Question 1178678

 {{{2a + 3b = 52}}} ...........eq.1
{{{3b + c = 41}}}...........eq.2
{{{bc = 60}}}...........eq.3
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Find {{{a+b+ c}}}

{{{3b + c = 41}}}...........eq.2, solve for {{{c}}}

{{{ c = 41-3b}}}...........eq1)

{{{bc = 60}}}...........eq.3, solve for {{{c}}}

{{{c = 60/b}}}................2)


from 1) and 2) we have


{{{  41-3b=60/b}}}........solve for {{{b}}}

{{{  41b-3b^2=60}}}

{{{  0=3b^2-41b+60}}}....factor

{{{   (b - 12) (3b - 5)=0}}}


solutions:

{{{b=12}}}
or
{{{b=5/3}}}


go to

{{{ c = 41-3b}}}...........eq1), substitute {{{b}}}

{{{ c = 41-3*12}}}
{{{ c = 5}}}

or

{{{ c = 41-3*(5/3)}}}
{{{ c = 36}}}


so we have two pairs of solutions for {{{b}}} and {{{c}}}

{{{b=12}}},{{{ c = 5}}}
and
{{{b=5/3}}},{{{ c = 36}}}

go to

{{{2a + 3b = 52}}} ...........eq.1, substitute {{{b=12}}} 

{{{2a + 3*12 = 52}}}
{{{2a + 36= 52}}}
{{{2a = 52-36}}}
{{{2a = 16}}}
{{{a = 8}}}


one solution is: {{{highlight(a = 8)}}},{{{highlight(b=12)}}} ,{{{ highlight(c = 5)}}}



then  {{{a+b+ c=8+12+5=25}}}


{{{2a + 3b = 52}}} ...........eq.1, substitute {{{b=5/3}}} 

{{{2a + 3*(5/3)= 52}}}
{{{2a + 5= 52}}}
{{{2a = 52-5}}}
{{{2a = 47}}}
{{{a = 47/2}}}


other solution is: {{{highlight(a = 47/2)}}}, {{{highlight(b=5/3)}}},{{{ highlight(c = 36)}}}


then  {{{a+b+ c=47/2+5/3+36=367/6=61&1/6}}}