Ex.3.4 Q1 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

Solve the following pair of linear equations by the elimination method and the substitution method:

(i) \(x + y = 5\) and \(2x-3y = 4\)

(ii) \(3x + 4y = 10\) and \(2x-2y = 2\)

(iii) \(3x-5y-4 = 0\) and \(9x = 2y + 7\)

(iv) \(\begin{align}\frac{x}{2} + \frac{{2y}}{3} =  - 1\end{align}\) and \( \begin{align}x - \frac{y}{3} = 3 \end{align}\)

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.4 | Question 1

Text Solution

What is the Unknown?

Solution for the linear pair of equations.

Reasoning:

Substitution method:

Pick either of the equations and write one variable in terms of the other then substitute the value of the obtained variable in other equation to solve.

Elimination method:

First multiply one or both the equations by some suitable non-zero constants to make the coefficients of one variable numerically equal then add or subtract one equation from the other so that one variable gets eliminated

(i) What is the Known?

\(x + y = 5 \\2x - 3 y= 4 \)

Steps:

Elimination method:

\[\begin{align}x + y = 5 \qquad \ldots \left( 1 \right)\\2x - 3 y= 4 \qquad \ldots \left( 2 \right)\end{align}\]

Multiplying equation \((1)\) by \(2\)

\[\begin{array}{l}
\left[ {x + {\rm{ }}y = 5} \right] \times 2\\
2x + 2y = 10 \qquad  \ldots \left( 3 \right)
\end{array}\]

By subtracting equation \((2)\) from equation \((3)\)

\[\begin{align}\left( {2x + 2y} \right) - \left( {2x - 3y} \right)& = 10 - 4\\2x + 2y - 2x + 3y &= 6\\
5y &= 6\\y &= \frac{6}{5}\end{align}\]

Substituteing \(\begin{align}y = \frac {{6}}{5}\end{align}\) in equation (1)

\[\begin{align}x + \frac{6}{5} &= 5\\x &= 5 - \frac{6}{5}\\x &= \frac{{25 - 6}}{5}\\x &= \frac{{19}}{5}\end{align}\]

Substitution method:

\[\begin{align}x + y &= 5 \qquad \ldots \left( 1 \right)\\2x - 3 y&= 4 \qquad \ldots \left( 2 \right)\end{align}\]

By solving equation \((1)\)

\[\begin{align}x + {\rm{ }}y &= 5\\y &= 5 - x \qquad \ldots \left( 3 \right)\end{align}\]

Substitute \(\begin{align}y = 5 - x\end{align}\) in equation \((2)\)

\[\begin{align}2x - 3(5 - x) &= 4\\2x - 15 + 3x &= 4\\5x &= 4 + 15\\x &= \frac{{19}}{5}\end{align}\]

Substituting \(\begin{align}x = \frac{{19}}{5}\end{align}\) in equation \((3)\)

\[\begin{align}y &= 5 - \frac{{19}}{5}\\y &= \frac{{25 - 19}}{5}\\y &= \frac{6}{5}\end{align}\]

The answer is,\(\begin{align}x = \frac{{19}}{5} \quad \text{and} \quad y = \frac{6}{5}\end{align}\)

(ii) What is the Known?

\(\begin{align}3x+4y&=10\\2x-2y&=2\end{align}\)

Steps:

Elimination method:

\[\begin{align}3x+4y&=10…………(1)\\2x-2y&=2…………. (2)\end{align}\]

Multiplying equation \((2)\) by \(2\)

\[\begin{align}&\left[ {2x - 2y = 2} \right] \times 2\\&4x - 4y = 4 \qquad \ldots \left( 3 \right)\end{align}\]

By adding equation \((1)\) and equation \((3)\)

\[\begin{align}\left( {3x + 4y} \right) + \left( {4x - 4y} \right) &= 10 + {4^{}}\\3x + 4y + 4x - 4y &= 14\\
7x &= 14\\x &= \frac{{14}}{7}\\x &= 2\end{align}\]

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

\[\begin{align}2  x\, 2 - 2y &= 2\\4 - 2y &= 2\\2y &= 4 - 2\\y &= \frac{2}{2}\\y &= 1\end{align}\]

Substitution method:

\[\begin{align}3x+4y&=10 \quad \dots(1)\\2x-2y&=2\quad \;\;\dots(2)\end{align}\]

By solving equation \((1)\)

\[\begin{align}3x + 4y & = 10\\4y &= 10 - 3x\\y &= \frac{{10 - 3x}}{4} \qquad \ldots \left( 3 \right)
\end{align}\]

Substituteing \(\begin{align}y = \frac{{10 - 3x}}{4}\end{align}\) in equation \((2)\)

\[\begin{align}2x - 2\left( {\frac{{10 - 3x}}{4}} \right) &= 2\\\frac{{4x - 10 + 3x}}{2} &= 2\\4x - 10 + 3x &= 4\\7x &= 4 + 10\\x &= \frac{{14}}{7}\\x &= 2\end{align}\]

Substituting \(\begin{align}x = 2\end{align}\) in equation (3)

\[\begin{align}y &= \frac{{10 - 3 \times 2}}{4}\\y &= \frac{{10 - 6}}{4}\\y &= \frac{4}{4}\\y &= 1\end{align}\]

The answer is,\(\begin{align}x = 2 \quad \text{and} \quad y = 1\end{align}\)

(iii) What is the Known?

\(\begin{align}3x - 5y - 4 &= 0\\9x &= 2y + 7\end{align}\)

Steps:

Elimination method:

\[\begin{align}3x - 5y - 4 &= 0 \qquad  \qquad \ldots \left( 1 \right)\\9x &= 2y + 7 \qquad \ldots \left( 2 \right)\end{align}\]

Multiplying equation \((1)\) by \(3\)

\[\begin{align}&\left[ {3x - 5y - 4 = 0} \right] \times 3\\&9x - 15y - 12 = 0 \qquad  \ldots \left( 3 \right)
\end{align}\]

By solving equation \((2)\)

\[9x - 2y - 7 = 0 \qquad  \ldots \left( 4 \right)\]

By subtracting equation \((4)\) from equation \((3)\)

\[\begin{align}\left( {9x - 15y - 12} \right) - \left( {9x - 2y - 7} \right) &= 0\\9x - 15y - 12 - 9x + 2y + 7 &= 0\\ - 13y - 5 &= 0\\ - 13y &= 5\\y &= - \frac{5}{{13}}\end{align}\]

Substituteting \(\begin{align} y = {-\frac {{5}}{13}}\end{align}\) in equation \((2)\)

\[\begin{align}9x &= 2\left( { - \frac{5}{{13}}} \right) + 7\\9x &= \frac{{ - 10 + 91}}{{13}}\\
x &= \frac{{81}}{{13}} \times \frac{1}{9}\\x& = \frac{9}{{13}}\end{align}\]

Substitution method:

\[\begin{align}&3x - 5y - 4 = 0 \qquad \ldots \left( 1 \right)\\&9x = 2y + 7 \qquad \qquad \ldots \left( 2 \right)\end{align}\]

By solving equation \((1)\)

\[\begin{align}3x - 5y - 4 &= 0\\5y &= 3x - 4\\y &= \frac{{3x - 4}}{5} \qquad \ldots \left( 3 \right)\end{align}\]

Substituteing \(\begin{align}y = \frac {{3x - 4}}{5} \end{align}\) in equation \((2)\)

\[\begin{align}9x &= 2\left( {\frac{{3x - 4}}{5}} \right) + 7\\9x &= \frac{{6x - 8 + 35}}{5}\\
45x &= 6x + 27\\45x - 6x &= 27\\39x &= 27\\x &= \frac{{27}}{{39}}\\x& = \frac{9}{{13}}\end{align}\]

Substitute \(\begin{align} x = \frac {{9}}{13}\end{align}\) in equation\( (3)\)

\[\begin{align}y &=\frac{{3\left( {\frac{9}{{13}}} \right) - 4}}{5}\\y& = \left( {\frac{{27 - 52}}{{13}}} \right) \times \frac{1}{5}\\y &= - \frac{{25}}{{13}} \times \frac{1}{5}\\y &= - \frac{5}{{13}}\end{align}\]

The answer is,\(\begin{align}y = \frac{{ - 5}}{{13}} \quad \text{and} \quad x = \frac{9}{{13}}\end{align}\)

(iv) What is the Known?

\(\begin{align}\frac{x}{2} + \frac{{2y}}{3} &= - 1\\x - \frac{y}{3} &= 3\end{align}\)

Steps:

Elimination method:

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

Multiplying equation \((1)\) by \(6\) and equation \((2)\) by \(3\)

\[\begin{align}\left[ {\frac{x}{2} + \frac{{2y}}{3} =  - 1} \right] \times 6\\
3x + 4y =  - 6& \qquad \dots (3)\\\\\left[ {x - \frac{y}{3} = 3} \right] \times 3\\3x - y = 9 &\qquad \dots(4)
\end{align}\]

By subtracting equation \((4)\) from equation \((3)\)

\[\begin{align}\left( {3x + 4y} \right) - \left( {3x - y} \right) &= - 6 - 9\\3x + 4y - 3x + y &= - 15\\5y &= - 15\\y &= - \frac{{15}}{5}\\y &= - 3\end{align}\]

Substitute \(y = - 3\) in equation (2)

\[\begin{align}x - \frac{{ - 3}}{3} &= 3\\x + 1 &= 3\\x &= 3 - 1\\x &= 2\end{align}\]

Substitution method:

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

By solving equation \((2)\)

\[\begin{align}x - \frac{y}{3} &= 3\\x &= \frac{y}{3} + 3\\x &= \frac{{y + 9}}{3} \qquad \dots(3)\end{align}\]

Substituteing \(\begin{align}x = \frac {{y+9}}{3} \end{align}\) in equation (1)

\[\begin{align}\frac{1}{2}\left( {\frac{{y + 9}}{3}} \right) + \frac{{2y}}{3} &= - 1\\\frac{{y + 9 + 4y}}{6} &= - 1\\5y + 9 &= - 6\\5y &= - 6 - 9\\y &= \frac{{ - 15}}{5}\\y &= - 3\end{align}\]

Substituiting \(y = -3\) in equation (3)

\[\begin{align}x &= \frac{{ - 3 + 9}}{3}\\x &= \frac{6}{3}\\x &= 2\end{align}\]

The answer is,\(\begin{align}x = 2 \quad \text{and} \quad y = - 3\end{align}\)

 

  
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