Solve the following pair of linear equations

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

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