Consider the following Pairs of Liner Equations: (i) 2x-3y=8, 4x-6y=9 (ii) 2x+3y-9=0, 4x+6y-18=0 Choose the Correct Alternative - Bzziii
(i) `2x-3y=8`, `4x-6y=9`
(ii) `2x+3y-9=0, 4x+6y-18=0`
Choose the correct alternative:
(a) (a) The pair in (1) has no solution, whereas the pair in (ii) has unique solution
(b) The pair in (i) has infinitely many solutions, whereas the pair in (ii) has no solution
(c) The pairs in (1) and (ii) have no solutions.
(d) The pair in (1) has no solution, whereas the pair in (ii) has infinitely many solution
Option (d) is the correct answer.
(i) `2x - 3y` = 8, `4x - 6y` = 9
⇒ `2x - 3y` = 8
⇒ `2x - 3y` - 8 = 0 -----------(a)
⇒ `4x - 6y` = 9
⇒ `4x - 6y` - 9 = 0 -----------(b)
from (a) and (b)
`a_{1}` = 2, `b_{1}` = -3, `c_{1}` = -8
`a_{2}` = 4, `b_{2}` = -6, `c_{2}` = -9
∴ `\frac{a_{1}}{a_{2}}` = `\frac{b_{1}}{b_{2}}` = `\frac{c_{1}}{c_{2}}`
⇒ `\frac{2}{4}` = `\frac{-3}{-6}` = `\frac{-8}{-9}`
⇒ `\frac{1}{2}` = `\frac{1}{2}` ≠ `\frac{-8}{-9}`
So,
`\frac{a_{1}}{a_{2}}` = `\frac{b_{1}}{b_{2}}` ≠ `\frac{c_{1}}{c_{2}}`
Thus, the pair has no Solution.
(ii) `2x + 3y - 9 = 0, 4x + 6y - 18 = 0`
⇒ `2x + 3y - 9 = 0 -----------(a)
⇒ 4x + 6y - 18 = 0 -----------(b)
from (a) and (b)
`a_{1}` = 2, `b_{1}` = 3, `c_{1}` = -9
`a_{2}` = 4, `b_{2}` = 6, `c_{2}` = -18
∴ `\frac{a_{1}}{a_{2}}` = `\frac{b_{1}}{b_{2}}` = `\frac{c_{1}}{c_{2}}`
⇒ `\frac{2}{4}` = `\frac{3}{6}` = `\frac{-9}{-18}`
⇒ `\frac{1}{2}` = `\frac{1}{2}` = `\frac{1}{2}`
So,
`\frac{a_{1}}{a_{2}}` = `\frac{b_{1}}{b_{2}}` = `\frac{c_{1}}{c_{2}}`
It has Infinity many solutions.
Therefor, the answer would be option (d) The pair in (1) has no solution, whereas the pair in (ii) has infinitely many solution.
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