4 and y = 5.

Substituting $\frac{1}{x + y} = a \text{ and } \frac{1}{x - y} = b$ in above eq. we get,

3a + 2b = 3 .......(i)

2a + 3b = $\frac{11}{3}$ .......(ii)

Multiplying eq. (i) by 2 and (ii) by 3 we get,

6a + 4b = 6 .......(iii)

6a + 9b = 11 .......(iv)

Subtracting eq. (iii) from iv we get,

⇒ 6a + 9b - (6a + 4b) = 11 - 6

⇒ 6a + 9b - 6a - 4b = 5

⇒ 5b = 5

⇒ b = 1.

$$\begin{aligned}\therefore \frac{1}{x - y} = 1 \\[1em] \Rightarrow x - y = 1 .......(v)\end{aligned}$$

Substituting value of b in eq. (iii) we get,

⇒ 6a + 4(1) = 6

⇒ 6a + 4 = 6

⇒ 6a = 2

⇒ a = $\frac{2}{6} = \frac{1}{3}.$

$\therefore \frac{1}{x + y} = \frac{1}{3}$

⇒ x + y = 3

⇒ x = 3 - y.

Putting above value of x in eq. (v) we get,

⇒ (3 - y) - y = 1

⇒ 3 - 2y = 1

⇒ 2y = 3 - 1

⇒ 2y = 2

⇒ y = 1.

⇒ x = 3 - y = 3 - 1 = 2.

Hence, x = 2 and y = 1.

Substituting $\frac{1}{2x + 3y} = a \text{ and } \frac{1}{3x - 2y} = b$ in above eq. we get,

$\frac{1}{2}a + \frac{12}{7}b = \frac{1}{2}$ .....(i)

7a + 4b = 2 .......(ii)

Multiplying eq. (i) by 14 we get,

7a + 24b = 7 ......(iii)

Subtracting eq. (ii) from (iii) we get,

⇒ 7a + 24b - (7a + 4b) = 7 - 2

⇒ 7a + 24b - 7a - 4b = 7 - 2

⇒ 20b = 5

⇒ b = $\frac{5}{20} = \frac{1}{4}$.

$\therefore \frac{1}{3x - 2y} = \frac{1}{4}$

⇒ 3x - 2y = 4 ........(iv)

Substituting value of b in eq. (iii) we get,

⇒ 7a + $24 \times \frac{1}{4}$ = 7

⇒ 7a + 6 = 7

⇒ 7a = 1

⇒ a = $\frac{1}{7}$.

$\therefore \frac{1}{2x + 3y} = \frac{1}{7}$

⇒ 2x + 3y = 7 .......(v)

Multiplying eq. (iv) by 2 and eq. (v) by 3 we get,

⇒ 6x - 4y = 8 .......(vi)

⇒ 6x + 9y = 21 .......(vii)

Subtracting eq. (vi) from (vii) we get,

⇒ (6x + 9y) - (6x - 4y) = 21 - 8

⇒ 13y = 13

⇒ y = 1.

Substituting value of y in eq. (vi) we get,

⇒ 6x - 4(1) = 8

⇒ 6x = 8 + 4

⇒ 6x = 12

⇒ x = 2.

Hence, x = 2 and y = 1.

Substitute $\frac{1}{x + 2y} = p \text{ and } \frac{1}{3x - 2y} = q$ in above equations,

$\frac{1}{2}p + \frac{5}{3}q = -\frac{3}{2}$ ........(i)

$\frac{5}{4}p - \frac{3}{5}q = \frac{61}{60}$ ........(ii)

Multiplying (i) by $\frac{3}{5}$ and (ii) by $\frac{5}{3}$ we get,

$\frac{3}{10}p + q = -\frac{9}{10}$ .......(iii)

$\frac{25}{12}p - q = \frac{61}{36}$ .......(iv)

Adding (iii) and (iv) we get,

$$\begin{aligned}\Rightarrow \frac{3}{10}p + q + \frac{25}{12}p - q = -\frac{9}{10} + \frac{61}{36} \\[1em] \Rightarrow \frac{18p + 125p}{60} = \frac{-162 + 305}{180} \\[1em] \Rightarrow \frac{143}{60}p = \frac{143}{180} \\[1em] \Rightarrow p = \frac{143 \times 60}{180 \times 143} \\[1em] \Rightarrow p = \frac{1}{3} \\[1em] \therefore \frac{1}{x + 2y} = \frac{1}{3} \\[1em] \Rightarrow x + 2y = 3 .......(v)\end{aligned}$$

Substituting value of p in (i) we get,

$$\begin{aligned}\Rightarrow \frac{1}{2} \times \frac{1}{3} + \frac{5}{3}q = -\frac{3}{2} \\[1em] \Rightarrow \frac{1}{6} + \frac{5}{3}q = -\frac{3}{2} \\[1em] \Rightarrow \frac{5}{3}q = -\frac{3}{2} - \frac{1}{6} \\[1em] \Rightarrow \frac{5}{3}q = \frac{-9 - 1}{6} \\[1em] \Rightarrow \frac{5}{3}q = -\frac{10}{6} \\[1em] \Rightarrow q = -\frac{10 \times 3}{6 \times 5} \\[1em] \Rightarrow q = -1 \\[1em] \therefore \frac{1}{3x - 2y} = -1 \\[1em] \Rightarrow 3x - 2y = -1 \\[1em] \Rightarrow 2y - 3x = 1 ......(vi)\end{aligned}$$

Subtracting (vi) from (v) we get,

⇒ x + 2y - (2y - 3x) = 3 - 1

⇒ x + 3x = 2

⇒ 4x = 2

⇒ x = $\frac{1}{2}$.

Substituting value of x from (v) we get,

$$\begin{aligned}\Rightarrow \frac{1}{2} + 2y = 3 \\[1em] \Rightarrow 2y = 3 - \frac{1}{2} \\[1em] \Rightarrow 2y = \frac{6 - 1}{2} \\[1em] \Rightarrow 2y = \frac{5}{2} \\[1em] \Rightarrow y = \frac{5}{4}.\end{aligned}$$

Hence, x = $\frac{1}{2} \text{ and } y = \frac{5}{4}.$

Substituting x = 3 and y = k in 3x - 4y + 7 = 0 we get,

⇒ 3(3) - 4(k) + 7 = 0

⇒ 9 - 4k + 7 = 0

⇒ 4k = 16

⇒ k = 4.

Hence, Option 3 is the correct option.

Given,

2x - y = 5 ......(i)

5x - y = 11 .....(ii)

Subtracting eq. (i) from (ii) we get,

⇒ 5x - y - (2x - y) = 11 - 5

⇒ 5x - y - 2x + y = 6

⇒ 3x = 6

⇒ x = 2.

Substituting value of x in eq. (i),

⇒ 2(2) - y = 5

⇒ 4 - y = 5

⇒ y = 4 - 5

⇒ y = -1.

Hence, Option 2 is the correct option.

Given,

x - y = 2 ......(i)

x + y = 4 ......(ii)

Adding both the equations we get,

⇒ x - y + x + y = 2 + 4

⇒ 2x = 6

⇒ x = 3.

Substituting value of x in (i),

⇒ 3 - y = 2

⇒ y = 3 - 2

⇒ y = 1.

Hence, Option 3 is the correct option.

Given,

$\frac{4}{x} + 5y = 7$ .......(i)

$\frac{3}{x} + 4y = 5$ .......(ii)

Multiplying eq. (i) by 4 and eq. (ii) by 5 we get,

$\frac{16}{x} + 20y = 28$ .......(iii)

$\frac{15}{x} + 20y = 25$ .......(iv)

Subtracting eq. (iv) from (iii) we get,

$$\begin{aligned}\Rightarrow \frac{16}{x} + 20y - \Big(\frac{15}{x} + 20y\Big) = 28 - 25 \\[1em] \Rightarrow \frac{16}{x} - \frac{15}{x} = 3 \\[1em] \Rightarrow \frac{1}{x} = 3 \\[1em] \Rightarrow x = \frac{1}{3}.\end{aligned}$$

Substituting value of x in (ii),

$$\begin{aligned}\Rightarrow \frac{3}{\frac{1}{3}} + 4y = 5 \\[1em] \Rightarrow 9 + 4y = 5 \\[1em] \Rightarrow 4y = 5 - 9 \\[1em] \Rightarrow 4y = -4 \\[1em] \Rightarrow y = -1.\end{aligned}$$

Hence, Option 1 is the correct option.

Substituting x = 2, y = -3 in x - 4y - 14 = 0 we get,

= 2 - 4(-3) - 14

= 2 + 12 - 14

= 0.

Substituting x = 2, y = -3 in 5x - y - 13 = 0 we get,

= 5(2) - (-3) - 13

= 10 + 3 - 13

= 13 - 13

= 0.

Hence, Option 4 is the correct option.

If we substitute x = 3 and y = 7 into the equation 5x - 2y = 1, we get

Taking L.H.S.

⇒ 5.(3) - 2.(7)

⇒ 15 - 14

⇒ 1.

Since, L.H.S. = R.H.S.

This confirms that (3, 7) is indeed a solution.

∴ Statement 1 is true.

A single linear equation with two variables (like x and y) represents a line in a coordinate plane.

A line has infinitely many points on it, and each point corresponds to a solution of the equation.

Therefore, a single linear equation has infinitely many solutions, not just one.

∴ Statement 2 is false.

∴ Statement 1 is true, and Statement 2 is false.

Hence, option 3 is the correct option.

One of the methods of solving a pair of linear equations is elimination method.

∴ Reason (R) is true.

Given, x - y = 1 .................(1)

2x + y = $\frac{7}{2}$ ..................(2)

Adding equation (1) and (2) we get

$$\begin{aligned}\Rightarrow (x - y) + (2x + y) = 1 + \frac{7}{2}\\[1em] \Rightarrow x - y + 2x + y = \frac{2}{2} + \frac{7}{2}\\[1em] \Rightarrow 3x = \frac{9}{2}\\[1em] \Rightarrow x = \frac{9}{2 \times 3}\\[1em] \Rightarrow x = \frac{3}{2}\end{aligned}$$

Substituting the value of x in equation (1), we get :

⇒ x - y = 1

$$\begin{aligned}\Rightarrow \frac{3}{2} - y = 1\\[1em] \Rightarrow \frac{3}{2} - 1 = y\\[1em] \Rightarrow \frac{3 - 2}{2} = y\\[1em] \Rightarrow y = \frac{1}{2}.\end{aligned}$$

So, x = $\frac{3}{2}$, y = $\frac{1}{2}$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).

Hence, option 3 is the correct option.

We can use cross-multiplication method to solve a pair of linear equations.

∴ Reason (R) is true.

Given, equations can be written as :

$\sqrt{2}x - \sqrt{3}y - 0$ = 0 .................(1)

$\sqrt{3}x + \sqrt{2}y$ - 5 = 0................(2)

By cross multiplication method,

$$\begin{aligned}\therefore \frac{x}{-\sqrt{3} \times (-5) - \sqrt{2} \times 0} = \frac{y}{0 \times \sqrt{3} - (-5) \times \sqrt{2}} = \frac{1}{\sqrt{2} \times \sqrt{2} - \sqrt{3} \times (-\sqrt{3})}\\[1em] \Rightarrow \frac{x}{5\sqrt{3} - 0} = \frac{y}{0 + 5 \sqrt{2}} = \frac{1}{2 + 3}\\[1em] \Rightarrow \frac{x}{5\sqrt{3}} = \frac{y}{5 \sqrt{2}} = \frac{1}{5}\\[1em] \Rightarrow \frac{x}{5\sqrt{3}} = \frac{1}{5} \text{ and } \frac{y}{5 \sqrt{2}} = \frac{1}{5} \\[1em] \Rightarrow x = \frac{5\sqrt{3}}{5} \text{ and } y = \frac{5\sqrt{2}}{5}\\[1em] \Rightarrow x = \sqrt{3} \text{ and } y = \sqrt{2}.\end{aligned}$$

So, the solution are x = $\sqrt{3}$, y = $\sqrt{2}$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given,

2x - $\frac{3}{4}y$ = 3 .........(i)

5x - 2y = 7 ..........(ii)

Multiplying eq. (i) by 5 and (ii) by 2 we get,

10x - $\frac{15}{4}y$ = 15 .......(iii)

10x - 4y = 14 ........(iv)

Subtracting eq. (iv) from (iii),

$$\begin{aligned}\Rightarrow 10x - \frac{15}{4}y - (10x - 4y) = 15 - 14 \\[1em] \Rightarrow 4y - \frac{15}{4}y = 1 \\[1em] \Rightarrow \frac{16y - 15y}{4} = 1 \\[1em] \Rightarrow \frac{y}{4} = 1 \\[1em] \Rightarrow y = 4.\end{aligned}$$

Substituting value of y in eq. (ii) we get,

⇒ 5x - 2(4) = 7

⇒ 5x - 8 = 7

⇒ 5x = 15

⇒ x = 3.

Hence, x = 3 and y = 4.

Given,

2(x - 4) = 9y + 2 ......(i)

x - 6y = 2 or x = 2 + 6y......(ii)

Substituting value of x from eq. (ii) in (i),

⇒ 2[(2 + 6y) - 4] = 9y + 2

⇒ 2[6y - 2] = 9y + 2

⇒ 12y - 4 = 9y + 2

⇒ 12y - 9y = 2 + 4

⇒ 3y = 6

⇒ y = 2.

x = 2 + 6y = 2 + 6(2) = 2 + 12 = 14.

Hence, x = 14 and y = 2.

Given,

97x + 53y = 177 .......(i)

53x + 97y = 573 .......(ii)

Multiplying eq. (i) by 53 and (ii) by 97,

5141x + 2809y = 9381 .......(iii)

5141x + 9409y = 55581 ......(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ 5141x + 9409y - (5141x + 2809y) = 55581 - 9381

⇒ 6600y = 46200

⇒ y = $\frac{46200}{6600}$ = 7.

Substituting value of y in (ii) we get,

⇒ 53x + 97(7) = 573

⇒ 53x + 679 = 573

⇒ 53x = 573 - 679

⇒ 53x = -106

⇒ x = -2.

Hence, x = -2 and y = 7.

Given,

Equations :

67x - 58y = 192 ........(1)

58x - 67y = 183 ........(2)

Multiplying equation (1) by 58, we get :

⇒ 58(67x - 58y) = 192 × 58

⇒ 3886x - 3364y = 11136 ........(3)

Multiplying equation (2) by 67, we get :

⇒ 67(58x - 67y) = 183 × 67

⇒ 3886x - 4489y = 12261 .........(4)

Subtracting equation (3) from (4), we get :

⇒ 3886x - 4489y - (3886x - 3364y) = 12261 - 11136

⇒ 3886x - 3886x - 4489y + 3364y = 1125

⇒ -1125y = 1125

⇒ y = $-\frac{1125}{1125}$ = -1.

Substituting value of y in equation (1), we get :

⇒ 67x - 58(-1) = 192

⇒ 67x + 58 = 192

⇒ 67x = 192 - 58

⇒ 67x = 134

⇒ x = $\frac{134}{67}$ = 2.

Hence, x = 2 and y = -1.

Given,

x + y = 7xy

2x - 3y + xy = 0

First we note that x = 0, y = 0 is a solution of equations.

Now when x ≠ 0 and y ≠ 0.

Dividing x + y = 7xy by xy,

$$\begin{aligned}\Rightarrow \frac{x}{xy} + \frac{y}{xy} = \frac{7xy}{xy} \\[1em] \Rightarrow \frac{1}{y} + \frac{1}{x} = 7 ......(i)\end{aligned}$$

Dividing 2x - 3y + xy = 0 by xy,

$$\begin{aligned}\Rightarrow \frac{2x}{xy} - \frac{3y}{xy} + \frac{xy}{xy} = 0 \\[1em] \Rightarrow \frac{2}{y} - \frac{3}{x} + 1 = 0 \\[1em] \Rightarrow \frac{2}{y} - \frac{3}{x} = -1 ......(ii)\end{aligned}$$

Substituting $\frac{1}{x}$ as p and $\frac{1}{y}$ as q we get,

q + p = 7 .......(iii)

2q - 3p = -1 ......(iv)

Multiplying (iii) by 3 we get,

3q + 3p = 21 .......(v)

Adding (iv) and (v) we get,

⇒ 2q - 3p + (3q + 3p) = -1 + 21

⇒ 2q + 3q = 20

⇒ 5q = 20

⇒ q = $\frac{20}{5}$

⇒ q = 4.

$\therefore \frac{1}{y} = 4 \text{ or } y = \frac{1}{4}$.

Substituting value of q in (iii) we get,

⇒ 4 + p = 7

⇒ p = 3.

$\therefore \frac{1}{x} = 3 \text{ or } x = \frac{1}{3}$.

Hence, x = 0, y = 0 and x = $\frac{1}{3}, y = \frac{1}{4}$.

Substituting $\frac{1}{x - y} = a \text{ and } \frac{1}{x + y} = b$ in above equations we get,

30a + 44b = 10 .......(i)

40a + 55b = 13 .......(ii)

Multiplying (i) by 4 and (ii) by 3 we get,

120a + 176b = 40 .......(iii)

120a + 165b = 39 .......(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ 120a + 165b - (120a + 176b) = 39 - 40

⇒ 165b - 176b = -1

⇒ -11b = -1

⇒ b = $\frac{1}{11}$.

$$\begin{aligned}\therefore \frac{1}{x + y} = \frac{1}{11} \\[1em] \Rightarrow x + y = 11 .........(v)\end{aligned}$$

Substituting value of b in (iv),

⇒ 120a + $165 \times \frac{1}{11}$ = 39

⇒ 120a + 15 = 39

⇒ 120a = 24

⇒ a = $\frac{24}{120} = \frac{1}{5}$.

$$\begin{aligned}\therefore \frac{1}{x - y} = \frac{1}{5} \\[1em] \Rightarrow x - y = 5 ........(vi)\end{aligned}$$

Adding eq. (v) and (vi) we get,

⇒ x + y + (x - y) = 11 + 5

⇒ 2x = 16

⇒ x = 8.

Substituting value of x in (v)

⇒ x + y = 11

⇒ 8 + y = 11

⇒ y = 11 - 8 = 3.

Hence, x = 8 and y = 3.

Given,

ax + by = a - b .......(i)

bx - ay = a + b .......(ii)

Multiplying eq. (i) by b and (ii) by a we get,

abx + b²y = ab - b² ......(iii)

abx - a²y = a² + ab .......(iv)

Subtracting eq. (iv) from (iii),

⇒ abx + b²y - (abx - a²y) = ab - b² - (a² + ab)

⇒ b²y + a²y = -b² - a²

⇒ y(b² + a²) = -(b² + a²)

⇒ y = -1.

Substituting value of y in (i),

⇒ ax + b(-1) = a - b

⇒ ax - b = a - b

⇒ ax = a - b + b

⇒ ax = a

⇒ x = 1.

Hence, x = 1 and y = -1.

Given,

3x + 2y = 2xy .......(i)

$\frac{1}{x} + \frac{2}{y} = 1\frac{1}{6}$ ......(ii)

Dividing eq. (i) by xy we get,

$\frac{3}{y} + \frac{2}{x} = 2$ .......(iii)

Substituting $\frac{1}{x}$ as p and $\frac{1}{y}$ as q in (ii) and (iii) we get,

p + 2q = $\frac{7}{6}$ ......(iv)

3q + 2p = 2 ........(v)

Multiplying (iv) by 6 and (v) by 3 we get,

6p + 12q = 7 ........(vi)

9q + 6p = 6 .......(vii)

Subtracting (vii) from (vi) we get,

6p + 12q - (9q + 6p) = 7 - 6

3q = 1

q = $\frac{1}{3}$.

$\therefore \frac{1}{y} = \frac{1}{3} \text{ or } y = 3.$

Substituting value of q in (iv) we get,

$$\begin{aligned}\Rightarrow p + 2 \times \frac{1}{3} = \frac{7}{6} \\[1em] \Rightarrow p + \frac{2}{3} = \frac{7}{6} \\[1em] \Rightarrow p = \frac{7}{6} - \frac{2}{3} \\[1em] \Rightarrow p = \frac{7 - 4}{6} \\[1em] \Rightarrow p = \frac{3}{6} \\[1em] \therefore \frac{1}{x} = \frac{3}{6} \\[1em] \Rightarrow x = \frac{6}{3} = 2.\end{aligned}$$

Hence, x = 2 and y = 3.

Substituting $\frac{1}{2x - y}$ = a and $\frac{1}{x + 2y}$ = b in above equations, we get :

⇒ $\frac{2}{3}a + \frac{1}{2}b = \frac{5}{12}$ ..........(1)

⇒ a - 2b = $\frac{1}{6}$ .........(2)

Multiplying equation (1) by $\frac{3}{2}$ we get :

$$\begin{aligned}\Rightarrow \frac{3}{2}\Big(\frac{2}{3}a + \frac{1}{2}b\Big) = \frac{3}{2} \times \frac{5}{12} \\[1em] \Rightarrow a + \frac{3}{4}b = \frac{5}{8} \text{ ...........(3)}\end{aligned}$$

Subtracting equation (2) from (3), we get :

$$\begin{aligned}\Rightarrow a + \frac{3}{4}b - (a - 2b) = \frac{5}{8} - \frac{1}{6} \\[1em] \Rightarrow a - a + \frac{3}{4}b + 2b = \frac{15 - 4}{24} \\[1em] \Rightarrow \frac{3b + 8b}{4} = \frac{11}{24} \\[1em] \Rightarrow \frac{11b}{4} = \frac{11}{24} \\[1em] \Rightarrow b = \frac{11}{24} \times \frac{4}{11} = \frac{1}{6}. \\[1em] \therefore \frac{1}{x + 2y} = \frac{1}{6} \\[1em] \Rightarrow x + 2y = 6 \\[1em] \Rightarrow x = 6 - 2y \text{ ............(4)}\end{aligned}$$

Substituting b = $\frac{1}{6}$ in equation (2), we get :

$$\begin{aligned}\Rightarrow a - 2 \times \frac{1}{6} = \frac{1}{6} \\[1em] \Rightarrow a - \frac{1}{3} = \frac{1}{6} \\[1em] \Rightarrow a = \frac{1}{6} + \frac{1}{3} \\[1em] \Rightarrow a = \frac{1 + 2}{6} \\[1em] \Rightarrow a = \frac{3}{6} \\[1em] \Rightarrow a = \frac{1}{2} \\[1em] \therefore \frac{1}{2x - y} = \frac{1}{2}\\[1em] \Rightarrow 2x - y = 2 \text{ ......(5)}\end{aligned}$$

Substituting value of x from equation (4) in (5), we get :

⇒ 2(6 - 2y) - y = 2

⇒ 12 - 4y - y = 2

⇒ 12 - 5y = 2

⇒ -5y = 2 - 12

⇒ -5y = -10

⇒ y = $\frac{-10}{-5}$ = 2.

Substituting y = 2 in equation (4), we get :

⇒ x = 6 - 2y

⇒ x = 6 - 2(2) = 6 - 4 = 2.

Hence, x = 2 and y = 2.

Given,

$2x - \frac{3}{y} = 9$ .......(i)

$3x + \frac{7}{y} = 2$ .......(ii)

Multiplying (i) by 3 and (ii) by 2 we get,

$6x - \frac{9}{y} = 27$ .......(iii)

$6x + \frac{14}{y} = 4$ .......(iv)

Subtracting (iii) from (iv) we get,

$$\begin{aligned}\Rightarrow 6x + \frac{14}{y} - \Big(6x - \frac{9}{y}\Big) = 4 - 27 \\[1em] \Rightarrow \frac{14}{y} + \frac{9}{y} = -23 \\[1em] \Rightarrow \frac{23}{y} = -23 \\[1em] \Rightarrow y = -1.\end{aligned}$$

Substituting value of y in (iv),

$$\begin{aligned}\Rightarrow 6x + \frac{14}{-1} = 4 \\[1em] \Rightarrow 6x - 14 = 4 \\[1em] \Rightarrow 6x = 18 \\[1em] \Rightarrow x = 3.\end{aligned}$$

Substituting values of x and y in x = ky + 5 we get,

⇒ 3 = k(-1) + 5

⇒ 3 = -k + 5

⇒ k = 5 - 3 = 2.

Hence, x = 3, y = -1 and k = 2.

Given,

$\frac{1}{x + y} - \frac{1}{2x} = \frac{1}{30}$

$\frac{5}{x + y} + \frac{1}{x} = \frac{4}{3}$

Substituting $\frac{1}{x + y} = a \text{ and } \frac{1}{x} = b$ in above equations,

$a - \frac{b}{2} = \frac{1}{30}$ ........(i)

$5a + b = \frac{4}{3}$ ........(ii)

Multiplying eq. (i) by 5 we get,

$5a - \frac{5b}{2} = \frac{1}{6}$ .......(iii)

Subtracting eq. (iii) from (ii) we get,

$$\begin{aligned}\Rightarrow 5a + b - \Big(5a - \frac{5b}{2}\Big) = \frac{4}{3} - \frac{1}{6} \\[1em] \Rightarrow b + \frac{5b}{2} = \frac{8 - 1}{6} \\[1em] \Rightarrow \frac{2b + 5b}{2} = \frac{7}{6} \\[1em] \Rightarrow \frac{7b}{2} = \frac{7}{6} \\[1em] \Rightarrow b = \frac{1}{3} \\[1em] \therefore \frac{1}{x} = \frac{1}{3} \\[1em] \Rightarrow x = 3.\end{aligned}$$

Substituting value of b in eq. (i) we get,

$$\begin{aligned}\Rightarrow a - \frac{\frac{1}{3}}{2} = \frac{1}{30} \\[1em] \Rightarrow a - \frac{1}{6} = \frac{1}{30} \\[1em] \Rightarrow a = \frac{1}{30} + \frac{1}{6} \\[1em] \Rightarrow a = \frac{1 + 5}{30} \\[1em] \Rightarrow a = \frac{6}{30} \\[1em] \Rightarrow a = \frac{1}{5} \\[1em] \therefore \frac{1}{x + y} = \frac{1}{5} \\[1em] \Rightarrow x + y = 5 \\[1em] \Rightarrow 3 + y = 5 \\[1em] \Rightarrow y = 2.\end{aligned}$$

Substituting values of x and y in 2x² - y² we get,

⇒ 2x² - y²

= 2(3)² - (2)²

= 2(9) - 4

= 18 - 4 = 14.

Hence, x = 3, y = 2 and 2x² - y² = 14.

Given,

$\frac{2}{y} + \frac{5}{x} = 19$ ......(i)

$\frac{5}{y} - \frac{3}{x} = 1$ .......(ii)

3x + 8y = 5 .......(iii)

Multiplying eq. (i) by 5 and eq. (ii) by 2 we get,

$\frac{10}{y} + \frac{25}{x} = 95$ ......(iii)

$\frac{10}{y} - \frac{6}{x} = 2$ .......(iv)

Subtracting (iv) from (iii) we get,

$$\begin{aligned}\Rightarrow \frac{10}{y} + \frac{25}{x} - \Big(\frac{10}{y} - \frac{6}{x}\Big) = 95 - 2 \\[1em] \Rightarrow \frac{25}{x} + \frac{6}{x} = 93 \\[1em] \Rightarrow \frac{31}{x} = 93 \\[1em] \Rightarrow x = \frac{31}{93} = \frac{1}{3}.\end{aligned}$$

Substituting value of x in (i) we get,

$$\begin{aligned}\Rightarrow \frac{2}{y} + \frac{5}{\frac{1}{3}} = 19 \\[1em] \Rightarrow \frac{2}{y} + 15 = 19 \\[1em] \Rightarrow \frac{2}{y} = 4 \\[1em] \Rightarrow y = \frac{1}{2}.\end{aligned}$$

Substituting values of x and y in eq. (iii),

$$\begin{aligned}\Rightarrow 3 \times \frac{1}{3} + 8 \times \frac{1}{2} = 5 \\[1em] \Rightarrow 1 + 4 = 5 \\[1em] \Rightarrow 5 = 5.\end{aligned}$$

Since, L.H.S. = R.H.S. hence, $x = \frac{1}{3} \text{ and } y = \frac{1}{2}$ satisfies the equation.

Hence, equations can be satisfied simultaneously with $x = \frac{1}{3} \text{ and } y = \frac{1}{2}$.