bearing in mind that perpendicular lines have negative reciprocal slopes, let's find the slope of the provided line then
![\bf y=\stackrel{\stackrel{m}{\downarrow }}{-15}x+3\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-15%7Dx%2B3%5Cqquad%20%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

well, we know the x-intercept is at x = 3, recall when a graph intercepts the x-axis y = 0, so this point is (3 , 0). Then we're really looking for the equation of a line whose slope is 1/5 and runs through (3 , 0).

The answer is really easy you can go through anything
Hello there!
4(3x - 11) + 23 = 5x - 14
Apply the distributive property to 4(3x - 11)
4(3x) + 4(-11)
12x - 44
We now have:
12x - 44 + 23 = 5x - 14
Combine like-terms on the left-hand side of the equation.
-44 + 23 = -21
12x - 21 = 5x - 14
Get x on one side by subtracting 5x from both sides..
12x - 5x = 7x
5x - 5x = 0
7x - 21 = -14
Add 21 to both sides to isolate 7x.
-21 + 21 = 0
-14 + 21 = 7
7x = 7
Divide both sides by 7 to solve for x.
7x / 7 = x
7 / 7 = 1
We are now left with the following solution:
x = 1
I hope this helps!