Question 154327
{{{8x=5y+2}}} Start with the given equation.



{{{8x-2=5y}}} Subtract 2 from both sides.



{{{(8x-2)/5=y}}} Divide both sides by 5.




{{{(8x)/5-(2)/5=y}}} Break up the fraction



{{{(8/5)x-2/5=y}}} Simplify



So after converting {{{8x=5y+2}}} into slope-intercept form, we get {{{y=(8/5)x-2/5}}}



We can see that the equation {{{y=(8/5)x-2/5}}} has a slope {{{m=8/5}}} and a y-intercept {{{b=-2/5}}}.



Since parallel lines have equal slopes, this means that we know that the slope of the unknown parallel line is {{{m=8/5}}}.

Now let's use the point slope formula to find the equation of the parallel line by plugging in the slope {{{m=8/5}}}  and the coordinates of the given point *[Tex \LARGE \left\(-6,5\right\)].



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



{{{y-5=(8/5)(x--6)}}} Plug in {{{m=8/5}}}, {{{x[1]=-6}}}, and {{{y[1]=5}}}



{{{y-5=(8/5)(x+6)}}} Rewrite {{{x--6}}} as {{{x+6}}}



{{{y-5=(8/5)x+(8/5)(6)}}} Distribute



{{{y-5=(8/5)x+48/5}}} Multiply



{{{y=(8/5)x+48/5+5}}} Add 5 to both sides. 



{{{y=(8/5)x+73/5}}} Combine like terms. note: If you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>.



So the equation of the line parallel to {{{8x=5y+2}}} that goes through the point *[Tex \LARGE \left\(-6,5\right\)] is {{{y=(8/5)x+73/5}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(8/5)x-2/5,(8/5)x+73/5),
circle(-6,5,0.08),
circle(-6,5,0.10),
circle(-6,5,0.12))}}} Graph of the original equation {{{y=(8/5)x-2/5}}} (red) and the parallel line {{{y=(8/5)x+73/5}}} (green) through the point *[Tex \LARGE \left\(-6,5\right\)].