Question 594498
If the y-intercept is 3, then this means that the line goes through (0,3)


So the line goes through the points (0,3) and (-6,7)



Now let's find the equation of the line that goes through the points (0,3) and (-6,7)



First let's find the slope of the line through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(-6,7\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(0,3\right)]. So this means that {{{x[1]=0}}} and {{{y[1]=3}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-6,7\right)].  So this means that {{{x[2]=-6}}} and {{{y[2]=7}}}.



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(7-3)/(-6-0)}}} Plug in {{{y[2]=7}}}, {{{y[1]=3}}}, {{{x[2]=-6}}}, and {{{x[1]=0}}}



{{{m=(4)/(-6-0)}}} Subtract {{{3}}} from {{{7}}} to get {{{4}}}



{{{m=(4)/(-6)}}} Subtract {{{0}}} from {{{-6}}} to get {{{-6}}}



{{{m=-2/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(-6,7\right)] is {{{m=-2/3}}}



Now let's use the point slope formula:



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y-3=(-2/3)(x-0)}}} Plug in {{{m=-2/3}}}, {{{x[1]=0}}}, and {{{y[1]=3}}}



{{{y-3=(-2/3)x+(-2/3)(-0)}}} Distribute



{{{y-3=(-2/3)x+0}}} Multiply



{{{y=(-2/3)x+0+3}}} Add 3 to both sides. 



{{{y=(-2/3)x+3}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(-6,7\right)] is {{{y=(-2/3)x+3}}}