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Linear equations are mathematical expressions that model relationships with a consistent rate of change. These equations represent straight lines on a graph and are of the first degree, meaning variables like \( x \) and \( y \) are raised only to the power of one. A linear equation usually takes the form of \( y = mx + b \), where:

• \( m \) represents the slope, which indicates the steepness or direction of the line.
• \( b \) is the y-intercept, the point where the line crosses the y-axis.

Linear equations are foundational in algebra and are used to solve real-world problems where quantities have a constant rate of change. In our example, the equation \( y = 15 - 2x \) is a simplified form of a linear equation. It provides an easy way to understand how changes in \( x \) affect \( y \). By interpreting the equation in this form, we can easily predict outcomes or analyze relationships between variables.

The slope-intercept form is a specific way of writing linear equations as \( y = mx + b \). This format is particularly useful because it clearly highlights two important features:

• The slope \( m \), which tells us how steep the line is.
• The y-intercept \( b \), which tells us where the line crosses the y-axis.

For instance, the equation \( y = 15 - 2x \) can be rearranged into \( y = -2x + 15 \), displaying a slope of \(-2\) and a y-intercept of \(15\).

The Importance of Slope

The slope is a crucial component as it indicates the rate of change. A negative slope, as in our example, suggests the line falls as it moves from left to right. Understanding the slope helps in predicting how one variable affects another.

Y-Intercept Explained

Meanwhile, the y-intercept shows the value of \( y \) when \( x \) is zero. This feature helps situate the line on a graph and serves as a starting point for plotting.

Solving linear equations involves finding the value of the unknown variable that makes the equation true. Substitution is a method used to achieve this by replacing one variable with a known value or expression.In our exercise, we are tasked with finding \( x \) when \( y \) equals 15. We substitute 15 into the equation \( y = 15 - 2x \) for \( y \) and solve for \( x \):

• Begin with the equation: \( 15 = 15 - 2x \).
• Subtract 15 from both sides to isolate terms with \( x \): \( 0 = -2x \).
• Finally, divide each side by \(-2\) to find \( x = 0 \).

This solution demonstrates the power of substitution, as it simplifies the process of finding unknowns by outsourcing known values into the equation. This basic yet sophisticated approach forms the backbone of solving many algebraic problems.