Question :
Solve the following pair of linear equations :
(i) px + qy = p – q and qx – py = p + q
(ii) ax + by = c and bx + ay = 1 + c
(iii) \(x\over a\) – \(y\over b\) = 0 and ax + by = \(a^2 + b^2\)
(iv) Solve for x and y :
(a – b)x + (a + b)y = \(a^2 – 2ab – b^2\)
(a + b)(x + y) = \(a^2 + b^2\)
(v) 152x – 378y = -74 and -378x + 152y = -604
Solution :
(i) The given linear equations are
px + qy = p – q \(\implies\) px + qy – (p – q) = 0 ………….(1)
qx – py = p + q \(\implies\) qx – py – (p + q) = 0 ………….(2)
Solving it by cross multiplication method, we get
\(x\over -q(p + q) – p(p – q)\) = \(y\over -q(p – q) + p(p + q)\) = \(1\over -p^2 – q^2\)
\(\implies\) \(x\over -pq – q^2 – p^2 + pq\) = \(y\over – pq + q^2 + p^2 + pq\) = \(1\over -(p^2 + q^2)\)
\(\implies\) \(x\over -(p^2 + q^2)\) = \(y\over p^2 + q^2\) = \(1\over -(p^2 + q^2)\)
\(\implies\) x = 1 and y = 1.
(ii) The given linear equations are
ax + by – c = 0 ………….(1)
bx + ay – (1 + c) = 0 ………….(2)
Solving it by cross multiplication method, we get
\(x\over -b(1 + c) + ac\) = \(y\over -bc + a(1 + c)\) = \(1\over a^2 – b^2\)
\(\implies\) \(x\over -b – bc + ac\) = \(y\over -bc + a + ac\) = \(1\over a^2 – b^2)\)
\(\implies\) \(x\over c(a – b) – b\) = \(y\over c(a – b) + a\) = \(1\over (a – b)(a + b)\)
\(\implies\) x = \(c\over a + b\) – \(b\over (a – b)(a + b)\)
y = \(c\over a + b\) + \(b\over (a – b)(a + b)\)
(iii) The given linear equations are
\(x\over a\) – \(y\over b\) = 0 \(\implies\) bx – ay = 0 ………..(1)
ax + by -\((a^2 + b^2)\) = 0 ………(2)
Solving it by cross multiplication method, we get
\(x\over a(a^2 + b^2) – 0\) = \(y\over 0 + b(a^2 + b^2)\) = \(1\over b^2 + a^2\)
\(\implies\) \(x\over a(a^2 + b^2)\) = \(y\over b(a^2 + b^2)\) = \(1\over a^2 + b^2\)
\(\implies\) x = a, y = b.
(iv) The given linear equations are
(a – b)x + (a + b)y = \(a^2 – 2ab – b^2\) ………..(1)
(a + b)(x + y) = \(a^2 + b^2\) ………(2)
Solving it by cross multiplication method, we get
\(x\over -(a + b)(a^2 + b^2) + (a + b)(a^2 – 2ab – b^2)\) = \(y\over -(a + b)(a^2 – 2ab – b^2) + (a – b)(a^2 + b^2)\) = \(1\over (a – b)(a + b) – {(a + b)}^2\)
\(\implies\) \(x\over (a + b)(-a^2 – b^2 + a^2 – 2ab – b^2)\) = \(y\over -a^3 + 2a^2b + ab^2 – a^2b + 2ab^2 – b^3 + a^3 + ab^2 – a^2b – b^3\) = \(1\over a^2 – b^2 – a^2 – b^2 – 2ab\)
\(\implies\) \(x\over (a + b)(-2ab – 2b^2)\) = \(y\over 4ab^2\) = \(1\over -2b^2 – 2ab\)
\(\implies\) \(x\over (a + b)(-2b)(a + b)\) = \(y\over 4ab^2\) = \(1\over -2b(a + b)\)
\(\implies\) x = a + b, y = \(-2ab\over a + b\)
(v) The given linear equations are
152x – 378y = -74 ………(1)
-378x + 152y = -604 ……….(2)
Adding equation (1) and (2), we get
-226x – 226y = -678 \(\implies\) x + y = 3 ……(3)
Subtracting equation (1) from (2), we get
-530x + 530y = -530 \(\implies\) x – y = 1 …….(4)
Adding equation (3) and (4), we get
2x = 4 \(\implies\) x = 2
Put the value of x = 2 in equation (4), we get
y = 1
