Practice Set 1.3 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables
ii. The given simultaneous equations are
4x + 3y – 4 = 0
∴ 4x + 3y = 4 …(i)
6x = 8 – 5y
∴ 6x + 5y = 8 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = c1and a2x + b2y = c2, we get
a1= 4, b1= 3, c1= 4 and
a2= 6, b2= 5, c2= 8
∴ (x, y) = (-2, 4) is the solution of the given simultaneous equations.
iii. The given simultaneous equations are
x + 2y = -1 …(i)
2x – 3y = 12 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = C1and a2x + b2y = c2, we get
a1= 1, b1= 2, c1= -1 and
a2= 2, b2= -3, c2= 12
∴ (x, y) = (3, -2) is the solution of the given simultaneous equations.
iv. The given simultaneous equations are
6x – 4y = -12
∴ 3x – 2y = -6 …(i) [Dividing both sides by 2]
8x – 3y = -2 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = c1and a2x + b2y = c2, we get
a1= 3, b1= -2, c1= -6 and
a2= 8, b2= -3, c2= -2
∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.
v. The given simultaneous equations are
4m + 6n = 54
2m + 3n = 27 …(i) [Dividing both sides by 2]
3m + 2n = 28 …(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with
a1m + b1n = c1and a2m + b2n = c2, we get
a1= 2, b1= 3, c1= 27 and
a2= 3, b2= 2, c2= 28
∴ (m, n) = (6, 5) is the solution of the given simultaneous equations.
vi. The given simultaneous equations are
2x + 3y = 2 …(i)
x = =
∴ 2x – y = 1 …(ii) [Multiplying both sides by 2]
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a1x + b1y = c1and a2x + b2y = c2, we get
a1= 2, b1= 3, c1= 2 and
a2= 2, b2= -1, c2= 1
ii. 2x – y = -1 and 2x – y = -4
Here, a1b2– b1a2= (2)(-1) – (-1) (2)
= -2 + 2 = 0
Graphically, we can check that these two lines are parallel and hence they do not have a solution.