Question

Match the following columns:

Column I	Column II
(a) Solution of $$\begin{gathered} ax+by=a-b, \\ bx-ay=a+b\mathrm{is}....... \end{gathered}$$	(p) $x=2,y=-2$
(b) Solution of $$\begin{gathered} 3x-2y=10, \\ 5x+3y=4\mathrm{is}....... \end{gathered}$$	(q) $x=a,y=b$
(c) Solution of $$\begin{gathered} 2x-y=5, \\ 3x+2y=11\mathrm{is}....... \end{gathered}$$	(r) $x=1,y=-1$
(d) Solution of $$\begin{gathered} x+y=a+b, \\ ax-by=a^2-b^2\mathrm{is}....... \end{gathered}$$	(s) $x=3,y=1$

Answer 1

The correct answer is:
(a) = (r)
(b) = (p)
(c) = (s)
(d) = (q)

(a)
ax + by = a − b ...(i)
bx − ay = a + b ...(ii)
On multiplying (i) by a and (ii) by b, we get:
a²x + aby = a² − ab ...(iii)
b²x − aby = ab + b² ...(iv)
On adding (iii) and (iv), we get:
(a² + b²)x = a² + b²
⇒ x = 1
On substituting x = 1 in (i), we get:
a + by = a − b
⇒ by = −b
⇒ y = −1
Hence, x = 1 and y = −1

(b)
3x − 2y = 10 ...(i)
5x + 3y = 4 ...(ii)
On multiplying (i) by 3 and (ii) by 2, we get:
9x − 6y = 30 ...(iii)
10x + 6y = 8 ...(iv)
On adding (iii) and (iv), we get:
19x = 38
⇒ x = 2
On substituting x = 2 in (i), we get:
6 − 2y = 10
⇒ −2y = 4
⇒ y = −2
Hence, x = 2 and y = −2

(c)
2x − y = 5 ...(i)
3x + 2y = 11 ...(ii)
On multiplying (i) by 2, we get:
4x − 2y = 10 ...(iii)
On adding (ii) and (iii), we get:
7x = 21
⇒ x = 3
On substituting x = 3 in (i), we get:
6 − y = 5
⇒ y = (6 − 5) = 1
Hence, x = 3 and y = 1

(d)
x + y = a + b ...(i)
ax − by = a² − b² ...(ii)
On multiplying (i) by b, we get:
bx + by = ab + b² ...(iii)
On adding (ii) and (iii), we get:
ax + bx = a² + ab
⇒ x(a + b) = a(a + b)
⇒ x = a
On substituting x = a in (i), we get:
a + y = a + b
⇒ y = b