Ex.3.3 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

Solve \(2x + 3y = 11\) and \(2x-4y = -24,\) hence find the value of \(m\) for which \(y = mx + 3.\)

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.3 | Question 2

Text Solution

Reasoning:

Solve the linear equations (1) and (2) by substitution method and substitute the values of \(x\) and \(y\) in \(y = mx + 3\) to get the value of \(m.\)

What is Known?

\[\begin{align}2x + 3y &= 11\\2x - 4y &= - 24\\y &= mx + 3\end{align}\]

What is Unknown?

Value of \(m\)

Steps:

\[\begin{align}2x + 3y &= 11 \qquad\;\;\; \dots(1)\\\\2x - 4y &= - 24 \qquad \dots(2)\end{align}\]

By solving the equation (1)

\[\begin{align}2x + 3y &= 11\\3y &= 11 - 2x\\y &= \frac{{11 - 2x}}{3} \qquad \dots(3)\end{align}\]

Substituting \(\begin{align} y = \frac{{11 - 2x}}{3}\end{align}\) in equation (2), we get

\[\begin{align}2x - 4\left( {\frac{{11 - 2x}}{3}} \right) &= - 24\\\frac{{6x - 44 + 8x}}{3} &= - 24\\14x - 44 &= - 72\\14x &= 44 - 72\\x &= - \frac{{28}}{{14}}\\x &= - 2\end{align}\]

Substituting \(x = - 2\) in equation (3)

\[\begin{align}y &= \frac{{11 - 2 \times \left( { - 2} \right)}}{3}\\y& = \frac{{11 + 4}}{3}\\y &= \frac{{15}}{3}\\y &= 5\end{align}\]

Now, Substituting \(x = - 2\) and \(y = 5\) in \(y = mx + 3\)

\[\begin{align}y &= mx + 3\\5 &= m( - 2) + 3\\5 - 3 &= - 2m\\2 &= - 2m\\m &= \frac{2}{{ - 2}}\\m &= - 1\end{align}\]

The answer is

\[\begin{align}x &= - 2\\y &= 5\\m &= - 1\end{align}\]

  
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