Question 188637
{{{1 + 3/(2 - i)}}} Start with the left side of the given equation.



{{{1 + (3(2 + i))/((2 - i)(2 + i))}}} Multiply the fraction by {{{(2+i)/(2+i)}}} (this is to make the denominator a real number)



{{{1 + (3(2 + i))/(4-i^2)}}} FOIL



{{{1 + (3(2 + i))/(4-(-1))}}} Plug in {{{i^2=-1}}}



{{{1 + (3(2 + i))/(4+1)}}} Rewrite the denominator



{{{1 + (3(2 + i))/(5)}}} Add




{{{1 + (6+3i)/(5)}}} Distribute



{{{5/5 + (6+3i)/(5)}}} Multiply the first term 1 by {{{5/5}}}



{{{(5+6+3i)/(5)}}} Combine the fractions



{{{(11+3i)/(5)}}} Combine like terms.



{{{11/5+(3i)/(5)}}} Break up the fraction



Since {{{11/5<>5}}} and {{{(3i)/(5)<>-i}}}, this means that {{{11/5+(3i)/(5)<>5-i}}}



So consequently {{{1 + 3/(2 - i)<>5-i}}}


So the statement is false.