If a-1x -b-1y - c-1- 0 - a-2x - b-2y - c-2- 0 are two-equations with an unique solution-then a-1-a-2- - - b-1-b-2- - - c-1-c-2 True-False-

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Byju's Answer

Standard VIII

Mathematics

Comparing the ratios of coefficients of LE

If a_1x +b_1y...

Question

If $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ are two equations with an unique solution, then
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

True

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False

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Solution

The correct option is B
False

If $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ are two equations with a unique solution(consistent), then
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

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Similar questions

If $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ are two equations with an unique solution, then
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

If $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ are two equations with no solutions , then $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Unique point is obtained for the pair of equations $a_{1}$ x + $b_{1}$ y + $c_{1}$ = 0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ = 0 if __________ .

The condition for $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ to be consistent and have a unique solution is

Q. If

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

is valid for the system of equations

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

, then these are ?

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If a1x+b1y+c1=0, a2x+b2y+c2=0 are two equations with an unique solution, then a1a2=b1b2=c1c2

The correct option is B False If a1x+b1y+c1=0, a2x+b2y+c2=0 are two equations with a unique solution(consistent), then a1a2≠b1b2

If $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ are two equations with an unique solution, then
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

The correct option is B
False

If $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ are two equations with a unique solution(consistent), then
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$