Write the equation of the line that passes through (7, 3) and is parallel to (y - 15) = -2(x + 5).

The given equation of the line is in point-slope form:

\[
y - 15 = -2(x + 5)
\]

This equation can be rewritten in slope-intercept form by simplifying:

\[
y - 15 = -2x - 10 \quad \Rightarrow \quad y = -2x + 5
\]

From this, the slope of the line is \(m = -2\).

Since the new line is parallel to the given line, it will have the same slope \(m = -2\). Now, we use the point-slope form of a line equation to find the equation of the line passing through the point \((7, 3)\):

\[
y - y_1 = m(x - x_1)
\]

Substitute \(m = -2\), \(x_1 = 7\), and \(y_1 = 3\):

\[
y - 3 = -2(x - 7)
\]

Simplify the equation:

\[
y - 3 = -2x + 14 \quad \Rightarrow \quad y = -2x + 17
\]

So, the equation of the line that passes through \((7, 3)\) and is parallel to the given line is:

\[
y = -2x + 17
\]

Would you like more details or have any questions?

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Here are 5 related questions to further explore the concept:

1. How would you find the equation of a line perpendicular to \(y = -2x + 5\) that passes through the same point?
2. What is the slope-intercept form of a line, and how is it different from the point-slope form?
3. How can you find the point where two lines intersect if their equations are given?
4. What are the conditions for two lines to be parallel in terms of their slopes?
5. Can you derive the equation of a line passing through two given points?

**Tip:** Parallel lines have the same slope but different y-intercepts.

Learn how to find the equation of a line passing through the point (7, 3) that is parallel to y - 15 = -2(x + 5). This solution explains key algebraic concepts, including point-slope form and the properties of parallel lines, suitable for students in Grades 8-10.

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