The solution of the equation ax - b - 0 is a a-b b -b c -b-a d b-a

Given equation is ax + b = 0 ⇒ ax = b [transposing b to RHS] ⇒ x = b/a [on dividing both sides by a] ...

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Standard VII

Mathematics

Methods of Solving an Equation

The solution ...

Question

Question 1

The solution of the equation ax + b = 0 is

(a) $\frac{a}{b}$

(b) -b

(c) $- \frac{b}{a}$

(d) $\frac{b}{a}$

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Solution

Given equation is ax + b = 0
$\Rightarrow a x = - b [t r a n s p o s i n g b t o R H S] \Rightarrow x = - \frac{b}{a} [o n d i v i d i n g b o t h s i d e s b y a]$

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Question 2

In the following question out of the four options only one is correct, write the correct answer:
(a) $x = \frac{a}{b}$
(b) $x = - b$
(c) $x = \frac{- b}{a}$
(d) $x = \frac{b}{a}$
The solution of the equation ax+b=0 is:

Q. In the following question out of the four options only one is correct, write the correct answer:
(a)

x = \frac{a}{b}

(b)

x = - b

(c)

x = \frac{- b}{a}

(d)

x = \frac{b}{a}

The solution of the equation ax + b = 0 is:

Q. Show that

x = \frac{- b c}{a d}

is a solution of the quadratic equation

a d^{2} (\frac{a x}{b} + \frac{2 c}{d}) x + b c^{2} = 0

The zero of the polynomial p(x) = ax+b is ______.

Q. The equations ax + b = 0 and cx + d = 0 are consistent, if

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Methods of Solving an Equation

Standard VII Mathematics

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Question 1 The solution of the equation ax + b = 0 is (a) ab (b) -b (c) −ba (d) ba

Given equation is ax + b = 0 ⇒ax=−b [transposing b to RHS]⇒x=−ba [on dividing both sides by a]

Question 1

The solution of the equation ax + b = 0 is

(a) $\frac{a}{b}$

(b) -b

(c) $- \frac{b}{a}$

(d) $\frac{b}{a}$

Given equation is ax + b = 0
$\Rightarrow a x = - b [t r a n s p o s i n g b t o R H S] \Rightarrow x = - \frac{b}{a} [o n d i v i d i n g b o t h s i d e s b y a]$