Linear Equations in Two Variables

Class 9 NCERT Maths

NCERT

1   The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs. x and that of a pen to be Rs. y.)

Solution :

$\\$$\bullet $ Let the cost of a notebook be Rs. x$\\$ $\bullet $ Let the cost of a pen be Rs. y$\\$ Given: Cost of Notebook is twice the cost of Pen$\\$ Therefore, we can write the required linear equation in the$\\$ form of, Cost of notebook $= 2 *$ Cost of pen$\\$ $x = 2y\\ x - 2y = 0$

2   Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b, c in each case:$\\$ (i) $2 x + 3 y = 9.35$$\\$ (ii) $x -\dfrac{y}{5}-10=0$$\\$ (iii)$ -2x + 3y = 6$$\\$ (iv) $x = 3y$$\\$ (v) $2x = - 5y$$\\$ (vi) $3x + 2 = 0$$\\$ (vii) $y - 2 = 0$$\\$ (viii) $5 = 2x$

Solution :

(i) $2 x + 3 y =9.35 --------------- \text{Equation }(1)$$\\$ $2 x + 3 y - 9.35 = 0$$\\$ Comparing this equation with standard form of the linear equation, $ax + by + c = 0 $we have,$\\$ $\bullet a = 2,$$\\$ $\bullet b = 3,$$\\$ $\bullet c = - 9.35$$\\$ (ii) $x -\dfrac{y}{5}-10=0--------------- \text{Equation} (1)$$\\$ Comparing Equation (1) with standard form of the linear equation, $ax + by + c = 0$ we have,$\\$ $\bullet a = 1,$$\\$ $\bullet b=\dfrac{-1}{5}$$\\$ $\bullet c=-10$$\\$ (iii)$ -2x + 3y = 6 --------------- \text{Equation }(1)$$\\$ $-2x + 3y - 6 = 0$$\\$ Comparing this equation with standard form of the linear equation,$ ax + by + c = 0$ we have,$\\$ $\bullet a = -2,$$\\$ $\bullet b = 3,$$\\$ $\bullet c = -6$$\\$ (iv)$ x = 3y ---------------\text{ Equation }(1)$$\\$ $1x - 3y + 0 = 0$$\\$ Comparing this equation with standard form of the linear equation, $ax + by + c = 0$ we have,$\\$ $\bullet a = 1,$$\\$ $\bullet b = -3,$$\\$ $\bullet c = 0$$\\$

(v) $2x = -5y ---------------\text{ Equation} (1)$$\\$ $2x + 5y + 0 = 0$$\\$ Comparing this equation with standard form of the linear equation, $ax + by + c = 0$ we have,$\\$ $\bullet a = 2,$$\\$ $\bullet b = 5,$$\\$ $\bullet c = 0$$\\$ (vi) $3x + 2 = 0 --------------- \text{Equation} (1)$$\\$ We can write Equation (1) as below,$\\$ $3x + 0y + 2 = 0$$\\$ Comparing this equation with $ax + by + c = 0,$$\\$ $\bullet a = 3,$$\\$ $\bullet b = 0,$$\\$ $\bullet c = 2$$\\$ (vii) $y -2 = 0 --------------- \text{Equation }(1)$$\\$ We can write Equation (1) as below,$\\$ $0x + 1y - 2 = 0$$\\$ Comparing this equation with standard form of the linear equation, $ax + by + c = 0 $ we have,$\\$ $\bullet a = 0,$$\\$ $\bullet b = 1,$$\\$ $\bullet c = -2$$\\$ (viii)$ 5 = 2x --------------- \text{Equation} (1)$$\\$ $-2x + 0y + 5 = 0$$\\$ Comparing this equation with standard form of the linear equation, $ax + by + c = 0 $ we have,$\\$ $\bullet a = -2,$$\\$ $\bullet b = 0,$$\\$ $\bullet c = 5$$\\$

3   Which one of the following options is true, and why? $y = 3x + 5$ has$\\$ (i) A unique solution,$\\$ (ii) only two solutions,$\\$ (iii) infinitely many solutions

Solution :

Given:$\\$ $\circ y = 3 x + 5$ is a linear Equation.$\\$ $\circ \text{For }x = 0, y = 0 + 5 = 5$ Therefore, $(0, 5)$ is another solution.$\\$ $\circ \text{For } x = 1, y = 3 × 1 + 5 = 8$ Therefore $(1, 8)$ is another solution.$\\$ $\circ \text{For } y = 0 , 3x + 5 = 0$ is the one solution$\\$ Clearly, for different values of $x$, we get another value of $y$.$\\$ Thus, the chosen value of $x$ together with this value of $y$ constitutes another solution of the given equation.$\\$ So, there is no end to different solutions of a linear equation in two variables. Therefore, a linear equation in two variables has infinitely many solutions$\\$ Hence (iii) is the correct answer

4   Write four solutions for each of the following equations:$\\$ (i) $2x + y = 7$$\\$ (ii) $\pi x + y = 9$$\\$ (iii)$ x = 4y$

Solution :

(i) $2x + y = 7$$\\$ Given,$\\$ $\bullet $ Linear Equation, $2x + y = 7$$\\$ We can write the above equation as below by simplifying$\\$ $y = 7- 2x ----------\text{Equation} (1)$$\\$ Let us now take different values of x and substituting in the Equation (1), we get$\\$ $\bullet \text{ For } x = 0,$$\\$ $2(0) + y = 7$$\\$ $ y = 7$$\\$ Hence, we get $(x, y) = (0, 7)$$\\$ $\bullet \text{ For} x = 1,$$\\$ $2(1) + y = 7$$\\$ $y = 5$$\\$ Hence, we get $(x, y) = (1, 5)$$\\$ $\bullet \text {For }x = 2,$$\\$ $2(2) + y = 7$$\\$ $ y = 3$$\\$ Hence, we get $(x, y) = (2, 3)$$\\$ $\bullet \text{ For }x = 3,$$\\$ $2(3) + y = 7$$\\$ $ y = 1$$\\$ Hence we get $(x, y) = (3, 1)$$\\$ Therefore the four solutions of the given equation are $(0,7) , (1,5), (2, 3), (3,1)$

(ii) $\pi x + y = 9$$\\$ Given,$\\$ $\bullet $ Linear Equation, $\pi x + y = 9$$\\$ We can write the above equation as below by Transposing$\\$ $y = 9- \pi x ----------\text{Equation }(1)$$\\$ Let us now take different values of x and substituting in the Equation (1), we get$\\$ $\bullet \text{For }x = 0,$$\\$ $y = 9 - \pi (0)$$\\$ $ y = 9$$\\$ Hence we get $(x, y) = (0, 9)$$\\$ $\bullet \text{For }x = 1,$$\\$ $y = 9- \pi(1)$$\\$ $= 9- \pi$$\\$ Hence we get $(x, y) = (1, 9- \pi )$$\\$ $\bullet \text{For }x = 2,$$\\$ $y = 9- \pi(2)$$\\$ Hence we get $(x, y) = (2, 9 -2 \pi )$$\\$ $\bullet \text{For }x = 3,$$\\$ $y = 9- \pi (3)$$\\$ Hence we get $(x, y) = (3, 9 - 3 \pi )$$\\$ Therefore the four solutions of the given equation are $(0, 9), (1, 9 - ), (2, 9 -2 ), (3, 9 - 3 )$

(iii) $x = 4y$$\\$ Given,$\\$ $\bullet $ Linear Equation, $x = 4y$$\\$ We can write the above equation as below by Transposing$\\$ $y = x/4$ ----------Equation (1)$\\$ Let us now take different values of x and substituting in the Equation (1), we get$\\$ $\bullet $For $x = 0,$$\\$ $y = 0/4 = 0$$\\$ Hence we get $(x, y) = (0, 0)$$\\$ $\bullet $For $x = 1,$$\\$ $y = 1/4$$\\$ Hence we get $(x, y) = (1, 1/4)$$\\$ $\bullet $ For $x = 2,$$\\$ $y = 2/4 = 1/2$$\\$ Hence we get $(x, y) = (2, 1/2)$$\\$ $\bullet $For $x = 3,$$\\$ $y = 3/4$$\\$ Hence we get $(x, y) = (3, 3/4)$$\\$ Therefore the four solutions of the given equation are $(0, 0), (1, 1/4), (2, 1/2), (3, 3/4)$

5   Check which of the following solutions of the equation are $x -2y = 4$ and which are not:$\\$ (i)$ (0, 2)$$\\$ (ii)$ (2, 0)$$\\$ (iii)$ (4, 0)$$\\$ (iv)$ ( 2, 4 2)$$\\$ (v)$ (1, 1)$$\\$

Solution :

Given : $x - 2y = 4$ is a Linear Equation----------Equation(1)$\\$ (i) $(0, 2)$ B Substituting $x = 0$ and $y = 2$ in the L.H.S of the given Equation (1)$\\$ $x - 2y\\ = (0) -(2) 2\\ = - 4 \neq 4 \neq RHS$$\\$ L.H.S $\neq $ R.H.S$\\$ Therefore, $(0, 2)$ is not a solution of this equation.$\\$ (ii) $(2, 0)$$\\$ By Substituting, $x = 2$ and $y = 0$ in the L.H.S of the given Equation (1),$\\$ $x - 2y\\ = 2 - 2(0)\\ = 2 \neq 4 \neq RHS$$\\$ L.H.S $\neq $ R.H.S $\\$ Therefore,$ (2, 0)$ is not a solution of this equation.

(iii)$ (4, 0)$$\\$ By Substituting, $x = 4$ and $y = 0$ in the L.H.S of the given Equation (1)$\\$ $x - 2y\\ = 4 - 2(0)\\ = 4$$\\$ L.H.S = R.H.S$\\$ Therefore, $(4, 0)$ is a solution of this equation.$\\$ (iv)$ ( 2, 4 2)$$\\$ By Substituting, $x = 2$ and $y = 4 \sqrt{2}$ in the L.H.S of the given Equation (1)$\\$ $x - 2y\\ = 2 - 8\sqrt{ 2}\\ =-7\sqrt{ 2} \neq 4 \neq RHS$$\\$ L.H.S $\neq$ R.H.S$\\$ Therefore, is not a solution of this equation.$\\$ (v)$ (1, 1)$$\\$ By Substituting, $x = 1$ and $y = 1$ in the L.H.S of the given Equation (1)$\\$ $x - 2y\\ = 1 - 2(1)\\ = 1 - 2\\ = - 1 \neq 4 \neq RHS$$\\$ L.H.S $\neq$ R.H.S$\\$ Therefore, $(1, 1)$ is not a solution of this equation.