Answer:
[tex]y=\frac{5}{3}x+\frac{10}{3}[/tex]
Step-by-step explanation:
Firstly we rewrite the equation that is parallel to the line we need because parallel lines have the same slope by the rule
[tex]m_1=m_2[/tex]
so we rewrite the equation in slope-intercept form:
[tex]5x-3y=17\\-3y=-5x+17\\y=\frac{-5x+17}{-3}\\\\y=\frac{-5x}{-3}+\frac{17}{-3} \\\\y=\frac{5}{3}x-\frac{17}{3}[/tex]
and now we compare it with our slope-intercept form equation:
[tex]y=mx+b[/tex]
the coefficient of x which is m is the slope and since parallel lines have equal slopes the line that we need would also have the same slope which is 5/3.
now the next step is the line that we need passes through the point (-2,0) so it means that it should satisfy that specific point so we insert that point into our slope-intercept form and insert the slope which is 5/3 which is m.
[tex]y=mx+b\\y=\frac{5}{3}x+b\\\\0=\frac{5}{3}(-2)+b\\\\0=\frac{-10}{3}+b\\\\b=\frac{10}{3}\\[/tex]
now we know that the y-intercept(b) of the line we need is 10/3 and slope(m) is 5/3 insert these values into our slope-intercept form:
[tex]y=mx+b\\y=\frac{5}{3}x+\frac{10}{3} \\[/tex]
so this is our given line.
