JEE Advanced 1980

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: For what values of $m$ , does the system of equations $3 x + m y = m; 2 x - 5 y = 20$ has solution satisfying the conditions $x > 0, y > 0$ .

Q: Why do we take the intersection of the two intervals at the end?

The problem demands that the solution $(x, y)$ satisfies $x > 0$ AND $y > 0$ simultaneously. If a value of $m$ makes $x$ positive but $y$ negative, the intersection point falls into the fourth quadrant, which violates our condition. We need the 'sweet spot' where both conditions are true.

Q: What happens if the denominator $15 + 2m$ equals zero?

If $15 + 2m = 0$, the lines are parallel or coincident, and the system does not have a unique intersection point in the finite plane. Since we are looking for a specific point $(x, y)$ in the first quadrant, we must exclude $m = -\frac{15}{2}$ from our domain.

Visualized Solution

Visualizing the System

Given system of equations:
1) $3 x + m y = m$
2) $2 x - 5 y = 20$
We require the solution $(x, y)$ to satisfy $x > 0$ and $y > 0$ .
This means the intersection point must lie strictly in the First Quadrant.

Expressing $x$ in terms of $y$

From equation (2): $2 x - 5 y = 20$
Rearranging to isolate $x$ :
$2 x = 20 + 5 y$
$x = 10 + \frac{5}{2} y$

Substituting into Equation 1

Substitute $x = 10 + \frac{5}{2} y$ into equation (1):
$3 (10 + \frac{5}{2} y) + m y = m$

Solving for $y$

Expand the equation:
$30 + \frac{15}{2} y + m y = m$
Group the $y$ terms:
$y (\frac{15}{2} + m) = m - 30$
$y (\frac{15 + 2 m}{2}) = m - 30$
Solve for $y$ :
$y = \frac{2 ( m - 30 )}{15 + 2 m}$

Solving for $x$

Substitute $y$ back into $x = 10 + \frac{5}{2} y$ :
$x = 10 + \frac{5}{2} \cdot \frac{2 ( m - 30 )}{15 + 2 m}$
$x = 10 + \frac{5 ( m - 30 )}{15 + 2 m}$
Simplify the expression:
$x = \frac{10 ( 15 + 2 m ) + 5 m - 150}{15 + 2 m}$
$x = \frac{25 m}{15 + 2 m}$

Analyzing the Condition $x > 0$

We require $x > 0$ :
$\frac{25 m}{15 + 2 m} > 0$
Identify the critical points:
Numerator: $m = 0$
Denominator: $15 + 2 m = 0 ⟹ m = - \frac{15}{2}$

Solving the Inequality for $x$

Using the Wavy Curve method for $\frac{m}{m + 7.5} > 0$ :
The expression is positive in the outer intervals.
Interval for $x > 0$ :
$m \in (- \infty, - \frac{15}{2}) \cup (0, \infty)$

Analyzing the Condition $y > 0$

We require $y > 0$ :
$\frac{2 ( m - 30 )}{15 + 2 m} > 0$
Identify the critical points:
Numerator: $m = 30$
Denominator: $m = - \frac{15}{2}$

Solving the Inequality for $y$

Using the Wavy Curve method for $\frac{m - 30}{m + 7.5} > 0$ :
The expression is positive in the outer intervals.
Interval for $y > 0$ :
$m \in (- \infty, - \frac{15}{2}) \cup (30, \infty)$

Finding the Final Range of $m$

We must find the intersection of both intervals:
1) $m \in (- \infty, - \frac{15}{2}) \cup (0, \infty)$
2) $m \in (- \infty, - \frac{15}{2}) \cup (30, \infty)$
Intersection:
$m \in (- \infty, - \frac{15}{2}) \cup (30, \infty)$

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The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Solution Diagram

Analyzing the Setup

We are given two lines in the Cartesian plane: 1.

2 x - 5 y = 20

3 x + m y = m

Our objective is to determine the values of the parameter

m

such that the intersection point

(x, y)

lies strictly in the first quadrant, meaning

x > 0

and

y > 0

The Algebraic Bridge

To find the intersection point, we first isolate

x

from the first equation:

x = 10 + \frac{5}{2} y

Substituting this expression for

x

into the second equation,

3 x + m y = m

, we get:

3 (10 + \frac{5}{2} y) + m y = m

Expanding and grouping the terms involving

y

30 + \frac{15}{2} y + m y = m

y (\frac{15}{2} + m) = m - 30

Solving for

y

, we obtain the vertical coordinate:

y = \frac{2 ( m - 30 )}{15 + 2 m}

Substituting this back into our expression for

x

, we find the horizontal coordinate:

x = \frac{25 m}{15 + 2 m}

Taming the Inequalities

For the intersection point to lie in the first quadrant, we must satisfy the conditions

x > 0

and

y > 0

. This leads to two rational inequalities:

1. For

x > 0

\frac{25 m}{15 + 2 m} > 0

The critical points are

m = 0

and

m = - \frac{15}{2}

. Using the wavy curve method, we find

x > 0

when

m \in (- \infty, - \frac{15}{2}) \cup (0, \infty)

2. For

y > 0

\frac{2 ( m - 30 )}{15 + 2 m} > 0

The critical points are

m = 30

and

m = - \frac{15}{2}

. Using the wavy curve method, we find

y > 0

when

m \in (- \infty, - \frac{15}{2}) \cup (30, \infty)

The Final Intersection

To satisfy both conditions simultaneously, we must find the intersection of the two solution sets:

(- \infty, - \frac{15}{2}) \cup (0, \infty) AND (- \infty, - \frac{15}{2}) \cup (30, \infty)

The common region for both inequalities is:

m \in (- \infty, - \frac{15}{2}) \cup (30, \infty)

Thus, the values of

m

that force the intersection point into the first quadrant are $m \in (- \infty, - 7.5) \cup (30, \infty)$ .

Common Traps & FAQs

Why can't I just cross-multiply the denominator to solve the inequality

\frac{25 m}{15 + 2 m} > 0

That is a classic trap! You cannot cross-multiply because you do not know the sign of the denominator

(15 + 2 m)

. If it is negative, the inequality sign would flip. Always use the Wavy Curve method for rational inequalities to account for all possible signs of the denominator.

Why do we take the intersection of the two intervals at the end?

The problem demands that the solution

(x, y)

satisfies

x > 0

AND

y > 0

simultaneously. If a value of

m

makes

x

positive but

y

negative, the intersection point falls into the fourth quadrant, which violates our condition. We need the 'sweet spot' where both conditions are true.

What happens if the denominator

15 + 2 m

equals zero?

15 + 2 m = 0

, the lines are parallel or coincident, and the system does not have a unique intersection point in the finite plane. Since we are looking for a specific point

(x, y)

in the first quadrant, we must exclude

m = - \frac{15}{2}

from our domain.

Animated Solution for Mathematics - Matrices and Determinants: For what values of $m$ , does the system of equations $3 x + m y = m; 2 x - 5 y = 20$ has solution satisfying the conditions $x > 0, y > 0$ .

Visualized Solution

Visualizing the System

Expressing $x$ in terms of $y$

Substituting into Equation 1

Solving for $y$

Solving for $x$

Analyzing the Condition $x > 0$

Solving the Inequality for $x$

Analyzing the Condition $y > 0$

Solving the Inequality for $y$

Finding the Final Range of $m$

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Algebraic Bridge

Taming the Inequalities

The Final Intersection

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Visualized Solution

Visualizing the System

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Substituting into Equation 1

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Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Analyzing the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>0

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Finding the Final Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Algebraic Bridge

Taming the Inequalities

The Final Intersection

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