Question

7x + 1 = 2/7 Mathdude500 please​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm \: 7x + 1 = \dfrac{2}{7}$

On transposition, we get

$\rm \: 7x = \dfrac{2}{7} - 1$

On taking LCM on RHS, we get

$\rm \: 7x = \dfrac{2 - 7}{7}$

$\rm \: 7x = \dfrac{- 5}{7}$

$\bf\implies \:x \: = \: - \: \dfrac{5}{49}$

Verification :-

$\rm \: \red{\: Consider\:LHS}$

$\rm \: 7x + 1$

On substituting the value of x, we get

$\rm \: = \: 7 \times \bigg( - \dfrac{5}{49} \bigg) + 1$

$\rm \: = \: - \: \dfrac{5}{7} + 1$

$\rm \: = \: \dfrac{ - 5 + 7}{7}$

$\rm \: = \: \dfrac{2}{7}$

$\rm \: \red{\:=\:RHS}$

Hence, Verified

$\rule{190pt}{2pt}$

In method of transposition,

$\red{\rm \: + \: changes \: to \: - \: }$

$\red{\rm \: - \: changes \: to \: + \: }$

$\red{\rm \: \times \: changes \: to \: \div \: }$

$\red{\rm \: \div \: changes \: to \: \times \: }$

Answer 2

$\rm {Given \: equation \: is}$

$\rm\blue{ 7x + 1 = \dfrac{2}{7}}$

$\rm{On \: transposition, we \: get}$

$\rm \blue{7x = \dfrac{2}{7} - 1}$

$\rm{On \: taking \: LCM \: on \: RHS, we \: get}$

$\rm \blue{ 7x = \dfrac{2 - 7}{7}} \\\rm \blue{ 7x = \dfrac{- 5}{7}} \\\rm \blue{\implies \:x \: = \: - \: \dfrac{5}{49}}$