Problem 63
Question
The values of \(m\) for which the system of equations \(3 x+m y=m\) and \(2 x-5 y=20\) has a solution satisfying the condition \(x>0, y>0\), are (A) \(m \in\left(-\infty, \frac{-15}{2}\right) \cup(0, \infty)\) (B) \(m \in\left(-\infty, \frac{-15}{2}\right) \cup(30, \infty)\) (C) \(m \in\left(-\infty, \frac{-15}{2}\right) \cup(0,30)\) (D) None of these
Step-by-Step Solution
Verified Answer
Option (B): \(m \in\left(-\infty, \frac{-15}{2}\right) \cup(30, \infty)\).
1Step 1: Write the Equations
Write the given equations of the system in a standard form:1. \(3x + my = m\)2. \(2x - 5y = 20\)
2Step 2: Express One Variable in Terms of the Other
Solve the second equation for one variable. Let's solve for \(x\):\[ 2x = 5y + 20 \]\[ x = \frac{5y + 20}{2} \]
3Step 3: Substitute into the First Equation
Substitute the expression for \(x\) from Step 2 into the first equation \(3x + my = m\):\[ 3 \left( \frac{5y + 20}{2} \right) + my = m \]
4Step 4: Simplify and Solve for \(y\)
Simplify and solve for \(y\):\[ \frac{15y + 60}{2} + my = m \]Multiply through by 2 to clear the fraction:\[ 15y + 60 + 2my = 2m \]Combine the terms involving \(y\):\[ (15 + 2m)y = 2m - 60 \]\[ y = \frac{2m - 60}{15 + 2m} \]
5Step 5: Solve for Positive Conditions
To satisfy the condition \(y > 0\), analyze:- \(\frac{2m - 60}{15 + 2m} > 0\)The fraction will be positive if both the numerator and denominator are either both positive or both negative.
6Step 6: Analyze Numerator and Denominator Separately
Conditions for positivity:1. \(2m - 60 > 0\) implies \(m > 30\).2. \(15 + 2m > 0\) implies \(m > -\frac{15}{2}\).Conditions for negativity:1. \(2m - 60 < 0\) implies \(m < 30\).2. \(15 + 2m < 0\) implies \(m < -\frac{15}{2}\).
7Step 7: Combine Solutions to Inequalities
Combine the conditions to satisfy \(y > 0\):- \(m > 30\) (both numerator and denominator positive), or- \(m < -\frac{15}{2}\) (both numerator and denominator negative).Therefore, possible values for \(m\) are the union:- \(m \in \left( -\infty, -\frac{15}{2} \right) \cup (30, \infty)\).
8Step 8: Choose the Correct Option
The conditions derived for \(x > 0, y > 0\) correspond to option (B):- \(m \in\left(-\infty, \frac{-15}{2}\right) \cup(30, \infty)\).
Key Concepts
InequalitiesSolution conditionsAlgebraic manipulation
Inequalities
Inequalities tell us about the relative size or order of two values. In this exercise, inequalities are crucial in determining the values of \(m\) that result in positive solutions for both \(x\) and \(y\).
The analysis of inequalities starts with recognizing the expression \( \frac{2m - 60}{15 + 2m} > 0 \). We assess positivity by ensuring either both the numerator and denominator are positive or negative. This set of inequalities helps pinpoint the intervals for \(m\) that satisfy the equation with positive values of \(y\).
Understanding inequalities allows us to perform algebraic operations such as dividing or multiplying both sides by a positive number without altering the inequality. However, the inequality direction reverses if multiplied or divided by a negative number.
In this context, let's see the splitting of our conditions into positive and negative cases to ensure all resulting expressions abide by these inequality rules. Such mastery helps students skillfully navigate through the process of evaluating conditions in systems of equations.
The analysis of inequalities starts with recognizing the expression \( \frac{2m - 60}{15 + 2m} > 0 \). We assess positivity by ensuring either both the numerator and denominator are positive or negative. This set of inequalities helps pinpoint the intervals for \(m\) that satisfy the equation with positive values of \(y\).
Understanding inequalities allows us to perform algebraic operations such as dividing or multiplying both sides by a positive number without altering the inequality. However, the inequality direction reverses if multiplied or divided by a negative number.
In this context, let's see the splitting of our conditions into positive and negative cases to ensure all resulting expressions abide by these inequality rules. Such mastery helps students skillfully navigate through the process of evaluating conditions in systems of equations.
Solution conditions
Solution conditions refer to the criteria necessary for a system of equations to maintain a specified solution set. Here, we are looking for solutions in which both \(x > 0\) and \(y > 0\).
First, there is the condition \(y > 0\), which was obtained by solving \( y = \frac{2m - 60}{15 + 2m} \) and ensuring the fraction remained greater than zero. For this fraction, remember:\[\begin{align*}&\text{Numerator:} && 2m - 60 \&\text{Denominator:} && 15 + 2m \\end{align*}\]Careful analysis of these parts indicates the intervals where both numerator and denominator are either positive or both negative.
Additionally, if you derive the value of \(x\) from the expression \( x = \frac{5y + 20}{2} \), if \(y\) is positive and large enough, \(x\) will follow suit remaining positive due to the additive portion \(20\).
Crafting the ideal solution requires combining these criteria into unified intervals for \(m\), capturing ranges that fulfill all conditions concurrently.
First, there is the condition \(y > 0\), which was obtained by solving \( y = \frac{2m - 60}{15 + 2m} \) and ensuring the fraction remained greater than zero. For this fraction, remember:\[\begin{align*}&\text{Numerator:} && 2m - 60 \&\text{Denominator:} && 15 + 2m \\end{align*}\]Careful analysis of these parts indicates the intervals where both numerator and denominator are either positive or both negative.
Additionally, if you derive the value of \(x\) from the expression \( x = \frac{5y + 20}{2} \), if \(y\) is positive and large enough, \(x\) will follow suit remaining positive due to the additive portion \(20\).
Crafting the ideal solution requires combining these criteria into unified intervals for \(m\), capturing ranges that fulfill all conditions concurrently.
Algebraic manipulation
Algebraic manipulation is an essential skill for deriving meaningful expressions from equations. It involves rearranging, simplifying, or even transforming equations to uncover variable relationships.
In the original problem, the first major move is solving for \(x\) using the second equation\[2x - 5y = 20\]to find \( x = \frac{5y + 20}{2} \). This substitution step into the first equation \(3x + my = m\) illustrates the power of algebraic manipulation: transforming a system into a new form where variables can be isolated and conditions set.
What follows is substituting back to simplify the relationships between \(y\) and \(m\), allowing us to identify how \(m\) influences \(y\) and therefore the solution set indicators like positivity.
Moreover, manipulation requires observing simplifications like clearing fractions, as multiplying the entire equation by 2 facilitated cleaner operations without fractions.
Overall, these steps not only simplify solving processes but reveal deeper insights into the numerical dependencies that dictate which values of \(m\) fulfill the initial conditions for \(x\) and \(y\).
In the original problem, the first major move is solving for \(x\) using the second equation\[2x - 5y = 20\]to find \( x = \frac{5y + 20}{2} \). This substitution step into the first equation \(3x + my = m\) illustrates the power of algebraic manipulation: transforming a system into a new form where variables can be isolated and conditions set.
What follows is substituting back to simplify the relationships between \(y\) and \(m\), allowing us to identify how \(m\) influences \(y\) and therefore the solution set indicators like positivity.
Moreover, manipulation requires observing simplifications like clearing fractions, as multiplying the entire equation by 2 facilitated cleaner operations without fractions.
Overall, these steps not only simplify solving processes but reveal deeper insights into the numerical dependencies that dictate which values of \(m\) fulfill the initial conditions for \(x\) and \(y\).
Other exercises in this chapter
Problem 61
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View solution Problem 65
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