Question

Your salary was \(\$ 30,200\) in 2007 and \(\$ 33,500\) in 2009 . Your salary follows a linear growth pattern. What salary will you be making in \(2012 ?\)

Short answer

The salary in 2012 will be $38,750.

Step-by-step solution

Find the Slope of the Linear Equation

First, calculate the slope of the linear equation. The slope \(m\) is given by the formula: \(m = (y_2 - y_1) / (x_2 - x_1)\). Here, \(x_1=2007\), \(y_1=30200\), \(x_2=2009\) and \(y_2=33500\). Substituting these values into the formula, we get slope \(m = (33500-30200) / (2009-2007) = 1650\).

Apply Point-Slope Form of Linear Equation

Now use the point-slope form of the linear equation: \(y-y_1 = m(x-x_1)\). Substituting the known values: \(y-30200=1650(x-2007)\). This will give us the equation that defines the salary as a function of the year.

Find the Salary in 2012

Now that we have the equation, substitute \(x=2012\) into the equation to get the salary in 2012. Hence, \(y-30200=1650(2012-2007)\), which simplifies to \(y=30200+1650*5=38750\). Therefore, the salary in 2012 will be $38,750.

Key concepts

Understanding Slope Calculation

When we're looking at linear growth, one of the essential things to determine is the slope. The slope tells us how steep a line is, giving us an idea of how quickly values are increasing or decreasing. It's calculated by comparing changes in the y-values over changes in the x-values.

Think about it like this: if you're driving up a hill, the steepness or slope would tell you how quickly you're gaining height compared to how much distance you're covering horizontally. In our salary problem, we

• have salary comparisons from two different years, which are essentially points on a graph
• utilize the formula for slope, which is practically: \( m = \frac{y_2 - y_1}{x_2 - x_1} \)

For this scenario:

• \(y_1 = 30,200\) (the salary in 2007) and \(y_2 = 33,500\) (the salary in 2009), where \(x_1 = 2007\) and \(x_2 = 2009\)
• Plug scenerio values into the formula: \( m = \frac{33,500 - 30,200}{2009 - 2007} = 1,650\)

This means your salary increases by $1,650 every year between these points.

Exploring Point-Slope Form

After determining the slope, the next step is using the point-slope form to find an equation of the line that represents this growth. The point-slope form is great because it starts with a point we know and a slope we've already calculated.

This mathematical form lets you write the line's equation very simply as:

• \(y - y_1 = m(x - x_1)\)

Here:

• \(y_1\) is 30,200 from the year 2007, \(m\) is 1,650 as calculated before, \(x_1\) is 2007 as well

Such re-formulation helps find any salary for a specific year (x) by plugging in numbers. Once rearranged and adjusted with given details from our scenario, it looks like:

• \(y - 30200 = 1650(x - 2007)\)

Plugging in a year into this equation directly lets us find out the projected salary.

Decoding Linear Equations

With both the slope and the point-slope form understood, we can see how these pieces come together to form a linear equation. Linear equations are straightforward, providing a direct line through data points, predicting future values based on past trends. The simplicity comes from its form, either from point-slope as derived, or when turned into slope-intercept form for clarity.

In our examples' refined point-slope form, you end up with a linear equation where:

• \(y = 30200 + 1650(x - 2007)\)
• which solves out logically to \(y = 1650x - 3206650\)

Applying this design:

• Plugging in 2012 for \(x\), unveils the salary by that year,
  calculated to \(y = 38,750\)

Linear equations like this are vital for predicting consistent growth, delivering a keen tool for analytical projections based on real-world linear patterns.