In the equation a x-b y-c-0- if y-c x-d-b then x-

The correct option is B ( d c)/(a+c) Given ax+by+c=0, Also given y= (cx d)/b, substitute this in ax+by+c=0 We get ax+ [b (cx d)/b]+c=0 So ax+cx d+c=0 (a+c)x= ...

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Standard X

Mathematics

Method of Substitution to Find the Solution of a Pair of Linear Equations

In the equati...

Question

In the equation ax+by+c=0, if $y = \frac{(c x - d)}{b}$ then x =

(d+c)/(a+c)

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(-d-c)/(a+c)

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(a+c)/(d+c)

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None of these

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Solution

The correct option is B
(-d-c)/(a+c)

Given $a x + b y + c = 0$ ,

Also given $y = \frac{(c x - d)}{b}$ , substitute this in $a x + b y + c = 0$

We get $a x + [b \frac{(c x - d)}{b}] + c = 0$

So $a x + c x - d + c = 0$

$(a + c) x = d - c$

Therefore $x = \frac{d - c}{a + c}$

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In the equation ax+by+c=0, if y=(cx−d)b then x =

The correct option is B (-d-c)/(a+c) Given ax+by+c=0, Also given y=(cx−d)b, substitute this in ax+by+c=0 We get ax+[b (cx−d)b]+c=0 So ax+cx−d+c=0 (a+c)x=d−c Therefore x=d−ca+c

In the equation ax+by+c=0, if $y = \frac{(c x - d)}{b}$ then x =

The correct option is B
(-d-c)/(a+c)

Given $a x + b y + c = 0$ ,

Also given $y = \frac{(c x - d)}{b}$ , substitute this in $a x + b y + c = 0$

We get $a x + [b \frac{(c x - d)}{b}] + c = 0$

So $a x + c x - d + c = 0$

$(a + c) x = d - c$

Therefore $x = \frac{d - c}{a + c}$