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)

Adding equations (3) and (4):
\(2x = 4\)
\(=> x = 2\)

Substituting the value of x in equation (3):
\(y = 1\)