Question 1161245
{{{y = (1/2)x - 6 }}}.........eq.1->already in slope intercept form
{{{-2x - y = -9}}}.............eq.2
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{{{-2x - y = -9}}}.............eq.2, write in slope intercept form

{{{y=-2x +9 }}}...........eq.2

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{{{y = (1/2)x - 6 }}}.........eq.1, slope is {{{1/2}}}, y-intercept is ({{{0}}},{{{-6}}})

{{{y=-2x +9 }}}...........eq.2, slope is {{{-2}}}, y-intercept is ({{{0}}},{{{9}}})

since a slope of eq.2 is negative reciprocal of slope of the eq.1, lines are {{{perpendicular}}} to each other

for each line find one more point to graph them


{{{y = (1/2)x - 6 }}}...let {{{y=0}}}

{{{0 = (1/2)x - 6 }}}

{{{6= (1/2)x }}}

{{{x=6/(1/2)}}}

{{{x=12}}}-> x-intercept is at ({{{12}}},{{{0}}})


{{{y=-2x +9 }}}....let {{{y=0}}}

{{{0=-2x +9 }}}


{{{-9=-2x  }}}

{{{x=-9/-2}}}

{{{x=4.5}}}-> x-intercept is at ({{{4.5}}},{{{0}}})


graph the line {{{y = (1/2)x - 6 }}} using points ({{{12}}},{{{0}}}) and ({{{0}}},{{{-6}}})

and
graph the line {{{y = -2x +9  }}} using points ({{{4.5}}},{{{0}}}) and ({{{0}}},{{{9}}})


{{{drawing( 600, 600, -15, 15, -15, 15,
circle(12,0,.13), locate(12,-0.5,p(12,0)),
circle(4.5,0,.13), locate(4.5,-0.5,p(4.5,0)),
circle(0,-6,.13), locate(0,-6,p(0,-6)),
circle(0,9,.13), locate(0,9,p(0,9)),
circle(6,-3,.13), locate(6,-3,p(6,-3)),
 graph( 600, 600, -15, 15, -15, 15, -2x +9 , (1/2)x - 6 )) }}}

as you can see from the graph, intersection point is at ({{{6}}},{{{-3}}}), therefore solution to the system of equations is: 

{{{x=6}}}
{{{y=-3}}}