# Ex.3.3 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 substitution method.

(i)

\begin{align}x + y &= 14\\x-y &= 4\end{align}

(ii)

\begin{align}s-t &= 3\\\frac{s}{3} + \frac{t}{2} &= 6\end{align}

(iii)

\begin{align}x-y = 3\\9x-3y = 9\end{align}

(iv)

\begin{align}0.2x + 0.3y& = 1.3\\0.4x + 0.5y &= 2.3\end{align}

(v)

\begin{align}\sqrt 2 x + \sqrt 3 y = 0\\\sqrt 3 x - \sqrt {8y} = 0\end{align}

(vi)

\begin{align}\frac{{3x}}{2} - \frac{{5y}}{3} &= - 2\\\frac{x}{3} + \frac{y}{2} &= \frac{{13}}{6} \end{align}

Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.3 | Question 1

## Text Solution

What is Known?

Pair of linear equations.

What is Unknown?

Solution for the given pair of linear equations.

Reasoning:

Pick any one of two equations, write one variable in terms of other. Now substituting this in other equation will result in one variable equation and easy to solve.

(i) Steps:

\begin{align}x + y &= 14 \qquad \dots(1)\\x - y &= 4 \qquad \;\;\dots(2)\end{align}

By solving the equation (1)

$y = 14 - x \qquad \dots(3)$

Substitute $$y = 14 - x$$ in equation (2), we get

\begin{align}x - \left( {14 - x} \right) &= 4\\2x - 14 &= 4\\2x &= 4 + 14\\2x &= 18\\x &= 9\end{align}

Substituting $$x = 9$$ in equation (3), we get

\begin{align}y &= 14 - 9\\y &= 5\end{align}

\begin{align}x &= 9\\y &= 5\end{align}

(ii) Steps:

\begin{align}s - t &= 3 \qquad \dots(1)\\\frac{s}{3} + \frac{t}{2} &= 6 \qquad \dots(2)\end{align}

By solving the equation (1)

\begin{align}s - t &= 3\\s &= 3 + t \qquad \dots(3)\end{align}

Substitute $$s = 3 + t$$ in equation (2), we get

\begin{align}\frac{{3 + t}}{3} + \frac{t}{2} &= 6\\\frac{{6 + 2t + 3t}}{6} &= 6\\6 + 5t &= 6 \times 6\\5t &= 36 - 6\\t &= \frac{{30}}{5}\\t &= 6\end{align}

Substituting $$t = 6$$ in equation (3), we get

\begin{align}s &= 3 + 6\\s &= 9\end{align}

\begin{align}s = 9\\ t= 6\end{align}

(iii) Steps:

\begin{align}3x - y &= 3 \qquad \dots(1)\\9x - 3y &= 9 \qquad \dots(2)\end{align}

By solving the equation (1)

\begin{align}3x - y &= 3\\y &= 3x - 3 \qquad \dots(1)\end{align}

Substitute $$y = 3x - 3$$ in equation (2), we get

\begin{align}9x - 3(3x - 3) &= 9\\9x - 9x + 9 &= 9\\9& = 9\end{align}

Shows that the lines are coincident and having infinitely many solutions.

The answer is $$y = 3x - 3$$

Where $$x$$ can take any value. i.e. Infinitely many Solutions.

(iv) Steps:

\begin{align}0.2x + 0.3y &= 1.3 \qquad \dots(1)\\0.4x + 0.5y &= 2.3 \qquad \dots(2)\end{align}

Multiply both the equations (1) and (2) by $$10,$$ to remove the decimal number and making it easier for calculation.

\begin{align}&[ 0.2x + 0.3y = 1.3 ] \times 10\\& \Rightarrow 2x + 3y = 13 \;\;\;\;\qquad \dots(3)\\\\&\left[ {0.4x + 0.5y = 23} \right] \times (10)\\& \Rightarrow 4x + 5y = 23\;\;\;\;\qquad \dots(4)\end{align}

By solving the equation (3)

\begin{align}2x + 3y &= 13\\3y &= 13 - 2x\\y &= \frac{{13 - 2x}}{3} \qquad \dots(5)\end{align}

Substitute \begin{align} y = \frac{{13 - 2x}}{3} \end{align} in equation (4), we get

\begin{align}4x + 5\left( {\frac{{13 - 2x}}{3}} \right) &= 23\\\frac{{12x + 65 - 10x}}{3} &= 23\\ 2x + 65 &= 23 \times 3\\2x &= 69 - 65\\x &= \frac{4}{2}\\x &= 2\end{align}

Substituting $$x = 2$$ in equation (5), we get

\begin{align}y &= \frac{{13 - 2 \times 2}}{3}\\y &= \frac{9}{3}\\y &= 3\end{align}

\begin{align}x = 2\\y = 3\end{align}

(v) Steps:

\begin{align}\sqrt 2 x + \sqrt 3 y &= 0\qquad \dots(1)\\\sqrt 3 x - \sqrt 8 y &= 0 \qquad \dots(2)\end{align}

By solving the equation (1)

\begin{align}\sqrt 2 x + \sqrt 3 y &= 0\\\sqrt 3 y& = - \sqrt 2 x\\y &= - \frac{{\sqrt 2 x}}{3} \qquad \dots(3)\end{align}

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

\begin{align}\sqrt 3 x - \sqrt 8 \left( {\frac{{ - \sqrt 2 x}}{3}} \right) &= 0\\ \sqrt 3 x + \frac{{\sqrt {16} x}}{3}& = 0\\\frac{{3\sqrt 3 x + 4x}}{3} &= 0\\x\left( {3\sqrt 3 x + 4} \right)& = 0\\x &= 0\end{align}

Substituting $$x = 0$$ in equation (3), we get

\begin{align}y &= \frac{{\sqrt 2 \times 0}}{3}\\y &= 0\end{align}

\begin{align}x &= 0\\y &= 0\end{align}

(vi) Steps:

\begin{align}\frac{{3x}}{2} - \frac{{5y}}{3} &= - 2 \qquad \dots(1)\\\frac{x}{3} + \frac{y}{2} &= \frac{{13}}{6} \qquad \dots\left( 2 \right)\end{align}

Multiply both the equations (1) and (2) by $$6,$$ to remove the decimal number and making it easier for calculation.

\begin{align}&\left[ {\frac{{3x}}{2} - \frac{{5y}}{3} = - 2} \right] \times 6\\&9x - 10y = - 12 \qquad \qquad \dots(3)\\ \\&\left[ {\frac{x}{3} + \frac{y}{2} = \frac{{13}}{6}} \right] \times 6\\&2x + 3y = 13 \qquad \qquad\dots(4)\end{align}

By solving the equation (3)

\begin{align}9x - 10y &= - 12\\10y &= 9x + 12\\y &= \frac{{9x + 12}}{{10}} \qquad \dots(5)\end{align}

Substituting \begin{align}y = \frac{{9x + 12}}{{10}} \end{align} in equation (4), we get

\begin{align}2x + 3\left( {\frac{{9x + 12}}{{10}}} \right) &= 13\\ \frac{{20x + 27x + 36}}{{10}} &= 13\\47x &= 130 - 36\\x& = \frac{{94}}{{47}}\\x &= 2 \end{align}

Substituting $$x = 2$$ in equation (5), we get

\begin{align}y &= \frac{{9 \times 2 + 12}}{{10}}\\y &= \frac{{30}}{{10}}\\y &= 3\end{align}

\begin{align}x &= 2\\y &= 3\end{align}