Question 1155472

{{{3x + 3 = 5y}}}......eq.1
{{{4x + 2y = -4}}} ......eq.2
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start with

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

{{{3x/5 + 3/5 = y}}}......substitute in eq.2


{{{4x + 2(3x/5 + 3/5)= -4}}} ......eq.2,...solve for {{{x}}}

{{{4x + 6x/5 + 6/5= -4}}}........both sides multiply by {{{5}}}

{{{20x + 6x + 6= -20}}}

{{{26x = -20-6}}}

{{{26x = -26}}}

{{{x = -1}}}


go to

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

{{{3(-1) + 3 = 5y}}}

{{{-3 + 3 = 5y}}}

{{{0 = 5y}}}

{{{y=0}}}


solution:

{{{x = -1}}}

{{{y=0}}}

intersection point is at: ({{{-1}}},{{{0}}})


{{{ graph( 600, 600, -10, 10, -10, 10, 3x/5 + 3/5 , -2x-2) }}}