Question 972509

{{{4x+y=6}}} ....eq.1
{{{x+y=3}}} ....eq.2

here you are given two equations of a line and you need to solve it as a system
that means these two lines might have intersection point in common, or they might have not common point at all ( in other words, they can be parallel), or they can have all points in common ( in other words, they can lie on top of each other, or simply be same line)

to find solution, you can decide whether you would use the graphing, substitution, or elimination method to solve the following system of equations 

it would be much easier to use the elimination method in this case because you can easily eliminate {{{y}}}

{{{4x+y=6}}} ....eq.1
{{{x+y=3}}} ....eq.2
-------------------------subtract eq.2 from eq.1

{{{4x+y-x-y=6-3}}} 

{{{4x+cross(y)-x-cross(y)=3}}}

{{{4x-x=3}}} 

{{{3x=3}}} 

{{{x=3/3}}}

{{{highlight(x=1)}}}

now, go back to eq.1 or eq.2, plug in {{{1}}} for {{{x}}} and solve for {{{y}}}:


{{{1+y=3}}} ....eq.2

{{{y=3-1}}} 

{{{highlight(y=2)}}} 

so, now we know that both lines intersect in one point which has coordinates {{{x=1}}} and {{{y=2}}}

({{{1}}},{{{2}}})-intersection point

we can check it on a graph:


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,2,.12), locate(1,2,p(1,2)),
 graph( 600, 600, -10, 10, -10, 10, 3-x,-4x+6)) }}}