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For what values of m does the system of equations $3x + my = m$ and $2x - 5y = 20$ have the solution satisfying the conditions $x > 0,y > 0$

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Last updated date: 17th Jun 2024
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Answer
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Hint: To solve this question, we will use the concept of a pair of linear equations in two variables. We will use the method of elimination by substitution in this system of equations to find out the values of m. We will also use some inequality rules.

Complete step-by-step answer:
Given that,
$ \Rightarrow 3x + my = m$ …… (i)
$ \Rightarrow 2x - 5y = 20$ …… (ii)
From equation (ii),
$ \Rightarrow 2x = 20 + 5y$
$ \Rightarrow x = \dfrac{{20 + 5y}}{2}$ …… (iii)
Putting the value of x from equation (iii) in equation (i), we will get
$ \Rightarrow 3\left( {\dfrac{{20 + 5y}}{2}} \right) + my = m$
$ \Rightarrow \dfrac{{60 + 15y}}{2} + my = m$
Taking 2 as L.C.M,
$ \Rightarrow \dfrac{{60 + 15y + 2my}}{2} = m$
Now, shifting 2 from L.H.S to R.H.S,
$ \Rightarrow 60 + 15y + 2my = 2m$
$ \Rightarrow 15y + 2my = 2m - 60$
Separating the equation to eliminate y, we will get
$ \Rightarrow y\left( {15 + 2m} \right) = 2m - 60$
$ \Rightarrow y = \dfrac{{2m - 60}}{{15 + 2m}}$ ……. (iv)
Here we get, $x = \dfrac{{20 + 5y}}{2}$ and $y = \dfrac{{2m - 60}}{{15 + 2m}}$
Now, according to the question
$ \Rightarrow x > 0$ …. (v)
$ \Rightarrow y > 0$ ….. (vi)
Putting the value of y in equation (vi), we will get
$ \Rightarrow \dfrac{{2m - 60}}{{15 + 2m}} > 0$
Multiplying and dividing both sides by 15 + 2m,
$ \Rightarrow \dfrac{{\left( {2m - 60} \right)\left( {15 + 2m} \right)}}{{{{\left( {15 + 2m} \right)}^2}}} > 0$
This can be written as:
$ \Rightarrow \left( {2m - 60} \right)\left( {15 + 2m} \right) > 0$
Taking 2 common from both brackets,
$ \Rightarrow 4\left( {m - 30} \right)\left( {\dfrac{{15}}{2} + m} \right) > 0$
From this, we can say that
$ \Rightarrow m - 30 > 0$ or ……. (vii)
$ \Rightarrow \dfrac{{15}}{2} + m > 0$ …… (viii)
Adding 30 both sides in equation (vii),
$ \Rightarrow m - 30 + 30 > 0 + 30$
$ \Rightarrow m > 30$
As we know that when both sides are divided or multiplied by a negative number then the inequality gets reversed.
So, the equation (viii) will become,
$ \Rightarrow m < \dfrac{{ - 15}}{2}$
Hence, the values of m form which it satisfies the given system of equations are $m > 30$ and $m < \dfrac{{ - 15}}{2}$

Note: Whenever we ask such types of questions, we have to remember the substitution method. In this method, one of the variables in terms of another variable from either of the two equations and then this expression is put in another equation to obtain an equation in one variable.