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Chapter 8: Problem 51

Graph the solution set of each inequality. $$3 x+5 y>10$$

Short Answer

Expert verified
The solution is a graph with a dashed line passing through points (0,2) and (2, 0.8) and shading above the line.

Step by step solution

01

Transform the inequality

First, the inequality needs to be rewritten to isolate y. You start by subtracting \(3x\) on both sides of the inequality \(3x + 5y > 10\), which gives you \(5y > 10 - 3x\). After that, divide all terms by 5 to get \(y > 2 - 0.6x\)
02

Plot line

The inequality resembles a linear equation in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept. In this case, the line will have a slope of -0.6 (downward) and the y-intercept is 2. So the line intersects the y-axis at point (0,2). Choose another suitable point for x, for example x = 2, and calculate y = \(2 - 0.6*2\) = 0.8. The second point is then (2, 0.8). Plot these points and draw the line through them.
03

Shading the solution region

The inequality is \(y > 2 - 0.6x\), which means all the y-values that are greater than \(2 - 0.6x\) satisfy the inequality. Thus, the solution region to the inequality is above the line. Specifically, the solution is all the points in the region above the line, not including the line itself because the inequality is 'greater than' and not 'greater than or equal to'. So, the line should be dashed to show that it is not included in the solution set. Then, shade the region above the line.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Inequalities
Linear inequalities are similar to linear equations but include inequality signs instead of equality signs. These inequalities can be expressed as:
  • Greater than (>)
  • Less than (<)
  • Greater than or equal to (鈮�)
  • Less than or equal to (鈮�)
In the inequality \(3x + 5y > 10\), we are looking for all the points
\((x, y)\) that make this statement true.
Instead of finding just one solution, with inequalities we find a whole region where all combinations of \(x\) and \(y\) satisfy the inequality condition.
Transforming these inequalities to a form like \(y > mx + b\) is crucial for graphing because it makes it easier to identify the boundary line, slope, and intercept, which are fundamental for visual representation.
Solution Set
The solution set of an inequality represents all the values that satisfy the inequality condition. For the inequality \(y > 2 - 0.6x\), the solution set is every point above the line \(y = 2 - 0.6x\) on the graph.Here's how to determine the solution set effectively:
  • First, solve the inequality for \(y\).
  • Plot the associated boundary line. Remember to use a dashed line if the inequality does not include equal to (like \(\gt\) or \(<\)).
  • Shade the region above the line for \(y > \) inequalities (and below the line for \(y < \) inequalities).
In this inequality, since the sign is \(\gt\), we shade the area above the dashed line.
These shaded areas represent all possible solutions or values of \(x\) and \(y\) that make the inequality true.
Slope-Intercept Form
The slope-intercept form is a commonly used way to represent linear equations and inequalities. It's written as \(y = mx + b\), where \(m\) is the slope, and \(b\) is the y-intercept of the line.In our inequality \(y > 2 - 0.6x\):
  • The slope \(m = -0.6\), which tells us that the line decreases by 0.6 units vertically for every unit it moves horizontally to the right.
  • The y-intercept \(b = 2\), meaning this line crosses the y-axis at the point (0,2).
This form helps in easily plotting the line on a graph as it provides both the direction (slope) and starting point (y-intercept). It鈥檚 intuitive because once you have these two details, drawing the line and finding additional points becomes straightforward.
Understanding this form is essential for graphing any linear inequality or equation accurately.
Graphing Techniques
Graphing inequalities involves creating a visual representation of the solution set or range that satisfies the inequality. Following systematic graphing techniques is vital for accuracy.Steps for graphing an inequality like \(y > 2 - 0.6x\) include:
  • Transform and Rearrange: Express the inequality in the slope-intercept form. Here, we transformed it to \(y > 2 - 0.6x\).
  • Plot the Boundary Line: Use the slope and y-intercept from the equation to plot the line. Begin at the intercept \((0,2)\) and use the slope to find another point, such as \((2,0.8)\).
  • Dashed Line: Use a dashed line to indicate that points on the line itself are not part of the solution set (since \(>\) doesn鈥檛 include equality).
  • Shade the Region: For \(y >\), shade above the line. This visually communicates all the combinations of \(x\) and \(y\) that make the inequality true.
These techniques help distinguish between inclusive or non-inclusive solutions and provide a clear graphical representation of all solutions.

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Most popular questions from this chapter

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. Privately owned, single-family homes in a small town were heated with gas, electricity, or oil. The percentage of homes heated with electricity was 9 times the percentage heated with oil. The percentage of homes heated with gas was 40 percentage points higher than the percentage heated with oil and the percentage heated with electricity combined. Find the percentage of homes heated with each type of fuel.

A family owns and operates three businesses. On their income-tax return, they have to report the depreciation deductions for the three businesses separately. In \(2004,\) their depreciation deductions consisted of use of a car, plus depreciation on 5 -year equipment (on which onc-fifth of the original value is deductible per year) and 10-year equipment (on which one- tenth of the original valuc is deductible per year). The car use (in miles) for cach business in 2004 is given in the following table, along with the original value of the depreciable 5 - and 10 year equipment used in each business that year. $$\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { Car } \\\\\text { Use }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 5-Year }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 10-Year }\end{array} \\\\\text { Business } & \text { (miles) } & \text { Equipment (S) } & \text { Equipment (S) } \\\\\hline 1 & 3200 & 9850 & 435 \\\2 & 8800 & 12,730 & 980 \\\3 & 6880 & 2240 & 615\\\\\hline\end{array}$$ The depreciation deduction for car use in 2004 was 37.5 cents per mile. Use matrix multiplication to determine the total depreciation deduction for each business in 2004.

Involve positive-integer powers of a square matrix \(A . A^{2}\) is defined as the product \(A A ;\) for \(n \geq 3, A^{n}\) is defined as the product \(\left(A^{n-1}\right) A\) Let \(A=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right] .\) Find the inverses of \(A^{2}\) and \(A^{3}\) without computing the matrices \(A^{2}\) and \(A^{3} .\)

For what value(s) of \(b\) does the following system of equations have two distinct, real solutions? $$\left\\{\begin{array}{l} y=-x^{2}+2 \\ y=x+b \end{array}\right.$$

Sarah can't afford to spend more than \(\$ 90\) per month on transportation to and from work. The bus fare is only \(\$ 1.50\) one way, but it takes Sarah 1 hour and 15 minutes to get to work by bus. If she drives the 15 -mile round trip, her one-way commuting time is reduced to 40 minutes, but it costs her S.40 per mile. If she works at least 20 days a month, how often does she have to drive in order to minimize her commuting time and keep within her monthly budget?
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