Tag: maths

We have,

$\sqrt{1+\dfrac{x}{289}}=1\dfrac{1}{17}$

$\sqrt{\dfrac{289+x}{289}}=\dfrac{18}{17}$

On squaring both sides, we get

$ {{\left( \sqrt{\dfrac{289+x}{289}} \right)}^{2}}={{\left( \dfrac{18}{17} \right)}^{2}} $

$ \dfrac{289+x}{289}=\dfrac{324}{289} $

$ 289+x=324 $

$ x=324-289 $

$ x=35 $

Hence, this is the answer.

$\dfrac{{2x - 1}}{2}\,\,\, - \dfrac{{x + 3}}{3}\,\, = \dfrac{{x - 2}}{5}\Rightarrow \dfrac{6x-3-2x-6}{6}=\dfrac{x-2}{5}\Rightarrow 20x-45=6x-12\Rightarrow 14x=33\Rightarrow x=\dfrac{33}{14}$

We have,

$
x(5-a)\quad =\quad 10-{ x }^{ 2 }\ x\quad =\quad 2\ So\ 2(5-a)\quad =\quad 10\quad -{ 2 }^{ 2 }\ 10\quad -\quad 2a\quad =\quad 10-4\ -2a\quad =\quad -4\ a\quad =\quad 2\ 
$

Given, $(x-2)^2=(x+1)(x-1)$

$\displaystyle \sqrt[3]{x}\,- 4\, =\, 0\, \Rightarrow\, \sqrt[3]{x}\, =\, 4$.
$\Rightarrow\, x\, =\, 4^{3}\, =\, 64$

On simplifying, we have

On simplified, we have

A linear equation can comprise of many variables. The most simple linear equation is in $one$ variable ,i.e, $ax+b=0$.