Chapter 5: Problem 282
In the following exercises, determine whether each ordered pair is a solution to the system. \(\left\\{\begin{array}{l}5 x-3 y<-2 \\ 10 x+6 y>4\end{array}\right.\) (a) \(\left(\frac{1}{5}, \frac{2}{3}\right)\) (b) \(\left(-\frac{3}{10}, \frac{7}{6}\right)\)
Short Answer
Expert verified
Neither (1/5, 2/3) nor (-3/10, 7/6) are solutions to the system.
Step by step solution
01
- Interpret the System of Inequalities
Given the system: 1) 5x - 3y < -2 2) 10x + 6y > 4
02
- Test (1/5, 2/3) for the first inequality
Substitute x = 1/5 and y = 2/3 into the first inequality. 5(1/5) - 3(2/3) < -2 Simplify to see if it is true: 1 - 2 < -2 -1 < -2 (False)
03
- Test (1/5, 2/3) for the second inequality
Substitute x = 1/5 and y = 2/3 into the second inequality. 10(1/5) + 6(2/3) > 4 Simplify to see if it is true: 2 + 4 > 4 6 > 4 (True)
04
- Conclusion for (1/5, 2/3)
Since the point (1/5, 2/3) does not satisfy both inequalities, it is not a solution to the system.
05
- Test (-3/10, 7/6) for the first inequality
Substitute x = -3/10 and y = 7/6 into the first inequality. 5(-3/10) - 3(7/6) < -2 Simplify to see if it is true: (-3/2) - (7/2) < -2 -10/2 < -2 -5 < -2 (True)
06
- Test (-3/10, 7/6) for the second inequality
Substitute x = -3/10 and y = 7/6 into the second inequality. 10(-3/10) + 6(7/6) > 4 Simplify to see if it is true: -3 + 7 > 4 4 > 4 (False)
07
- Conclusion for (-3/10, 7/6)
Since the point (-3/10, 7/6) does not satisfy both inequalities, it is not a solution to the system.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Inequalities
Graphing inequalities on a coordinate plane helps visualize solutions for a set of equations. Each inequality divides the plane into two regions: one that satisfies the inequality and one that doesn't. To graph an inequality, follow these steps:
- Start by graphing the inequality as if it were an equation (use a solid line for \(≤\) or \(≥\) inequalities and a dashed line for \(<\) or \(>\) inequalities).
- Pick a test point (not on the boundary line) and substitute its coordinates into the inequality to see if it makes the inequality true.
- Shade the region where the inequality holds true.
For example, to graph \(5x - 3y < -2\), graph the line \(5x - 3y = -2\) using a dashed line, then shade below the line if the test point satisfies the inequality.
- Start by graphing the inequality as if it were an equation (use a solid line for \(≤\) or \(≥\) inequalities and a dashed line for \(<\) or \(>\) inequalities).
- Pick a test point (not on the boundary line) and substitute its coordinates into the inequality to see if it makes the inequality true.
- Shade the region where the inequality holds true.
For example, to graph \(5x - 3y < -2\), graph the line \(5x - 3y = -2\) using a dashed line, then shade below the line if the test point satisfies the inequality.
Solution to System of Inequalities
A solution to a system of inequalities is an ordered pair \(x, y\) that satisfies all inequalities in the system simultaneously. To determine if an ordered pair is a solution:
- Substitute the coordinates of the ordered pair into each inequality.
- Simplify the inequalities. If the pair satisfies each inequality, it is a solution.
For instance, consider the system:
\(5x - 3y < -2\)
and \(10x + 6y > 4\).
Testing the point \((1/5, 2/3)\):
- Substitute into \(5x - 3y < -2\), we get \(-1 < -2\) (False).
- Substitute into \(10x + 6y > 4\), we get \(6 > 4\) (True).
Because the point does not satisfy the first inequality, it is not a solution.
- Substitute the coordinates of the ordered pair into each inequality.
- Simplify the inequalities. If the pair satisfies each inequality, it is a solution.
For instance, consider the system:
\(5x - 3y < -2\)
and \(10x + 6y > 4\).
Testing the point \((1/5, 2/3)\):
- Substitute into \(5x - 3y < -2\), we get \(-1 < -2\) (False).
- Substitute into \(10x + 6y > 4\), we get \(6 > 4\) (True).
Because the point does not satisfy the first inequality, it is not a solution.
Substitution Method
The substitution method is a way to determine if an ordered pair is a solution to a system of inequalities. It involves substituting the values of \(x\) and \(y\) into each inequality and simplifying to check their truth values. Steps to follow:
- Choose an ordered pair to test.
- Substitute the values of \(x\) and \(y\) from the pair into each inequality.
- Simplify each inequality to see if it holds true.
For the ordered pair \(\left(\frac{1}{5}, \frac{2}{3}\right)\), in the system of inequalities:
- For the first inequality \(5x - 3y < -2\), substitute \(x = \frac{1}{5}\) and \(y = \frac{2}{3}\), then simplify: \(1 - 2 < -2\). This is false, so the pair doesn't satisfy this inequality.
- Choose an ordered pair to test.
- Substitute the values of \(x\) and \(y\) from the pair into each inequality.
- Simplify each inequality to see if it holds true.
For the ordered pair \(\left(\frac{1}{5}, \frac{2}{3}\right)\), in the system of inequalities:
- For the first inequality \(5x - 3y < -2\), substitute \(x = \frac{1}{5}\) and \(y = \frac{2}{3}\), then simplify: \(1 - 2 < -2\). This is false, so the pair doesn't satisfy this inequality.
Ordered Pairs
Ordered pairs \(x, y\) represent points on a coordinate plane. They are essential in checking solutions for systems of inequalities. Here's how they work:
- The first number in the pair is the \(x\)-coordinate, representing the horizontal position.
- The second number is the \(y\)-coordinate, representing the vertical position.
For example, to determine if \(\left(-\frac{3}{10}, \frac{7}{6}\right)\) is a solution, substitute:
- Into \(5x - 3y < -2\), giving \( -5 < -2\) (True).
- Into \(10x + 6y > 4\), giving \(4 > 4\) (False).
Since the pair does not satisfy both inequalities simultaneously, it is not a solution.
- The first number in the pair is the \(x\)-coordinate, representing the horizontal position.
- The second number is the \(y\)-coordinate, representing the vertical position.
For example, to determine if \(\left(-\frac{3}{10}, \frac{7}{6}\right)\) is a solution, substitute:
- Into \(5x - 3y < -2\), giving \( -5 < -2\) (True).
- Into \(10x + 6y > 4\), giving \(4 > 4\) (False).
Since the pair does not satisfy both inequalities simultaneously, it is not a solution.