Worksheet on Simultaneous Linear Equations is available here. Students can get various methods to solve the system of linear equations from this page. One can learn how to solve two simultaneous equations in two variables using comparison, substitution, elimination, and cross-multiplication methods. You can even find different types of problems on linear equations in two variables. So, practice as many questions as possible for a better understanding of the concept.
This System of Linear Equations in Two Variables Worksheet page includes the questions with detailed step by step solution. Therefore, have a look at these problems and practice them to score good marks in the examination easily.
1. Use the comparison method to solve the following simultaneous linear equations:
(i) x + y = 1, x 2y = 5
(ii) 6x + 7y = 5, 2x 3y 8 = 0
(iii) 9x 6y = 12, 4x + 6y = 14
(iv) x y = -1, 2y + 3x = 12
Solution:
(i) x + y = 1, x 2y = 5
Given pair of the system of linear equations are
x + y = 1 —— (1)
x 2y = 5 ——- (2)
Express x in terms of y
From equation (1) x + y = 1, we get
x = 1 y
From equation (2), we get
x = 5 + 2y
Equate the values of x obtained from both equations.
1 y = 5 + 2y
2y + y = 1 5
3y = -4
y = -4/5
Substitute y = -4/5 in equation (1)
x 4/5 = 1
x = 1 + 4/5
x = (5 + 4)/5 = 9/5
Therefore, the required solution is x = 9/5, y = -4/5.
(ii) 6x + 7y = 5, 2x 3y 8 = 0
Given simultaneous linear equations are
6x + 7y = 5 ——- (1)
2x 3y 8 = 0 —— (2)
Express x in terms of y
From equation (1) 6x + 7y = 5, we get
6x = 5 7y
x = (5 7y)/6
From equation (2) 2x 3y 8 = 0, we get
2x = 3y + 8
x = (3y + 8)/2
Equate the values of x obtained from both equations.
(5 7y)/6 = (3y + 8)/2
2(5 7y) = 6(3y + 8)
10 14y = 18y + 48
10 48 = 18y + 14y
32y = -38
y = -38/32 = -19/16
Putting y = -19/16 in equation (2)
2x 3(-19/16) 8 = 0
2x + 57/16 8 = 0
2x + (57 128)/16 = 0
2x -71/16 = 0
2x = 71/16
x = 71/32
Therefore, the required solution set is x = 71/32, y = -19/16.
(iii) 9x 6y = 12, 4x + 6y = 14
Given the system of linear equations are
9x 6y = 12 —— (i)
4x + 6y = 14 —— (ii)
Express y in terms of x
From equation (i), we get
9x 12 = 6y
y = (9x 12)/6 —— (iii)
From equation (ii), we get
6y = (14 4x)
y = (14 4x)/6 ——- (iv)
Equation equation (iii) and equation (iv)
(9x 12)/6 = (14 4x)/6
Cross multiply the fractions
6(9x 12) = 6(14 4x)
9x 12 = 14 4x
9x + 4x = 14 + 12
13x = 26
x = 26/13
x = 2
Putting x = 2 in equation (ii)
4(2) + 6y = 14
8 + 6y = 14
6y = 14 8
6y = 6
y = 6/6
y = 1
Therefore, the required solution set is x = 2, y = 1.
(iv) x y = -1, 2y + 3x = 12
Given pair of linear equations are
x y = -1 —— (i)
2y + 3x = 12 —— (ii)
Express y in terms of x
From equation (i), we get
x + 1 = y
From equation (ii), we get
2y = (12 3x)
y = (12 3x)/2
Equate the values of x obtained from both equations.
x + 1 = (12 3x)/2
2(x + 1) = (12 3x)
2x + 2 = 12 3x
2x + 3x = 12 2
5x = 10
x = 10/5
x = 2
Substituting x = 2 in equation (ii)
2y + 3(2) = 12
2y + 6 = 12
2y = 12 6
2y = 6
y = 6/2
y = 3
Therefore, the required solution is x = 2, y = 3.
2. Solve the following simultaneous equations by using the Elimination Method:
(i) 2x y = 5, x + 3y 9 = 0
(ii) 2x 3y = 1, 3x 4y = 1
(iii) (2x/3) + (y/2) = -1, (-x/3) + y = 3
Solution:
(i) 2x y = 5, x + 3y 9 = 0
Given the system of linear equations are
2x y = 5 — (i)
x + 3y = 9 —— (ii)
Multiply the equation (i) by 3.
3(2x y) = 3(5)
6x 3y = 15 — (iii)
Add equation (ii) and equation (iii)
x + 3y + 6x 3y = 9 + 15
7x = 24
x = 24/7
Substitute x = 24/7 in equation (ii)
24/7 + 3y = 9
3y = 9 24/7
3y = (63 24)/7
y = 39/21
Therefore, the required solution is x = 24/7, y = 39/21.
(ii) 2x 3y = 1, 3x 4y = 1
Given the system of linear equations are
2x 3y = 1 —- (i)
3x 4y = 1 —- (ii)
Multiply equation (i) by 3
3(2x 3y) = 3(1)
6x 9y = 3 —— (iii)
Multiply equation (ii) by 2
2(3x 4y) = 2(1)
6x 8y = 2 — (iv)
Subtract equation (iii) from equation (iv)
(6x 9y) (6x 8y) = 3 2
6x 9y 6x + 8y = 1
-y = 1
y = -1
Putting y = -1 in equation (iv)
6x 8(-1) = 2
6x + 8 = 2
6x = 2 8
6x = -6
x = -1
Therefore, the required solution set is x = -1, y = -1.
(iii) (2x/3) + (y/2) = -1, (-x/3) + y = 3
Given pair of linear equations are
(2x/3) + (y/2) = -1
(4x + 3y)/6 = -1
(4x + 3y) = -6 — (i)
(-x/3) + y = 3
(-x + 3y)/3 = 3
-x + 3y = 9 — (ii)
Subtract equation (i) from equation (ii)
-x + 3y (4x + 3y) = 9 (-6)
-x + 3y 4x 3y = 9 + 6
-5x = 15
x = -15/5
x = -3
Put x = -3 in equation (ii)
-(-3) + 3y = 9
3 + 3y = 9
3y = 9 3
3y = 6
y = 6/3
y = 2
Therefore, the required solution is x = -3, y = 2.
3. Solve each other pair of the equation given below using the Substitution Method:
(i) x + y = 12, x €“ y = 2
(ii) 2x 5y = 9, 4x y = 9
(iii) x y = 5, 2x y = 11
Solution:
(i) x + y = 12, x y = 2
Given pair of simultaneous linear equations are
x + y = 12 —— (i)
x y = 2 —— (ii)
From equation (ii)
x = y + 2
Substitute the obtained x value in equation (i)
(y + 2) + y = 12
2y + 2 = 12
2y = 12 2
2y = 10
y = 10/2
y = 5
Putting y = 5 in equation (ii)
x 5 = 2
x = 2 + 5
x = 7
Therefore, the required solution is x = 7, y = 5.
(ii) 2x 5y = 9, 4x y = 9
Given pair of linear equations are
2x 5y = 9 — (i)
4x y = 9 — (ii)
From equation (ii), we can write
4x 9 = y
Substituting the obtained y value in equation (i)
2x 5(4x 9) = 9
2x 20x + 45 = 9
-18x = 9 45
-18x = -36
x = 36/18
x = 2
Putting x = 2 in equation (ii)
4(2) y = 9
8 9 = y
y = -1
Therefore, the required solution is x = 2, y = -1.
(iii) x y = 5, 2x y = 11
Given simultaneous linear equations are
x y = 5 —— (i)
2x y = 11 —— (ii)
From equation (i), we can write as
x = 5 + y
Substituting the new x value in equation (ii)
2(5 + y) y = 11
10 + 2y y = 11
10 + y = 11
y = 11 10
y = 1
Substitute y = 1 in equation (i)
x 1 = 5
x = 5 + 1
x = 6
Therefore, the solution is x = 6, y = 1.
4. Solve the below-mentioned simultaneous linear equations by the method of cross-multiplication:
(i) 2x + y = 4, x + y = 6
(ii) 4x 3y = 7, 3x y = 4
(iii) 3x + y = 8, 4x + 3y = 14
Solution:
(i) 2x + y = 4, x + y = 6
The transposition of the given simultaneous linear equations are
2x + y 4 = 0 —— (i)
x + y 6 = 0 ———- (ii)
Multiply equation (i) by 1 and equation (ii) by 1, we get
1(2x + y 4) = 1 x 0
2x + y 4 = 0 ——- (iii)
1(x + y 6) = 1x 0
x + y 6 = 0 —R12;- (iv)
Subtract equation (iv) from equation (iii)
[2x + y 4 = 0] [x + y 6 = 0]
2x + y 4 x y + 6 = 0
x + 2 = 0
x = -2
Substitute x = -2 in equation (ii)
-2 + y 6 = 0
y 8 = 0
y = 8
Therefore, the required solution set is x = -2, y = 8.
(ii) 4x 3y = 7, 3x y = 4
The transposition of the given system of linear equations are
4x 3y 7 = 0 —— (i)
3x y 4 = 0 —— (ii)
Multiply the equation (i) by -1 and equation (ii) by -3, we get
-1(4x 3y 7) = -1 x 0
-4x + 3y + 7 = 0 ——— (iii)
-3(3x y 4) = -3 x 0
-9x + 3y + 12 = 0 ——- (iv)
Subtract equation (iv) from equation (iii)
[-4x + 3y + 7] [-9x + 3y + 12] = 0
-4x + 3y + 7 + 9x 3y 12 = 0
5x 5 = 0
5x = 5
x = 5/5
x = 1
Putting x = 1 in equation (i)
4(1) 3y 7 = 0
4 3y 7 = 0
-3y 3 = 0
-3y = 3
y = -3/3
y = -1
Therefore, the required solution set is x = 1, y = -1.
(iii) 3x + y = 8, 4x + 3y = 14
The transposition of the given linear equations are
3x + y 8 = 0 — (i)
4x + 3y 14 = 0 —— (ii)
Multiplying the equation (i) by 3, we get
3(3x + y 8) = 0 x 3
9x + 3y 24 = 0 — (iii)
Multiplying the equation (ii) by 1, we get
1(4x + 3y 14) = 1 x 0
4x + 3y 14 = 0 —— (iv)
Subtracting equation (iv) from equation (iii)
[9x + 3y 24 = 0] [4x + 3y 14 = 0]
(9x + 3y 24) (4x + 3y 14) = 0
9x + 3y 24 4x 3y + 14 = 0
5x 10 = 0
5x = 10
x = 10/5
x = 2
Putting x = 2 in equation (ii)
4(2) + 3y 14 = 0
8 + 3y 14 = 0
3y 6 = 0
3y = 6
y = 6/3
y = 2
Therefore, the required solution set is x = 2, y = 2.
5. Solve the following simultaneous equations:
(i) 4/(x 3) + 6/(y 4) = 5, 5/(x 3) 3/(y 4) = 1
(ii) (y/6) (x/15) = 4, (y/3) (x/12) = 19/4
Solution:
(i) 4/(x 3) + 6/(y 4) = 5, 5/(x 3) 3/(y 4) = 1
Given simultaneous linear equations are
4/(x 3) + 6/(y 4) = 5 — (i)
5/(x 3) 3/(y 4) = 1
5/(x 3) 1 = 3/(y 4) —— (ii)
Putting equation (ii) in equation (i)
4/(x 3) + 2(5/(x 3) 1) = 5
4/(x 3) + 10/(x 3) 2 = 5
14/(x 3) = 5 + 2
14/(x 3) = 7
2/(x 3) = 1
2 = (x 3)
x = 2 + 3
x = 5
Substituting x = 5 in equation (i)
4/(5 3) + 6/(y 4) = 5
4/2 + 6/(y 4) = 5
2 + 6/(y 4) = 5
6/(y 4) = 5 2
6/(y 4) = 3
2/(y 4) = 1
2 = y 4
y = 2 + 4
y = 6
Therefore, the required solution set is x = 5, y = 6
(ii) (y/6) (x/15) = 4, (y/3) (x/12) = 19/4
Given linear equations are
(y/6) (x/15) = 4
(15y 6x) /90 = 4
15y 6x = 90 x 4
15y 6x = 360 ——- (i)
(y/3) (x/12) = 19/4
(12y 3x)/36 = 19/4
4(12y 3x) = 19 x 36
(12y 3x) = 171 —- (ii)
Express y in terms of x
12y = 171 + 3x
y = (171 + 3x) / 12
Substitute the obtained y value in equation (i)
15[(171 + 3x) / 12] 6x = 360
3(5[(171 + 3x) / 12] 2x) = 360
5[(171 + 3x) / 12] 2x = 120
(855 + 15x)/12 2x = 120
855 + 15x 24x = 120 x 12
855 9x = 1440
9x = 855 1440
9x = -585
x = -585/9
x = -65
Putting the value of x in equation (ii)
(12y 3(-65)) = 171
12y + 195 = 171
12y = 171 195
12y = -24
y = -24/12
y = -2
Therefore, the required solution is x = -65, y = -2.