Question 114482
how to get the values of x here in this equation (clue: there are 3 answers, 1 fraction and 2 whole numbers), will you please help me. This is the equation: 

{{{(x^2+2x+1)/(x^2+3x+2) = (x^2+5x+6)/ (x^2+4x+3)}}}

We will try to factor what ever is possible to factor in order to simplify this equation: 

first this quadratic trinomial where we recognize the square of sum 

{{{x^2+2x+1 =  (x+1)^2 = (x+1)(x+1)}}}…………………….(1)

Now factor this one
{{{x^2+3x+2}}}…….we can replace {{{3x}}} with {{{1x + 2x}}}
	
{{{x^2+1x + 2x +2}}}…….group first two terms and last two terms together
	
{{{(x^2+1x) + (2x +2)}}}…….notice that {{{x}}} is common for first group and {{{2}}} is common for second group; so factor them out

{{{x(x+1) + 2(x +1)}}}……. Factor out common {{{x+1}}}, combine like terms

{{{ (x+1) (x + 2) }}}……………………………………………….(2)

Now factor this one

{{{x^2 + 5x + 6}}}……….. replace {{{5x }}}with {{{2x+3x}}}
	
{{{x^2 + 2x + 3x + 6}}}……….. group first two terms and last two terms together

{{{(x^2 + 2x )+( 3x + 6)}}}……….. factor a {{{1x}}} out of the first group and factor a {{{3}}} out of the second group

{{{x(x + 2 ) + 3(x + 2)}}}……….. combine like terms
	
{{{ (x + 2 ) (x + 3) }}}……………………………………………..(3)

Now factor this one
	
{{{x^2+4x+3}}}………………. replace {{{4x }}}with {{{1x+3x}}}

{{{x^2+1x+3x +3}}}……. group first two terms and last two terms together

{{{(x^2+1x) + (3x +3)}}}……. factor a {{{1x}}} out of the first group and factor a {{{3}}} out of the second group

{{{x(x+1) + 3(x + 1)}}}……. combine like terms
{{{ (x+1) (x + 3) }}}……………………………………………….(4)

Now substitute (1), (2), (3), and (4) in original equation:


{{{(x^2+2x+1)/(x^2+3x+2) = (x^2+5x+6)/ (x^2+4x+3)}}}

{{{ (x+1)(x+1)/ (x+1) (x + 2) =  (x + 2 ) (x + 3) / (x+1) (x + 3) }}}….. simplify

{{{ (x+1)/ (x + 2) =  (x + 2 )  / (x+1)  }}}….....here we have denominators {{{x+2}}} and {{{x+1}}}; they cannot be equal to {{{0}}} ( to divide by {{{0}}} doesn't make sense

so we will find values of {{{x}}} for which it will be equal to {{{0}}}


they would be equal to {{{0}}} if:

{{{x+2 = 0}}}

if {{{x = -2}}}.........................one {{{solution}}}}

and

{{{x+1=0}}}

if {{{x=-1}}}........................second {{{solution}}}}


now find third solution:


{{{ (x+1)(x+1) =  (x + 2 )(x + 2) }}}…..multiply

{{{ x^2+ 2x + 1 =  x^2 + 4x + 4 }}}….....simplify (eliminate {{{x^2}}} on both sides

{{{ 2x + 1 = 4x + 4 }}}….....move {{{2x}}} to the right and {{{4}}} to the left

{{{1 - 4 = 4x -2x }}}….....

{{{- 3 = 2x }}}….....divide both sides by {{{2}}}

{{{- 3/2 = 2x/2 }}}….....

{{{- 3/2 = x }}}….....


or

{{{x = -3/2}}}….......................third {{{solution}}}}