Question 1050937
substitution method

{{{x=8y-4}}}....eq.1
{{{x=4y-4}}}....eq.2
------------------------substitute {{{x}}} from eq.1 in eq.2

{{{8y-4=4y-4}}}....eq.2...solve for {{{y}}}

{{{8y-4y=4-4}}}

{{{4y=0}}}
{{{y=0}}}

{{{x=8y-4}}}....eq.1...substitute {{{y}}}

{{{x=8*0-4}}}

{{{x=-4}}}

solutions:

{{{x=-4}}}
{{{y=0}}}

or, two lines intersect at point ({{{-4}}},{{{0}}})



{{{ graph( 600, 600, -10, 10, -10, 10, x/8+4/8, x/4+4/4) }}}