Elimination method questions and practice questions are provided here for students to understand the concept very well. The elimination method is generally used to solve the system of equations. It means finding the values of variables in the given equation. Students can find a number of elimination method questions to enhance their knowledge and score high in exams. To know more about the elimination method, click here.
| What is an Elimination Method?
When solving linear equations with two or three variables, the elimination method helps a lot. According to the definition of the elimination method, it involves eliminating one of the terms containing any of the variables to make the computations easier. This is performed by multiplying or dividing a number into the equation(s) until one of the variable terms’ coefficients is the same. Then, to eliminate or remove that term from the result, we add or subtract both equations. The elimination method is also known as the addition method. Also, read: System of Linear Equations. |
Elimination Method Questions with Solutions
| The elimination method is used to solve a system of equations. The steps are as follows:
Step 1: To acquire a common coefficient of any of the variables in both equations, multiply or divide both linear equations by a non-zero value. Step 2: Solve both equations by adding or subtracting the same terms. Step 3: Simplify the result such that we only get an answer in the form of y = c, where c is any constant. Step 4: To obtain the value of the other specified variable, replace this value with either of the provided equations. |
1. Solve the equations by elimination method.
3a = b + 5
5a – b = 11
Solution:
Given:
3a = b + 5
The above equation can be written as:
3a – b = 5 …(1)
5a – b = 11 …(2)
Using the elimination method, the second term of both equations can be eliminated by subtracting the equations.
Therefore,
3a – b = 5
5a – b = 11
(-) (+) (-)
__________
-2a + 0 = -6
Hence, -2a = -6
a = -6/-2
a = 3
Now, substitute a = 3 in equation (1), we get the value of b.
3(3) – b = 5
9 – b = 5
-b = 5 – 9
-b = -4
b = 4
Therefore, the values of a and b are 3 and 4, respectively.
2. Find the values of a and b from the system of equation 2a + b = -4 and 5a – 3b = 1 using the method of elimination.
Solution:
Given equations are:
2a + b = -4 …(1)
5a – 3b = 1 …(2)
Now, multiply the equation (1) on both sides by 3, and we get
6a + 3b = -12 …(3)
Now, solve the equation (2) and (3) to get the value of a:
6a + 3b = -12
5a – 3b = 1
__________
11a + 0 = -11
Therefore, 11a = -11
a = -11/11
a = -1
Hence, the value of a is -1.
Now, substitute a = -1 in equation (1),
2(-1) + b = -4
-2 + b = -4
b = -4 + 2
b = -2
Hence, the values of a and b are -1 and -2, respectively.
3. Solve the following pair of linear equations using the elimination method: a + b = 5, 2a – 3b = 4
Solution:
Given equations are:
a + b = 5 …(1)
2a – 3b = 4 …(2)
Now, multiply both sides of the equation (1) by 2, and we get
2a + 2b = 10 …(3)
Now, solving equation (2) and (3) using elimination method, we get
2a – 3b = 4
2a + 2b = 10
(-) (-) (-)
_____________
-5b = -6
-5b = -6
Hence, the value of b is:
b = -6/-5
b = 6/5
Now, substitute b = 6/5 in equation (1), we get
a + (6/5) = 5
5a + 6 = 25
5a = 25 – 6
5a = 19
a = 19/5
Hence, the values of a and b are 19/5 and 6/5, respectively.
4. Solve the following equations using the elimination method: p + q – 40 = 0 and 7p + 3q = 180.
Solution:
Given pair of linear equations are:
p + q – 40 = 0
It can be written as:
p + q = 40 …(1)
7p + 3q = 180 …(2)
Now, multiply the equation (1) by 7, we get
7p + 7q = 280 …(3)
Now, solve the equations (2) and (3) using elimination method, we get the following:
7p + 3q = 180
7p + 7q = 280
(-) (-) (-)
____________
-4q = – 100
Hence,
q = -100/-4
q = 100/4
q = 25
Now, substitute q = 25 in equation (1), we get
p + 25 = 40
p = 40 – 25
p = 15
Therefore, p = 15 and q = 25.
5. Using the elimination method, solve the following equations: a/2 + 2/3 b = -1 and a – 1/3b = 3.
Solution:
Given equations are:
a/2 + 2/3 b = -1 …(1)
a – 1/3b = 3 …(2)
Now, multiply equation (1) by 6 and equation (2) by 3, we get;
3a + 4b = -6 …(3)
3a – b = 9 …(4)
Now, solve the equations (3) and (4), we get
3a + 4b = -6
3a – b = 9
(-) (+) (-)
__________
0 + 5b = -15
Thus, 5b = -15
b = -15/5
b = -3
Now, substitute b = -3 in equation (4), we get;
3a – (-3) = 9
3a + 3 = 9
3a = 9 – 3
3a = 6
a = 6/3
a = 2
Hence, the solution of the given equations a/2 + 2/3 b = -1 and a – 1/3b = 3 are:
a = 2 and b = -3.
Also, read:
6. If p = a and q = b are the solutions of the equations p – q = 2 and p + q = 4, then find the values of a and b.
Solution:
Given equations are:
p – q = 2 …(1)
p + q = 4 …(2)
Now, solve the equations (1) and (2)
I.e.,
p – q = 2
p + q = 4
________
2p + 0 = 6
Therefore,
2p = 6
p = 6/2
p = 3
Now, substitute p = 3 in equation (1), we get
3 – q = 2
-q = 2 – 3
-q = -1
Hence, q = 1
As given, p = a and q = b,
The values of a and b are 3 and 1, respectively.
7. Determine the values of a and b which satisfies both the equations 47a + 31b = 63 and 31a + 47b = 15.
Solution:
Given equations:
47a + 31b = 63 …(1)
31a + 47b = 15 …(2)
To solve the equations, multiply equation (1) by 31 and equation (2) by 47.
1457a + 961b = 1953 …(3)
1457a + 2209b = 705 …(4)
Now, solve the equations (3) and (4), we get
1457a + 961b = 1953
1457a + 2209b = 705
(-) (-) (-)
___________________
0 – 1248b = 1248
Thus, -1248b = 1248
b = -1248/1248
b = -1
Now, substitute b = -1 in equation (2), we get
31a + 47(-1) = 15
31a – 47 = 15
31a = 15 + 47
31a = 62
Thus, a = 62/31
a = 2
Therefore, a = 2 and b = -1.
8. A book shop was selling 3 books and 5 notebooks for Rs. 309 and 6 books and 2 notebooks for Rs. 282 during a closing down deal. What would be the cost of a book and a notebook?
Solution:
From the given conditions, we can frame 2 equations:
Let’s take book = a and notebook = b.
Condition 1: 3 books and 5 notebooks for Rs. 309
I.e., 3a + 5b = 309 …(1)
Condition 2: 6 books and 2 notebooks for Rs. 282
I.e., 6a + 2b = 282 …(2)
To solve the equations, multiply equation (1) by 2:
6a + 10b = 618 …(3)
Now, solve the equations (2) and (3)
6a + 2b = 282
6a + 10b = 618
(-) (-) (-)
_____________
0 – 8b = – 336
Hence, -8b = -336
b = -336/-8
b = 42
Therefore, the cost of a notebook is Rs. 42.
Now, substitute b = 42 in equation (2)
6a + 2(42) = 282
6a +84 = 282
6a = 282 – 84
6a = 198
a = 198/6
a = 33
Therefore, the cost of a book is Rs. 33.
Hence, the cost of a book and a notebook = Rs. 33 + Rs. 42
= Rs. 75.
9. Solve the following system of equations using the elimination method:
3p + 4q = 10 and 2p – 2q = 2.
Solution:
Given equations:
3p + 4q = 10 …(1)
2p – 2q = 2 …(2)
Now, multiply equation (2) by 2:
4p – 4q = 4 …(3)
Now, solve the equations (1) and (3),
3p + 4q = 10
4p – 4q = 4
____________
7p + 0 = 14
Hence, 7p = 14
p = 14/7
p = 2
Substituting p = 2 in equation (2), we get
2(2) – 2q = 2
4 – 2q = 2
-2q = 2 – 4
-2q = -2
q = -2/-2
q = 1
Hence, the solutions of the system of equations are p = 2 and q = 1.
10. Find two numbers if the addition of two numbers is 14, and their difference is 2.
Solution:
Let the two numbers be p and q.
According to the given condition, we can frame the equation as follows:
p + q = 14 …(1)
p – q = 2 …(2)
Now, solve the equations (1) and (2),
p + q = 14
p – q = 2
________
2p + 0 = 16
Thus, 2p = 16
Therefore, p = 16/2 = 8
p = 8
Now, substitute p = 8 in equation (2), we get
8 – q = 2
-q = 2 – 8
-q = -6
Hence, q = 6
Therefore, the two numbers are:
p = 8 and q = 6.
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Practice Questions
- Use the elimination method to solve the following system of linear equations: 3a – b – 7, 2a + 5b + 1 = 0.
- Find the value of p and q for the given system of linear equations using the elimination method: 2p + 3q = 11, p + 2q = 7.
- Solve the following system of linear equations by the elimination method: 3a – 5b – 4 = 0 and 9a = 2b + 7.
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