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# Solve the following pair of linear equations: (i) $$px + py = p - q$$ $$qx –py = p + q$$ (ii) $$ax + by = c$$ $$bx + ay = 1 + c$$ (iii) $${{x}\over{a}} - {{y}\over{b}} = 0$$ $$ax + by = a ^2 + b^2$$ (iv) $$(a - b)x + (a + b)y = a^2 -2ab - b^2$$ $$(a + b)(x + y) = a^2 + b^2$$ (v) $$152x - 378y = -74$$ $$-378x + 152y = -604$$

(i)$$px + py = p - q$$ … (1)
$$qx –py = p + q$$… (2)

Multiplying equation (1) by p and equation (2) by q:
$$p^2 x + pqy = p^2 - pq$$… (3) $$q^2 x - pqy = pq + q^2$$… (4)

Adding equations (3) and (4), :
=>$$p^2 x + q^2 x = p^2 + q^2$$
=>$$(p^2 + q^2)x = p^2 + q^2$$

=> x = $${{p^x + q^x} \over {p^2 + q^2}}$$= 1

Substituting the value of in equation (1):
$$p(1) + qy = p - q$$
=>$$qy = -q => y = -1$$

Hence the required solution is x = 1 and y = –1.

(ii)$$ax + by = c$$… (1)
$$bx + ay = 1 + c$$ … (2)

Multiplying equation (1) by a and equation (2) by b:
$$a^2x + aby = ac$$… (3)
$$b^2x + aby = b + bc$$ … (4)

Subtracting equation (4) from equation (3),
$$(a^2 - b^2)x = ac - bc -b$$
=>$$x = {{c(a - b) - b} \over {a^2 - b^2}}$$

Substituting the value of x in equation (1):
$$a ({{c(a - b) - b} \over {a^2 - b^2}}) + by = c$$
=>$${{ac(a - b) - ab} \over {a^2 - b^2}} + by = c$$
=>$$by = c -{{ac(a - b) - ab} \over {a^2 - b^2}}$$
=>$$by = {{a^2c - b^2c - a^2c + abc + ab} \over {a^2 - b^2}}$$
=> $$by = {{ abc - b^2 c+ ab} \over {a^2 - b^2}}$$
=>$$y = {{ c(a - b)+ a} \over {a^2 - b^2}}$$

(iii)$${{x}\over{a}} - {{y}\over{b}} = 0$$
=>$$bx - ay = 0$$……..(1)
=>$$ax + by = a^2 - b^2$$……..(2)

Multiplying equation (1) and (2) by b and a respectively:
=> $$xb^2 - aby = 0$$……..(3)
=> $$xa^2 + aby = a^3 - ab^2$$……..(4)

Adding equations (3) and (4), we obtain:
$$xb^2 + xa^2 = a^3 + ab^2$$
=>$$x(b^2 + a^2) = a (b^2 + a^2)$$
=> $$x = a$$

Substituting the value of in equation (1):
$$b(a) - ay = 0$$
=>$$ab - ay = 0$$
=> $$y = b$$

(iv) $$(a - b)x + (a + b)y = a^2 -2ab - b^2$$ … (1)
$$(a + b)(x + y) = a^2 + b^2$$……..(2)

Subtracting equation (2) from (1):
$$(a - b)x - (a + b)x = (a^2 - 2ab - b^2) - ( a^2 + b^2)$$
=>$$(a - b - a - b)x = -2ab - 2b^2$$
=>$$-2bx = -2b(a + b)$$
=> $$x = a + b$$

Substituting the value of in equation (1):
$$(a - b)(a + b) + (a + b)y = a^2 -2ab - b^2$$
=>$$a^2 - b^2 + (a + b)y = a^2 -2ab - b^2$$
=>$$(a + b)y = -2ab$$
=>$$y = {{-2ab} \over {a + b}}$$

(v)$$152x – 378y = –74$$… (1)
$$–378x + 152y = –604$$ … (2)

Adding the equations (1) and (2):
$$–226x – 226y = –678$$
$$=> x + y = 3$$ ………(3)

Subtracting the equation (2) from equation (1):
$$530x – 530y = 530$$
$$=> x – y = 1$$ ……..(4)

$$2x = 4$$
$$=> x = 2$$
$$y = 1$$