Question 1187055

{{{x^2+y^2=5}}}.........eq.1
{{{3x-y=5}}}..............eq.2
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{{{3x-y=5}}}..............eq.2, solve for {{{y}}}

{{{3x-5=y}}}..............substitute in eq.1

{{{x^2+(3x-5)^2=5}}}.........eq.1, solve for {{{x}}}

{{{x^2+9x^2 - 30x + 25=5}}}

{{{10x^2 - 30x + 25-5=0}}}

{{{10x^2 - 30x + 20=0}}}.........divide by {{{10}}}

{{{x^2 - 3x + 2=0}}}.........factor

{{{(x - 2) (x - 1) = 0}}}

solutions:

if {{{(x - 2)  = 0}}}=>{{{x=2}}}
if {{{  (x - 1) = 0}}}=>{{{x=1}}}

go to

{{{3x-y=5}}}..............eq.2, substitute {{{x=2}}}

{{{3*2-y=5}}}

{{{6-5=y}}}

{{{y=1}}}

{{{3x-y=5}}}..............eq.2, substitute {{{x=1}}}

{{{3*1-y=5}}}

{{{3-5=y}}}

{{{y=-2}}}

so, solutions to the system are:

{{{x=2}}}, {{{y=1}}}

or

{{{x=1}}}, {{{y=-2}}}