SATHEE: Matrices and Determinants 4 Question 30

Matrices and Determinants 4 Question 30

31. For what values of $m$ , does the system of equations $3 x + m y = m$ and $2 x - 5 y = 20$ has a solution satisfying the conditions $x > 0, y > 0$ ?

(1979, 3M )

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Answer:

Correct Answer: 31. (1)

Solution:

Given system of equations are

$3 x + m y = m and 2 x - 5 y = 20$

Here,

$\begin{aligned} Δ = | \begin{array}{cc} 3 & m \\ 2 & - 5 \end{array} | = - 15 - 2 m \\ Δ_{x} = | \begin{array}{cc} m & m \\ 20 & - 5 \end{array} | = - 25 m \\ Δ_{y} = | \begin{array}{cc} 3 & m \\ 2 & 20 \end{array} | = 60 - 2 m \end{aligned}$

If $Δ = 0$ , then system is inconsistent, i.e. it has no solution.

If $Δ \neq 0$ , i.e. $m \neq \frac{15}{2}$ , the system has a unique solution for any fixed value of $m$ .

We have, $x = \frac{Δ_{x}}{Δ} = \frac{- 25 m}{- 15 - 2 m} = \frac{25 m}{15 + 2 m}$

and

$y = \frac{Δ_{y}}{Δ} = \frac{60 - 2 m}{- 15 - 2 m} = \frac{2 m - 60}{15 + 2 m}$

For $x > 0, \frac{25 m}{15 + 2 m} > 0$

$\Rightarrow m > 0$

or $m < - \frac{15}{2}$

and $y > 0, \frac{2 m - 60}{2 m + 15} > 0 \Rightarrow m > 30$ or $m < - \frac{15}{2}$

From Eqs. (i) and (ii), we get $m < - \frac{15}{2}$ or $m > 30$

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अगला

Matrices and Determinants 4 Question 30

31. For what values of $m$ , does the system of equations $3 x + m y = m$ and $2 x - 5 y = 20$ has a solution satisfying the conditions $x > 0, y > 0$ ?

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Matrices and Determinants 4 Question 30

31. For what values of m, does the system of equations 3x+my=m and 2x−5y=20 has a solution satisfying the conditions x>0,y>0 ?

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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31. For what values of $m$ , does the system of equations $3 x + m y = m$ and $2 x - 5 y = 20$ has a solution satisfying the conditions $x > 0, y > 0$ ?