Maharashtra Board Class 10 Math Part 1 Solution Chapter 1 Practice set 1.2 – Linear Equations in Two Variables
Balbharati Maharashtra Board Class 10 Math Part 1 Solution Chapter 1: Linear Equations in Two Variables. Marathi or Math Part 1 Medium Students of Class 10 get here Linear Equations in Two Variables full Practice set Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 1 Solution |
Chapter No- |
1 |
Chapter Name- |
Linear Equations in Two Variables |
Practice set |
1.2 |
Linear Equations in Two Variables
Practice set 1.2
(1) x + y = 3 —- (i)
x – y = 4 —- (ii)
x + y = 3
X | 3 | -2 | 0 |
Y | 0 | 5 | 3 |
(x, y) | (3, 0) | (-2, 5) | (0, 3) |
x – y = 4
X | 4 | -1 | 0 |
Y | 0 | – 5 | – 4 |
(x, y) | (4, 0) | (-1, -5) | (0,-4) |
Plotting graph above given tables
From the graph we can conclude the values of x and y
Solution is (x, y) = (3.5, – 0.5)
(2)
(1) x + y = 6 —– (i)
x – y = 4 (ii)
x + y = 6
X | 0 | 2 | 4 |
Y | 6 | 4 | (4, 2) |
(x, y) | (0, 6) | (2, 4) | (4, 2) |
x – y = 4
X | 0 | -2 | 2 |
Y | -4 | -6 | 2 |
(x, y) | (0, 4) | (-4, -6) | (2, -2) |
Plotting the points of above tables on graph we get,
From the graph we can conclude that the values of
x & y is
Solution is (x, y) = (5, 1)
(2) x + y = 5 —— (i)
x – y = 3 —– (ii)
x + y = 5
x | 0 | 2 | 5 |
y | 5 | 3 | 0 |
(x, y) | (0, 5) | (2, 3) | (5, 0) |
x – y = 3
x | 0 | 3 | 6 |
Y | -3 | 0 | 3 |
(x, y) | (0, -3) | (3, 0) | (6, 3) |
Plotting the above point on graph sheet,
From the graph we can conclude the value of x and y.
Solution is (x, y) = (4, 1)
(3) x + y = 0 —– (i)
2x – y = 9 —— (ii)
x + y = 0
X | 1 | 3 | 6 |
Y | -1 | -3 | -6 |
(x, y) | (1, -1) | (3, -3) | (6, -6) |
2x – y = 9
X | 2 | 5 | 4 |
Y | -5 | 1 | -1 |
(x, y) | (2, -5) | (5, 1) | (4, -1) |
Plotting the above point on graph sheet
From the graph we can conclude that the value of x & y (i).
Solution is (x, y) = (3, -3)
(4) 3x – y = 2 —— (i)
2x – y = 3 —– (ii)
3x – y = 2
X | 0 | 1 | 2 |
Y | -2 | 1 | 4 |
(x, y) | (0, -2) | (1, 1) | (2, 4) |
2x – y = 3
X | 0 | 2 | 4 |
Y | -3 | 1 | 5 |
(x, y) | (0, -3) | (2, 1) | (4, 5) |
Plotting the point on graph sheet
From the graph we can conclude that the value of x & y
Solution is (x, y) = (-1, -5)
(5) 3x – 4y = -7 —- (i)
5x – 2y = 0 —— (ii)
3x – 4y = -7
X | 3 | -1 | 7 |
Y | 4 | 1 | 7 |
(x, y) | (3, 4) | (-1, 1) | (7, 7) |
5x – 2y = 0
X | 2 | -2 | 0 |
Y | 5 | -5 | 0 |
(x, y) | (2, 5) | (-2, -5) | (0, 0) |
Plotting the point on graph sheet
From the graph we can conclude the values of x & y
So in is (x, y) = (1, 2.5)
(6) 2x – 3y = 4 —— (i)
3y – x = 4 —— (ii)
2x – 3y = 4
X | -1 | -4 | 5 |
Y | -2 | -4 | 2 |
(x, y) | (-1, -2) | (-4, -4) | (5, 2) |
3y – x = 4
X | 2 | -1 | -4 |
Y | 2 | 1 | 0 |
(x, y) | (2, 2) | (-1, 1) | (-4, 0) |
Plotting the above point on graph sheet
From the graph we can conclude the values of x and y
Solution is (x, y) = (8, 4)
Here is your solution of Maharashtra Board Class 10 Math Part 1 Solution Chapter 1 Practice set 1.2 – Linear Equations in Two Variables
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