8 x-7-0- 8 x -7-0- 8 x -7- x -8-7The error in solving linear equation 8 x -7-0 isA- x-8-7B- 8 x-7C- 8 x -7-0

The correct option is A x= 8 +7In the steps below, 8x + 7 = 0 ⇒ 8x= 7+0 ⇒ 8x= 7 ⇒ x= 8 +7, nbsp;x= 8+7 is incorrect . It is supposed to be nbsp;x= 7/ ...

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Byju's Answer

Standard VIII

Mathematics

Reducing Equations to Simpler Form

8 x+7=0⇒ 8 x ...

Question

$8 x + 7 = 0$
$\Rightarrow 8 x = - 7 + 0$
$\Rightarrow 8 x = - 7$
$\Rightarrow x = 8 + 7$
The error in solving linear equation $8 x + 7 = 0$ is .

x = 8 + 7

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8 x = - 7

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8 x = - 7 + 0

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Solution

The correct option is A $x = 8 + 7$
In the steps below,
$8 x + 7 = 0$
$\Rightarrow 8 x = - 7 + 0$
$\Rightarrow 8 x = - 7$
$\Rightarrow x = 8 + 7$ ,

$x = 8 + 7$ is incorrect .

It is supposed to be $x = \frac{- 7}{8}$ .

Suggest Corrections

Similar questions

Q. Solve:

x^{2} + 8 x + 7

=0.

Q. The solution of

\frac{d^{3} y}{d x^{3}} - 8 \frac{d^{2} y}{d x^{2}} = 0

satisfying

y (0) = \frac{1}{8}, y^{1} (0) = 0

and

y^{2} (0) = 1

is [here

y^{n} (x) = \frac{d^{n} y}{d x^{n}}]

Q. If x² +1/x² = 7 and x≠0 ; find the value of :7x² + 8x - 7/x³ - 8/x.

Q. If

x^{2} + \frac{1}{x^{2}} = 7

and

x \neq 0

; find the value of:

7 x^{3} + 8 x + \frac{7}{x^{3}} + \frac{8}{x} .

Q. If

x^{2} + \frac{1}{x^{2}} = 7

and

x \neq 0

; find the value of :

7 x^{3} + 8 x - \frac{7}{x^{3}} - \frac{8}{x}

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8x+7=0 ⇒8x=−7+0 ⇒8x=−7 ⇒x=8+7 The error in solving linear equation 8x+7=0 is .

The correct option is A x=8+7In the steps below, 8x+7=0 ⇒8x=−7+0 ⇒8x=−7 ⇒x=8+7, x=8+7 is incorrect . It is supposed to be x=−78.

$8 x + 7 = 0$
$\Rightarrow 8 x = - 7 + 0$
$\Rightarrow 8 x = - 7$
$\Rightarrow x = 8 + 7$
The error in solving linear equation $8 x + 7 = 0$ is .

The correct option is A $x = 8 + 7$
In the steps below,
$8 x + 7 = 0$
$\Rightarrow 8 x = - 7 + 0$
$\Rightarrow 8 x = - 7$
$\Rightarrow x = 8 + 7$ ,

$x = 8 + 7$ is incorrect .

It is supposed to be $x = \frac{- 7}{8}$ .