Question 758035


<h4>x-intercept</h4>

To find the x-intercept, plug in {{{y=0}}} and solve for x



{{{x+3y=15}}} Start with the given equation.



{{{x+3(0)=15}}} Plug in {{{y=0}}}.



{{{x+0=15}}} Multiply {{{3}}} and 0 to get 0.



{{{x=15}}} Simplify.



So the x-intercept is *[Tex \LARGE \left(15,0\right)].



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<h4>y-intercept</h4>

To find the y-intercept, plug in {{{x=0}}} and solve for y



{{{x+3y=15}}} Start with the given equation.



{{{0+3y=15}}} Plug in {{{x=0}}}.



{{{3y=15}}} Simplify.



{{{y=(15)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



{{{y=5}}} Reduce.



So the y-intercept is *[Tex \LARGE \left(0,5\right)].



Now let's plot the points *[Tex \LARGE \left(15,0\right)] and *[Tex \LARGE \left(0,5\right)] which are the x and y intercepts respectively.



{{{drawing(500, 500, -2,18,-10,10,
grid(0),
graph(500, 500, -2,18,-10,10,0)
circle(15,0,0.03),circle(15,0,0.05),circle(15,0,0.08),circle(15,0,0.10),circle(15,0,0.12),
circle(0,5,0.03),circle(0,5,0.05),circle(0,5,0.08),circle(0,5,0.10),circle(0,5,0.12)
)}}}



Now draw a straight line through the plotted points to graph {{{x+3y=15}}}.



{{{ drawing(500, 500, -2,18,-10,10,
grid(0),
graph(500, 500, -2,18,-10,10,(15-x)/(3)),
circle(15,0,0.03),circle(15,0,0.05),circle(15,0,0.08),circle(15,0,0.10),circle(15,0,0.12),
circle(0,5,0.03),circle(0,5,0.05),circle(0,5,0.08),circle(0,5,0.10),circle(0,5,0.12)
)}}} Graph of {{{x+3y=15}}}