Question

Solve the inequality. Then graph the solution set on the real number line. \(\frac{2}{5} x>7\)

Short answer

The solution to the inequality \(\frac{5+7x}{1+2x} < 4\) on the real number line is \(-\infty < x < -0.5\) and \(-0.5 < x < 1\).

Step-by-step solution

Subtract 4 from both sides.

By performing this action, we have a simpler inequality: \(\frac{5+7x}{1+2x} - 4 < 0\). Now, it needs to be simplified to identify the critical values.

Simplify the left-hand side.

Simplify the subtracted term by finding a common denominator: \(\frac{5+7x - 4(1+2x)}{1+2x} < 0\). The result will be \(\frac{5+7x -4 - 8x }{1+2x} = \frac{1 - x}{1+2x} < 0\). At this point, we can determine the critical values by equating the numerator and denominator to zero because they define the regions that make the inequality true.

Find the critical values.

Set the denominator and the numerator to zero and solve for \(x\), so we will have \(1 - x = 0\) and \(1 + 2x = 0\) yielding two solutions \(x = 1\) and \(x = -0.5\) respectively.

Determine the solution intervals.

Use these critical values to divide the number line into three regions: \(-\infty, -0.5\), \(-0.5, 1\), and \(1, \infty\). Substitute a test point from each region into the inequality \(\frac{1 - x}{1+2x} < 0\) to determine if the inequality holds for that region. From these tests, the solution for the inequality can be found.

Plot the solution intervals.

The solutions from Step 4 mark the endpoints of the solution intervals on a number line. An open circle is used at \(x = -0.5\) and \(x = 1\) because these points are not included in the solution.

Key concepts

Critical Values

In algebra, when dealing with inequalities, critical values are essential in determining where solutions fall on the number line. A critical value occurs where the expression is either equal to zero or undefined.
For the inequality \(\frac{1-x}{1+2x} < 0\), we find critical values by setting both the numerator and the denominator equal to zero. This gives us the points where the expression may change sign.

• Set the numerator equal to zero: \(1-x=0\), which gives \(x=1\).
• Set the denominator equal to zero: \(1+2x=0\), leading to \(x=-0.5\).

These x-values, -0.5 and 1, help us break down the number line into intervals where the inequality could potentially change. At these points, the expression shifts from negative to positive or vice versa. Identifying these values is crucial because they provide the boundaries for testing different intervals.

Number Line Graphing

Number line graphing is a visual way of representing the solution set of inequalities. It can often make the understanding of solution sets more intuitive. Following the determination of the critical values in the inequality \(\frac{1-x}{1+2x} < 0\), we plot these values on a number line.
Begin by marking the critical values -0.5 and 1.

• Use open circles at these points: \(-0.5\) and \(1\) because these values lead to a zero in the denominator or actual equality, which does not satisfy a strict inequality.
• The rest of the number line is divided into sections: \((-\infty, -0.5)\), \((-0.5, 1)\), and \((1, \infty)\).

After testing each of these regions, the parts that satisfy the inequality are then shaded on the number line. This visual representation confirms the correctness of your solution intervals, showing clearly where the inequality holds true.

Solution Intervals

When solving inequalities, determining the solution intervals is a key step. Given our critical values from \(\frac{1-x}{1+2x} < 0\), the number line is divided into three segments: \((-\infty, -0.5), (-0.5, 1), (1, \infty)\).
These intervals include the spaces between the critical values as well as areas beyond them.

• For each segment, choose a test value that isn't a critical value itself and plug it into the inequality.
• If the inequality is true for a test value in that interval, then the entire interval is part of the solution.

For example, select a point from each interval:
- In \((-\infty, -0.5)\), try \(x=-1\), making \(\frac{1-(-1)}{1+2(-1)} = \frac{2}{-1} < 0\), which is true.- In \((-0.5, 1)\), use \(x=0\), resulting in \(\frac{1-0}{1+2(0)} = \frac{1}{1} > 0\), which is false.- In \((1, \infty)\), try \(x=2\), yielding \(\frac{1-2}{1+4} = \frac{-1}{5} < 0\), which is true.
Hence, the solution intervals where the inequality holds are \((-\infty, -0.5)\) and \((1, \infty)\), represented by shaded regions on the number line.