Problem 67
Question
Solve and graph the inequality. $$10(1-y)<-4(y-2)$$
Step-by-Step Solution
Verified Answer
The solution to the inequality is \(y > 1/3\). In the graph, this is represented by an open circle at \(1/3\) with an arrow pointing to the right, indicating all numbers greater than \(1/3\).
1Step 1: Distribute the terms
Start by distributing on both sides of the inequality: \(10 - 10y < -4y + 8\).
2Step 2: Group like terms
Bring like terms together, in this case the y terms on one side and the constant on the other. This is done by adding \(10y\) to both sides and subtracting \(8\) from both sides to give: \(6y > 2\).
3Step 3: Solve for y
Now, we need to isolate y, which we can do through dividing by 6 on both sides. However, remember to reverse the inequality because we are dividing by a positive number: \(y > 2/6\). Simplifying gives \(y > 1/3\).
4Step 4: Graph the solution
On a number line, mark the point \(1/3\). Draw an arrow pointing to the right from \(1/3\) to indicate all the numbers greater than \(1/3\). Since \(y\) is strictly greater than \(1/3\) but does not equal \(1/3\), the point \(1/3\) will remain an open circle.
Key Concepts
Solving InequalitiesGraphing InequalitiesLinear Inequalities
Solving Inequalities
Solving inequalities is an important aspect of understanding algebra. Inequalities are similar to equations but involve a range of solutions rather than a single solution. In this exercise, we began with \(10(1-y) < -4(y-2)\). The goal was to find the range of values for which the inequality holds true.
We started by expanding both sides: distributing \(10\) and \(-4\) across their respective terms, resulting in \(10 - 10y < -4y + 8\). Solving inequalities involves similar steps to solving equations, like combining like terms and isolating the variable.
Next, we rearranged the inequality to group variables on one side and constants on the other, yielding \(6y > 2\). This step requires careful attention to signs, as moving terms across the inequality could involve reversing the sign if negative numbers are involved, but in this case, it wasn’t necessary.
Finally, we solved for \(y\) by dividing both sides by 6, leading to \(y > \frac{1}{3}\). This shows that any number greater than \(\frac{1}{3}\) satisfies the inequality, therefore, the solution set is all real numbers greater than \(\frac{1}{3}\).
We started by expanding both sides: distributing \(10\) and \(-4\) across their respective terms, resulting in \(10 - 10y < -4y + 8\). Solving inequalities involves similar steps to solving equations, like combining like terms and isolating the variable.
Next, we rearranged the inequality to group variables on one side and constants on the other, yielding \(6y > 2\). This step requires careful attention to signs, as moving terms across the inequality could involve reversing the sign if negative numbers are involved, but in this case, it wasn’t necessary.
Finally, we solved for \(y\) by dividing both sides by 6, leading to \(y > \frac{1}{3}\). This shows that any number greater than \(\frac{1}{3}\) satisfies the inequality, therefore, the solution set is all real numbers greater than \(\frac{1}{3}\).
Graphing Inequalities
Graphing helps visualize the solution of inequalities, especially when dealing with a range of values. Once we determined that \(y > \frac{1}{3}\), we could express this visually on a number line.
Begin by marking the critical value \(\frac{1}{3}\) on the number line. Because the inequality is strict (meaning \(y\) cannot equal \(\frac{1}{3}\)), we use an open circle to indicate that this endpoint is not included.
To complete the graph, draw an arrow going to the right from \(\frac{1}{3}\). This arrow signifies that all numbers greater than \(\frac{1}{3}\) are solutions to the inequality. Graphing in this way provides a clear, visual representation of the infinite solutions beyond just symbolic expression.
Begin by marking the critical value \(\frac{1}{3}\) on the number line. Because the inequality is strict (meaning \(y\) cannot equal \(\frac{1}{3}\)), we use an open circle to indicate that this endpoint is not included.
To complete the graph, draw an arrow going to the right from \(\frac{1}{3}\). This arrow signifies that all numbers greater than \(\frac{1}{3}\) are solutions to the inequality. Graphing in this way provides a clear, visual representation of the infinite solutions beyond just symbolic expression.
Linear Inequalities
Linear inequalities, like linear equations, represent the relationships between variables that correspond to linear expressions. The only difference is that these relationships describe a range or region rather than a single line.
In our problem, \(10(1-y) < -4(y-2)\), we initially transformed it into \(10 - 10y < -4y + 8\), showing how these inequalities involve linear expressions. Linear inequalities are characterized by having no exponent on the variable higher than one.
Key aspects to remember include that adding or subtracting the same number from both sides doesn't change the inequality. Similarly, multiplying or dividing by a positive number keeps the inequality direction unchanged. However, a crucial rule with inequalities is that if you multiply or divide both sides by a negative number, the inequality sign must be flipped. This exercise provides insight into how simple manipulations can lead to understanding the range of solutions in linear contexts.
In our problem, \(10(1-y) < -4(y-2)\), we initially transformed it into \(10 - 10y < -4y + 8\), showing how these inequalities involve linear expressions. Linear inequalities are characterized by having no exponent on the variable higher than one.
Key aspects to remember include that adding or subtracting the same number from both sides doesn't change the inequality. Similarly, multiplying or dividing by a positive number keeps the inequality direction unchanged. However, a crucial rule with inequalities is that if you multiply or divide both sides by a negative number, the inequality sign must be flipped. This exercise provides insight into how simple manipulations can lead to understanding the range of solutions in linear contexts.
Other exercises in this chapter
Problem 66
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