Question 233407
{{{system(2x^2-6x+4=0,3y-12=-4x)}}}
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The first equation contains only one unknown, x, so we
can solve it for x separately:

{{{2x^2-6x+4=0}}}

Divide every term by 2

{{{(2x^2)/2-(6x)/2+4/2=0/2}}}

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

Factor the left side:

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

Set each factor equal to 0:

{{{x-2=0}}}
{{{x=2}}}

{{{x-1=0}}}
{{{x=1}}}

Now we substitute in the other equation

{{{3y-12=-4x)}}}

Substituting {{{x=2}}}

{{{3y-12=-4x)}}}
{{{3y-12=-4(2))}}}
{{{3y-12=-8}}}
{{{3y=4)}}}
{{{y=4/3}}}

So one solution is (2,{{{4/3}}})

Substituting {{{x=1}}}

{{{3y-12=-4x)}}}
{{{3y-12=-4(1))}}}
{{{3y-12=-4}}}
{{{3y=8)}}}
{{{y=8/3}}}

So the other solution is (1,{{{8/3}}})

Edwin</pre>