Given fractional expression:
[tex]{\sf \longmapsto \dfrac{x-2}{3} = \dfrac{x+1}{4}}[/tex]
Cross multiply the numbers.
[tex]{\sf \longmapsto 4 (x-2) = 3 (x +1)}[/tex]
Multiply the number outside the bracket with the numbers in the bracket.
[tex]{\sf \longmapsto 4x - 8 = 3x +3}[/tex]
Shift all variables on LHS and constants on RHS.
[tex]{\sf \longmapsto 4x - 3x = 3+8}[/tex]
Subtract the values on LHS and Add the values on RHS.
[tex]{\sf \longmapsto 7x = (11)}[/tex]
Shift the number 7 from LHS to RHS.
[tex]{\sf \longmapsto x = \dfrac{11}{7}}[/tex]
[tex]\underline{\boxed{\bf So, \: the \: value \: of \: x \: is \: \dfrac{11}{7}.}}[/tex]
Verification :
[tex]{\sf \longmapsto \dfrac{x-2}{3} = \dfrac{x+1}{4}}[/tex]
Substitute the value of the x.
[tex]{\sf \longmapsto \dfrac{\frac{11}{7}-2}{3}= \dfrac{\frac{11}{7}+1}{4}}[/tex]
[tex]{\sf \longmapsto \dfrac{\frac{11-14}{7}}{3} = \dfrac{\frac{11+7}{7}}{4}}[/tex]
[tex]{\sf \longmapsto \dfrac{\frac{-3}{7}}{3}= \dfrac{\dfrac{18}{7}}{4}}[/tex]
Cancel the number 7 on numerator Of LHS and RHS.
[tex]{\sf \longmapsto \dfrac{-3}{3} = \dfrac{18}{4}}[/tex]
Write the fraction in lowest form by cancellation method.
[tex]{\sf \longmapsto \dfrac{-3}{3} = \dfrac{9}{2}}[/tex]
So,
[tex]{\sf \longmapsto LHS ≠ RHS}[/tex]
