Practice Set 1.2 Algebra 10th Standard Maths Part 1 Chapter 1 Linear Equations in Two Variables Solutions Maharashtra Board
Practice Set 1.2 Solutions
Question 1. Complete the following table to draw graph of the equations.
i. $x+y=3$
| $x$ | 3 | -2 | 0 |
| $y$ | 0 | 5 | 3 |
| $(x,y)$ | (3,0) | (-2,5) | (0,3) |
ii. $x-y=4$
| $x$ | 4 | -1 | 0 |
| $y$ | 0 | -5 | -4 |
| $(x,y)$ | (4,0) | (-1,-5) | (0,-4) |
Question 2. Solve the following simultaneous equations graphically.
i. $x+y=6$; $x-y=4$
The given simultaneous equations are $x+y=6 \Rightarrow y=6-x$ and $x-y=4 \Rightarrow y=x-4$.
Table for $x+y=6$:
| $x$ | 0 | 6 | 3 | 4 |
| $y$ | 6 | 0 | 3 | 2 |
| $(x,y)$ | (0,6) | (6,0) | (3,3) | (4,2) |
Table for $x-y=4$:
| $x$ | 0 | 4 | 2 | 5 |
| $y$ | -4 | 0 | -2 | 1 |
| $(x,y)$ | (0,-4) | (4,0) | (2,-2) | (5,1) |
The two lines intersect at point (5, 1).
$\therefore x=5$ and $y=1$ is the solution of the simultaneous equations $x+y=6$ and $x-y=4$.
ii. $x+y=5$; $x-y=3$
The given simultaneous equations are $x+y=5 \Rightarrow y=5-x$ and $x-y=3 \Rightarrow y=x-3$.
Table for $x+y=5$:
| $x$ | 0 | 5 | 2 | 3 |
| $y$ | 5 | 0 | 3 | 2 |
| $(x,y)$ | (0,5) | (5,0) | (2,3) | (3,2) |
Table for $x-y=3$:
| $x$ | 0 | 3 | 2 | 1 |
| $y$ | -3 | 0 | -1 | -2 |
| $(x,y)$ | (0,-3) | (3,0) | (2,-1) | (1,-2) |
The two lines intersect at point (4, 1).
$\therefore x=4$ and $y=1$ is the solution of the simultaneous equations $x+y=5$ and $x-y=3$.
iii. $x+y=0$; $2x-y=9$
The given simultaneous equations are $x+y=0 \Rightarrow y=-x$ and $2x-y=9 \Rightarrow y=2x-9$.
Table for $x+y=0$:
| $x$ | 0 | 2 | -2 | -1 |
| $y$ | 0 | -2 | 2 | 1 |
| $(x,y)$ | (0,0) | (2,-2) | (-2,2) | (-1,1) |
Table for $2x-y=9$:
| $x$ | 0 | 2 | 5 | 4 |
| $y$ | -9 | -5 | 1 | -1 |
| $(x,y)$ | (0,-9) | (2,-5) | (5,1) | (4,-1) |
The two lines intersect at point (3, -3).
$\therefore x=3$ and $y=-3$ is the solution of the simultaneous equations $x+y=0$ and $2x-y=9$.
iv. $3x-y=2$; $2x-y=3$
The given simultaneous equations are $3x-y=2 \Rightarrow y=3x-2$ and $2x-y=3 \Rightarrow y=2x-3$.
Table for $3x-y=2$:
| $x$ | 0 | 1 | -1 | 2 |
| $y$ | -2 | 1 | -5 | 4 |
| $(x,y)$ | (0,-2) | (1,1) | (-1,-5) | (2,4) |
Table for $2x-y=3$:
| $x$ | 0 | 1 | 2 | 3 |
| $y$ | -3 | -1 | 1 | 3 |
| $(x,y)$ | (0,-3) | (1,-1) | (2,1) | (3,3) |
The two lines intersect at point (-1, -5).
$\therefore x=-1$ and $y=-5$ is the solution of the simultaneous equations $3x-y=2$ and $2x-y=3$.
v. $3x-4y=-7$; $5x-2y=0$
The given simultaneous equations are $3x-4y=-7 \Rightarrow 4y=3x+7 \Rightarrow y=\frac{3x+7}{4}$ and $5x-2y=0 \Rightarrow 2y=5x \Rightarrow y=\frac{5}{2}x$.
Table for $3x-4y=-7$:
| $x$ | -1 | -5 | 3 | 5 |
| $y$ | 1 | -2 | 4 | 5.5 |
| $(x,y)$ | (-1,1) | (-5,-2) | (3,4) | (5,5.5) |
Table for $5x-2y=0$:
| $x$ | 0 | 2 | -2 | 1 |
| $y$ | 0 | 5 | -5 | 2.5 |
| $(x,y)$ | (0,0) | (2,5) | (-2,-5) | (1,2.5) |
The two lines intersect at point (1, 2.5).
$\therefore x=1$ and $y=2.5$ is the solution of the simultaneous equations $3x-4y=-7$ and $5x-2y=0$.
vi. $2x-3y=4$; $3y-x=4$
The given simultaneous equations are $2x-3y=4 \Rightarrow 3y=2x-4 \Rightarrow y=\frac{2x-4}{3}$ and $3y-x=4 \Rightarrow 3y=x+4 \Rightarrow y=\frac{x+4}{3}$.
Table for $2x-3y=4$:
| $x$ | 2 | -1 | 5 | 8 |
| $y$ | 0 | -2 | 2 | 4 |
| $(x,y)$ | (2,0) | (-1,-2) | (5,2) | (8,4) |
Table for $3y-x=4$:
| $x$ | 2 | -4 | 5 | -1 |
| $y$ | 2 | 0 | 3 | 1 |
| $(x,y)$ | (2,2) | (-4,0) | (5,3) | (-1,1) |
The two lines intersect at point (8, 4).
$\therefore x=8$ and $y=4$ is the solution of the simultaneous equations $2x+3y=4$ and $3y-x=4$.
Textbook Pg. No. 8 Questions
Question 1: Solve the following simultaneous equations by graphical method.
Equations: $x-y=1$ and $5x-3y=1$
Table for $x-y=1$:
| $x$ | 0 | 1 | 3 | -2 |
| $y$ | -1 | 0 | 2 | -3 |
| $(x,y)$ | (0,-1) | (1,0) | (3,2) | (-2,-3) |
Table for $5x-3y=1$:
| $x$ | 2 | 5 | -1 | -4 |
| $y$ | 3 | 8 | -2 | -7 |
| $(x,y)$ | (2,3) | (5,8) | (-1,-2) | (-4,-7) |
The two lines intersect at point (-1, -2).
$\therefore (x,y)=(-1,-2)$ is the solution of the given simultaneous equations.
Solve the above equations by method of elimination. Check your solution with the solution obtained by graphical method:
The given simultaneous equations are:
- $x-y=1$ ... (i)
- $5x-3y=1$ ... (ii)
Multiplying equation (i) by 3, we get $3x-3y=3$ ... (iii)
Subtracting equation (iii) from (ii), we get:
$5x-3y=1$
$3x-3y=3$
$\Rightarrow 2x = -2 \Rightarrow x = \frac{-2}{2} = -1$
Substituting $x=-1$ in equation (i), we get:
$-1-y=1 \Rightarrow -y=1+1 \Rightarrow -y=2 \Rightarrow y=-2$
$\therefore (x,y)=(-1,-2)$ is the solution of the given simultaneous equations.
$\therefore$ The solution obtained by elimination method and by graphical method is the same.
Question 2: The following table contains the values of x and y co-ordinates for ordered pairs to draw the graph of $5x-3y=1$.
| x | 0 | $\frac{1}{5}$ | 1 | -2 |
| y | $-\frac{1}{3}$ | 0 | $\frac{4}{3}$ | $-\frac{11}{3}$ |
| (x,y) | $(0,-\frac{1}{3})$ | $(\frac{1}{5},0)$ | $(1,\frac{4}{3})$ | $(-2,-\frac{11}{3})$ |
i. Is it easy to plot these points?
No. Here, $-\frac{1}{3}=-0.33..., \frac{4}{3}=1.33..., -\frac{11}{3}=-3.66...$
The above numbers are non-terminating and recurring decimals.
$\therefore$ It is not easy to plot the given points.
ii. Which precaution is to be taken to find ordered pairs so that plotting of points becomes easy?
While finding ordered pairs, numbers should be selected in such a way that the co-ordinates obtained will be integers.
Textbook Pg. No. 9 Question 3
To solve simultaneous equations $x+2y=4$; $3x+6y=12$ graphically, following are the ordered pairs.
Table for $x+2y=4$:
| $x$ | -2 | 0 | 2 |
| $y$ | 3 | 2 | 1 |
| $(x,y)$ | (-2,3) | (0,2) | (2, 1) |
Table for $3x+6y=12$:
| $x$ | -4 | 1 | 8 |
| $y$ | 4 | 1.5 | -2 |
| $(x,y)$ | (-4,4) | (1, 1.5) | (8,-2) |
i. Are the graphs of both the equations different or same?
The graphs of both the equations are same.
ii. What are the solutions of the two equations $x+2y=4$ and $3x+6y=12$? How many solutions are possible?
The solutions of the given equations are (-2, 3), (0, 2), (1, 1.5), etc.
$\therefore$ Infinite solutions are possible.
iii. What are the relations between coefficients of x, coefficients of y and constant terms in both the equations?
Ratio of coefficients of $x = \frac{1}{3}$.
Ratio of coefficients of $y = \frac{2}{6} = \frac{1}{3}$.
Ratio of constant terms $= \frac{4}{12} = \frac{1}{3}$.
$\therefore$ Ratios of coefficients of x = ratio of coefficients of y = ratio of the constant terms.
iv. What conclusion can you draw when two equations are given but the graph is only one line?
When two equations are given but the graph is only one line, the equations will have infinite solutions.
Textbook Pg. No. 10 Question 4
Draw graphs of $x-2y=4$, $2x-4y=12$ on the same co-ordinate plane. Observe it. Think of the relation between the coefficients of x, coefficients of y and the constant terms and draw the inference.
Table for $x-2y=4 \Rightarrow 2y=x-4 \Rightarrow y=\frac{x-4}{2}$:
| $x$ | 0 | 2 | -2 | 4 |
| $y$ | -2 | -1 | -3 | 0 |
| $(x,y)$ | (0,-2) | (2,-1) | (-2,-3) | (4,0) |
Table for $2x-4y=12 \Rightarrow x-2y=6 \Rightarrow 2y=x-6 \Rightarrow y=\frac{x-6}{2}$:
| $x$ | 0 | -2 | 2 | 4 |
| $y$ | -3 | -4 | -2 | -1 |
| $(x,y)$ | (0,-3) | (-2,-4) | (2,-2) | (4,-1) |
Ratio of coefficients of $x = \frac{1}{2}$.
Ratio of coefficients of $y = \frac{-2}{-4} = \frac{1}{2}$.
Ratio of constant terms $= \frac{4}{12} = \frac{1}{3}$.
$\therefore$ Ratio of coefficients of x = ratio of coefficients of y $\neq$ ratio of constant terms.
If ratio of coefficients of x = ratio of coefficients of y $\neq$ ratio of constant terms, then the graphs of the two equations will be parallel to each other.
Condition of consistency in Equations:
| Sr. No. | Simultaneous Equations | $\frac{a_1}{a_2}$ | $\frac{b_1}{b_2}$ | $\frac{c_1}{c_2}$ | Comparison of ratios | Graphical interpretation | Algebraic interpretation |
|---|---|---|---|---|---|---|---|
| 1. | $x+y=3$; $x-y=1$ | $\frac{1}{1}$ | $\frac{1}{-1}$ | $\frac{3}{1}$ | $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ | Intersecting lines | Unique solution (OR) Only one common solution |
| 2. | $2x-y=-1$; $2x-y=4$ | $\frac{2}{2}$ | $\frac{-1}{-1}$ | $\frac{-1}{4}$ | $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ | Parallel lines | No solution |
| 3. | $x-y=-2$; $2x-2y=-4$ | $\frac{1}{2}$ | $\frac{-1}{-2}$ | $\frac{-2}{-4}$ | $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ | Coincident lines | Infinitely many solutions |