Question

what is the equation of the line parallel to 3x 5y = 11 that passes through the point (8, -3)?

Answer 1

Answer:

$y=-\frac{3x}{5} +\frac{9}{5}$

Step-by-step explanation:

$3x + 5y = 11$

Step 1: Re-write in slope-intercept form:

Slope-intercept formula: $y=mx+b$

$3x+5y=11$

Subtract  $3x$  from both sides of the equation.

$5y=11-3x$

Divide each term in  $5y=11-3x$  by  $5$  and simplify.

$y=\frac{11}{5} -\frac{3x}{5}$

Now, write in  $y=mx+b$  form:

Reorder  $\frac{11}{5}$  and  $-\frac{3x}{5}$ .

$y=-\frac{3x}{5} +\frac{11}{5}$

Step 2: Substitute the point for x and y, then solve:

(8, -3)

x = 8

y = -3

Plug them into the equation:

$\bold{-3}=-\frac{3(\bold{8})}{5} +b$

Solve:

$-\frac{3\left(8\right)}{5}+b=-3$

$=-\frac{3\cdot \:8}{5}+b=-3$

$=-\frac{24}{5}+b=-3$

$=-\frac{24}{5}+b+\frac{24}{5}=-3+\frac{24}{5}$

$b=\frac{9}{5}$