Question 916395

given:

{{{Y= -(1/2)x - 5}}}......eq.1

{{{Y= (3/4)x -10}}}......eq.2

Points (x,y): (4, -7), (0, -10), (3, -9/2)

to determine which points are solutions to the given system, solve the system


{{{Y= -(1/2)x - 5}}}......eq.1

{{{Y= (3/4)x -10}}}......eq.2.. ..substitute {{{y}}} in eq.1
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{{{(3/4)x -10= -(1/2)x - 5}}} ......eq.1............solve for {{{x}}}

{{{(3/4)x +(1/2)x = 10 - 5}}}

{{{(3/4)x +(2/4)x = 10 - 5}}}

{{{(5/4)x  = 5}}}

{{{5x  = 5*4}}}

{{{x=20/5}}}

{{{highlight(x=4)}}}.....plug in eq. 1 or 2 and find {{{y}}}

{{{Y= -(1/2)*4 - 5}}} ......eq.1

{{{Y= -(1/cross(2))*cross(4)2 - 5}}}

 {{{Y= -(1*2) - 5}}} 

{{{Y= -2 - 5}}} 

{{{highlight(Y= -7)}}} 


so, given point (4, -7), is a solution to the given system


{{{drawing( 600, 600, -10, 10, -10, 10,circle(4,-7,.2),locate(4,-7,p(4,-7)),  graph( 600, 600, -10, 10, -10, 10, -(1/2)x - 5, (3/4)x -10)) }}}