Question

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

Explanation:

Step1: Find the slope of the given line

First, rewrite the equation \(5x - 3y = 3\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
Starting with \(5x-3y = 3\), we solve for \(y\):
\(-3y=-5x + 3\)
Divide each term by \(-3\): \(y=\frac{5}{3}x - 1\)
The slope of the line \(5x - 3y = 3\) is \(m=\frac{5}{3}\). Since parallel lines have the same slope, the line we want to find also has a slope of \(\frac{5}{3}\).

Step2: Use the point - slope form to find the equation of the line

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)\) is a point on the line and \(m\) is the slope.
We know that \(m = \frac{5}{3}\) and the point \((x_1,y_1)=(3,-3)\). Substitute these values into the point - slope form:
\(y-(-3)=\frac{5}{3}(x - 3)\)
Simplify the left - hand side: \(y + 3=\frac{5}{3}(x - 3)\)

Step3: Convert to slope - intercept form (optional, but to get a standard form)

Distribute the \(\frac{5}{3}\) on the right - hand side: \(y+3=\frac{5}{3}x-5\)
Subtract 3 from both sides: \(y=\frac{5}{3}x-5 - 3\)
\(y=\frac{5}{3}x-8\)

If we want to write it in standard form \(Ax+By = C\) (where \(A\), \(B\), and \(C\) are integers and \(A\geq0\)):
Multiply through by 3 to get rid of the fraction: \(3y = 5x-24\)
Rearrange to get \(5x-3y=24\)

Answer:

\(y=\frac{5}{3}x - 8\) (or \(5x - 3y = 24\))