Question

Look at the statements given below:

I. $(3x+4y=5\mathrm{and}2\mathrm{x}+3\mathrm{y}=9)\Rightarrow (x=-21\mathrm{and}\mathrm{y}=17)$
II. $(x-y+1=0\mathrm{and}3\mathrm{x}-5\mathrm{y}+1=0)\Rightarrow (x=-2\mathrm{and}y=-1)$
III. $\left(\frac{x}{a}+\frac{y}{b}=2\mathrm{and}\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=4\right)\Rightarrow (x=2a\mathrm{and}y=2b)$
Which of the above statements are true?
(a) I only
(b) II only
(c) I and II
(d) I and III

Answer 1

Option (c) is correct.

For I:
3x + 4y = 5 ...(i)
2x + 3y = 9 ...(ii)
On multiplying (i) by 2 and (ii) by 3 and subtracting, we get:
y = 17
On substituting y = 17 in (i), we get:
3x + 68 = 5
⇒ 3x = (5 − 68) = −63
⇒ x = −21
Thus, x = −21 and y = 17

For II:
x − y = −1 ...(i)
3x − 5y = −1 ...(ii)
On multiplying (i) by 3 and subtracting, we get:
y = −1
On substituting y = −1 in (i), we get:
x + 1 = −1
⇒ x = −2
Thus, x = −2 and y = −1

For III:
$\frac{x}{a}+\frac{y}{b}=2$ ...(i)
$\frac{x}{a}-\frac{y}{b}=4$ ...(ii)
On adding (i) and (ii), we get:
$\frac{2x}{a}=6\Rightarrow x=3a$
Putting x = 3a in (i), we get:
y = −b
Thus, x = 3a and y = −b
So, x = 2a and y = 2b is not true.