Elimination Method Questions

The elimination method is a fundamental algebraic technique used to solve systems of linear equations by eliminating one variable through addition or subtraction. It is commonly taught in high school math and appears frequently in competitive exams like JEE. By aligning coefficients and performing simple operations, this method simplifies solving two or more equations efficiently. Whether in basic algebra or advanced Gauss elimination for matrices, elimination method questions help build strong problem-solving skills in linear systems.

1.0What Is the Elimination Method?

The elimination method involves adding or subtracting equations to eliminate one of the variables, allowing you to solve for the other. It’s especially effective for solving systems of two or three linear equations.

2.0Steps to Solve Elimination Method Questions

1. Arrange equations in standard form: Ax+By=CAx + By = C
2. Multiply one or both equations (if necessary) to get equal coefficients of one variable.
3. Add or subtract the equations to eliminate that variable.
4. Solve for the remaining variable.
5. Substitute the found value back into one of the original equations to find the other variable.

3.0Elimination Method Questions with Answers 

Question 1: Solve the system:

$3x+4y=10$... (1)

$2x-4y=6$.....(2)

Solution:

Add (1) and (2):

$(3x+4y)+(2x-4y)=10+6$

$\Rightarrow 5x=16\Rightarrow x=\frac{16}{5}$

Substitute into (1):

$3\cdot \frac{16}{5}+4y=10$

$\frac{48}{5}+4y=10$

$4y=10-\frac{48}{5}=\frac{2}{5}$

$y=\frac{1}{10}$

Answer: $x=\frac{16}{5},y=\frac{1}{10}$

Question 2: Solve the system using elimination:

$x+2y=8$.......(1)

$2x-3y=-1$...(2)

Solution:

Multiply equation (1) by 2:

$2x+4y=16$......(3)

Subtract (2) from (3):

$(2x+4y)-(2x-3y)=16+1$

$\Rightarrow 7y=17$

$\Rightarrow y=\frac{17}{7}$

Substitute into (1):

$x+2.\frac{17}{7}=8$

$\Rightarrow x=8-\frac{34}{7}=\frac{22}{7}$

Answer: $x=\frac{22}{7},y=\frac{17}{7}$

Example 3: Solve: 

$4x+5y=9$

$6x-5y=1$

Solution: 

$4x+5y=9$....(1)

$6x-5y=1$.....(2)

Step 1: Add equations (1) and (2):

$(4x+5y)+(6x-5y)=9+1$

$\Rightarrow 10x=10$

$\Rightarrow x=1$

Step 2: Substitute into (1):

$4(1)+5y=9$

$\Rightarrow 5y=5$

$\Rightarrow y=1$

Answer: x = 1, y = 1

Example 4: Solve: 

$3x+2y=7$

$5x+4y=13$ 

Solution:

3x + 2y = 7 .....(1)

5x + 4y = 13.....(2)  

Step 1: Multiply (1) by 2 to match the coefficients of yy: 

$6x+4y=14(3)$

Step 2: Subtract (2) from (3):

$(6x+4y)-(5x+4y)=14-13$

$\Rightarrow x=1$ 

Step 3: Substitute into (1):

$3(1)+2y=7$

$\Rightarrow 2y=4$

$\Rightarrow y=2$

Answer: x = 1, y = 2

 Example 5: Solve:

$x+y+z=3$

$2x+3y+z=7$

$x+2y+3z=8$

Solution:  

Augmented Matrix: $\left[\begin{array}{cccc}1& 1& 1& 3\\ 2& 3& 1& 7\\ 1& 2& 3& 8\end{array}\right]$

Row operations to eliminate variables (not shown in full here):

After transforming to upper triangular form and back-substitution:

Answer: x = 1, y = 1, z = 1

4.0Practice Questions on Elimination Method

1. Solve using elimination:

$2x+3y=13$

$3x-y=5$

2. Solve using elimination:

$x+2y=10$

$2x+4y=20$

(Hint: What happens if the equations are dependent?)

3. Use Gauss elimination to solve:

$x+y+z=6$

$x+2y+3z=14$

$2x+3y+z=10$

4. Solve the system:

$5x+2y=12$

$3x-y=4$

5. Solve using Gauss elimination:

$2x+y-z=3$

$x-y+2z=0$

$3x+2y+z=5$