Question 461005


{{{a = 1 + b}}},.......1

{{{b = 5-2a}}}...........2
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{{{a = 1 + b}}},.......1.substitute in 2

{{{b = 5-2(1 + b)}}}...........2.....solve for {{{b}}}

{{{b = 5-2 -2b)}}}

{{{b +2b= 3}}}


{{{3b= 3}}}

{{{b= 1}}}.....find {{{a}}}


{{{a = 1 + b}}},.......1

{{{a = 1 + 1}}}

{{{a = 2}}}


check:

{{{a = 1 + b}}},.......1

{{{2 = 1 + 1}}},.......1

{{{2 = 2}}},.......1


{{{b = 5-2a}}}...........2

{{{1 = 5-2*2}}}...........2

{{{1 = 5-4}}}...........2

{{{1= 1}}}...........2