Question

Write an equation of the line that passes through (5, 2) and is parallel to the line defined by 3x = 5y + 3. Write the answer in slope-intercept form (if possible)
and In standard form (Ax + By = C) with smallest Integer coefficients. Use the "Cannot be written" button, If applicable.

Answer 1

First, we'll start by finding the slope of the given line defined by the equation \(3x = 5y + 3\). We can rearrange it to slope-intercept form \(y = mx + b\).

1. Rearranging the equation:
\[
3x - 3 = 5y
\]
\[
5y = 3x - 3
\]
\[
y = \frac{3}{5}x - \frac{3}{5}
\]

From the equation, we see that the slope \(m\) of the given line is \(\frac{3}{5}\).

Since we want a line that is parallel to this one, it will have the same slope, \(\frac{3}{5}\). Now we use the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point the line passes through. Here, \((x_1, y_1) = (5, 2)\).

2. Using the point-slope form:
\[
y - 2 = \frac{3}{5}(x - 5)
\]

Now let's simplify this into slope-intercept form:
\[
y - 2 = \frac{3}{5}x - \frac{15}{5}
\]
\[
y - 2 = \frac{3}{5}x - 3
\]
\[
y = \frac{3}{5}x - 1
\]

This is the equation in slope-intercept form:
\[
y = \frac{3}{5}x - 1
\]

3. Now, we need to convert this to standard form \(Ax + By = C\) with integer coefficients.

Starting from the slope-intercept form:
\[
y = \frac{3}{5}x - 1
\]
Multiply through by 5 to eliminate the fraction:
\[
5y = 3x - 5
\]

Rearranging gives:
\[
-3x + 5y = -5
\]

To write it in standard form with positive coefficients, we can multiply the entire equation by -1:
\[
3x - 5y = 5
\]

So, the final answers are:

**In slope-intercept form:**
\[
y = \frac{3}{5}x - 1
\]

**In standard form:**
\[
3x - 5y = 5
\]