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Point-slope form is a way to write the equation of a straight line. It is particularly useful for finding the equation of a line when you know one point on the line and the slope. The general formula is:

• \(y - y_1 = m(x - x_1)\)

This format is based on a known point \((x_1, y_1)\) and the slope \(m\) of the line. Understanding this form helps to easily set up and visualize linear equations using points and slopes from given data.
Let's take the equation from the exercise, \(y = 25 - 2(x + 5)\). In this expression, you can identify the rearranged point-slope form by first expanding and rearranging it. When expanded, it becomes \(y = -2x + 15\) which indicates the point-slope equivalent was derived using the point \((-5, 25)\) before the transformation. The equation gives us the slope \(m=-2\) and helps to identify point-slope features.

Solving equations is a fundamental skill in algebra that involves finding unknown values that satisfy the equation. In our example, we identify that the equation \(y = -2x + 15\) needs rearranging to solve for unknown variables.

• Determine the value you want to solve for: For example, when \(y = 15\), what is \(x\)?
• Substitute known values into the equation: Plug \(y = 15\) into \(y = -2x + 15\): \(15 = -2x + 15\).
• Work through solving the equation: Subtract 15 from each side to isolate terms containing \(x\).

This results in \(0 = -2x\). Divide each side by \(-2\) to solve for \(x\): \(x = 0\).
Solving equations requires logical manipulation to isolate variables, enabling you to solve real-world problems and other algebraic questions. Always check your results by substituting found values back into the original equation to ensure it holds true.

Linear equations express a direct relationship between two variables usually in the form of \(y = mx + b\), which is known as the slope-intercept form. The variables represent:

• \(y\) as the dependent variable
• \(x\) as the independent variable

The \(m\) is the slope, showing the steepness of the line, and \(b\) is the y-intercept, indicating where the line crosses the y-axis.
Linear equations, such as \(y = -2x + 15\) from our exercise, graph as straight lines. These lines can be adjusted by changing their slopes or intercepts. For instance, a more negative \(m\) would make the line steeper downward.
Linear equations are easy to work with and a great introduction to more complex algebraic concepts. They can be used to model many real-world scenarios where relationships between variables are simple and direct. Recognizing their structure lets you convert between forms like point-slope and slope-intercept, enhancing problem-solving flexibility.