NCERT Solutions for Class 9 Maths Chapter 13 — Exercise Set 13.2#
A solution of a linear equation in two variables is an ordered pair $(x, y)$ that makes the equation true. Every linear equation in two variables has infinitely many solutions, and they are exactly the points of its graph — a straight line.
To find a solution: choose any value of $x$, substitute it, and solve for $y$.
1. Verify if the ordered pair (4, 3) is a solution of 5x − 6y = 2. Explain your reasoning. #
Solution
We substitute $x = 4$, $y = 3$ in the left-hand side:
$$5x - 6y = 5(4) - 6(3)$$$$5x - 6y = 20 - 18$$
$$5x - 6y = 2$$
This equals the right-hand side, 2.
∴ Yes, $(4, 3)$ is a solution of $5x - 6y = 2$.
2. Find any two solutions for each of the following equations: (i) 7x − 3y = 21 (ii) 2x + 3y = 5 #
(i) 7x − 3y = 21
Putting $x = 0$:
$$-3y = 21$$$$\Rightarrow y = -7$$
Putting $y = 0$:
$$7x = 21$$$$\Rightarrow x = 3$$
∴ Two solutions are $(0, -7)$ and $(3, 0)$. (Any others, such as $(6, 7)$, are also correct.)
(ii) 2x + 3y = 5
Putting $x = 1$:
$$2 + 3y = 5$$$$\Rightarrow y = 1$$
Putting $x = 4$:
$$8 + 3y = 5$$$$\Rightarrow y = -1$$
∴ Two solutions are $(1, 1)$ and $(4, -1)$.
3. In the equations given below, m and n are unknown constants: 2mx + 3y = 7; 4x + ny = −10. If (2, −1) is the solution of both equations, find the values of m and n. #
Solution
Substituting $x = 2$, $y = -1$ in the first equation:
$$2m(2) + 3(-1) = 7$$$$\Rightarrow 4m - 3 = 7$$
$$\Rightarrow 4m = 10$$
$$\Rightarrow m = \dfrac{5}{2}$$
Substituting $x = 2$, $y = -1$ in the second equation:
$$4(2) + n(-1) = -10$$$$\Rightarrow 8 - n = -10$$
$$\Rightarrow n = 18$$
∴ $m = \dfrac{5}{2}$ and $n = 18$.
4. Find two solutions which lie in different quadrants for each of the following linear equations. Identify the quadrants in which the points lie. Verify your solutions by representing the linear equations on a graph paper. #
(i) 5x + 3y = 7
$x = 2$: $10 + 3y = 7$, so $y = -1$. The point $(2, -1)$ is in Quadrant IV.
$x = -1$: $-5 + 3y = 7$, so $y = 4$. The point $(-1, 4)$ is in Quadrant II.
(ii) 5x − 3y = 7
$x = 2$: $10 - 3y = 7$, so $y = 1$. The point $(2, 1)$ is in Quadrant I.
$x = -1$: $-5 - 3y = 7$, so $y = -4$. The point $(-1, -4)$ is in Quadrant III.
(iii) −5x + 3y = 7
$x = 1$: $-5 + 3y = 7$, so $y = 4$. The point $(1, 4)$ is in Quadrant I.
$x = -2$: $10 + 3y = 7$, so $y = -1$. The point $(-2, -1)$ is in Quadrant III.
(iv) −5x − 3y = 7
$x = 1$: $-5 - 3y = 7$, so $y = -4$. The point $(1, -4)$ is in Quadrant IV.
$x = -2$: $10 - 3y = 7$, so $y = 1$. The point $(-2, 1)$ is in Quadrant II.
Verification on a graph
Each pair of points lies on its line, in the quadrants stated.
5. Consider the graph of the equation 3x − 7y = 21 shown below. Does the point C (2, 3) lie on the line? Does it satisfy the equation? Can points that do not lie on the line satisfy the equation? #
Solution
From the graph, $C(2, 3)$ is above the line, which passes through $A(7, 0)$ and $B(0, -3)$, so C does not lie on the line.
Checking the equation:
$$3x - 7y = 3(2) - 7(3)$$$$3x - 7y = 6 - 21$$
$$3x - 7y = -15$$$$\Rightarrow -15 \neq 21$$
∴ $C$ does not satisfy the equation.
No, a point that is not on the line can never satisfy the equation: the graph of a linear equation is made of exactly its solutions. Every solution is on the line, and every point on the line is a solution.
6. State whether the following sentences are True or False. Justify your answer. #
Solution
| Statement | True/False | Reason |
|---|---|---|
| (i) A linear equation in two variables has only one solution. | False | It has infinitely many: for every value of $x$ there is a matching $y$ (when $b \neq 0$). |
| (ii) The graph of a linear equation in two variables always passes through the origin. | False | $x + y = 1$ does not: $(0, 0)$ gives $0 \neq 1$. Only equations with $c = 0$ pass through the origin. |
| (iii) A linear equation in two variables can never have rational solutions. | False | $x + y = 1$ has the rational solution $(1, 0)$. |
| (iv) $x = 3$ is a valid linear equation in two variables. | True | It is $1 \cdot x + 0 \cdot y - 3 = 0$, with $a = 1$, $b = 0$. Its graph is a vertical line. |
| (v) The equation $2x + 3y = 7$ has infinitely many solutions. | True | Every value of $x$ gives a value of $y = \dfrac{7 - 2x}{3}$. |
| (vi) The point $(1, 2)$ is a solution of the equation $2x + 3y = 7$. | False | $2(1) + 3(2) = 8 \neq 7$. |
7. (i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other. (ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same set of solutions. #
(i) 3x + 4y = 7 and 6x + 8y = 14
The second equation is the first multiplied by 2.
- If $(x, y)$ satisfies $3x + 4y = 7$, then multiplying by 2 gives $6x + 8y = 14$.
- If $(x, y)$ satisfies $6x + 8y = 14$, then dividing by 2 gives $3x + 4y = 7$.
So every solution of one is a solution of the other — they have the same solutions (the same line). For example, $(1, 1)$ satisfies both.
(ii) ax + by = c and kax + kby = kc
If $ax + by = c$, then multiplying both sides by $k$:
$$k(ax + by) = kc$$$$\Rightarrow kax + kby = kc$$
If $kax + kby = kc$, then since $k \neq 0$ we may divide both sides by $k$:
$$ax + by = c$$So the two equations have exactly the same solutions. (The condition $k \neq 0$ is needed: with $k = 0$, the second equation becomes $0 = 0$, which every pair satisfies.)