Question 922541
{{{Y=2x+1}}}............eq.1
{{{Y=4x+5}}}............eq.2
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since left sides in both eq.1 and eq.2are same, then right sides must be same too

{{{2x+1=4x+5}}} .........solve for {{{x}}}

{{{1-5=4x-2x}}} 

{{{-4=2x}}}

{{{-4/2=x}}} 

{{{x=-2}}}

go to

{{{y=2x+1}}}............eq.1 plug in {{{x=-2}}} and solve for {{{y}}}

{{{y=2(-2)+1}}}

{{{y=-4+1}}}

{{{y=-3}}}


solution is: a point of intersection at ({{{-2}}},{{{-3}}})


graph it:

for linear function you need two points to graph it

one point is ({{{-2}}},{{{-3}}})

go to {{{y=2x+1}}}............eq.1 and find one more point

{{{y=2x+1}}}....if {{{x=0}}}

{{{y=2*0+1}}}

{{{y=1}}}

so, second point is ({{{0}}},{{{1}}})

do same for second line

one point is ({{{-2}}},{{{-3}}})

go to {{{Y=4x+5}}}............eq.2 and find one more point

{{{Y=4x+5}}}....if {{{x=0}}}

{{{y=4*0+5}}}

{{{y=5}}}

so, second point is ({{{0}}},{{{5}}})


plot these points and draw a lines through:

{{{drawing( 600, 600, -10, 10, -10, 10,circle(0,5,.2),locate(0,5,p(0,5)),circle(0,1,.2),locate(0,1,p(0,1)),circle(-2,-3,.2),locate(-2,-3,p(-2,-3)), graph( 600, 600, -10, 10, -10, 10, 4x+5, 2x+1)) }}}