Elimination Method

In mathematics, a system of linear equations consists of two or more linear equations with two or more unknown variables. Solving a system of linear equations involves determining the values of the unknown variables that meet the conditions set by all the equations at the same time. Various methods can be employed to find these solutions, including the graphical method, substitution method, elimination method, cross-multiplication method, and more. In this article, we will focus on the Elimination Method, providing a detailed explanation and step-by-step guide to solving systems of linear equations using this technique, along with illustrative examples.

1.0What is Elimination Method?

The Elimination Method is an algebraic approach for solving systems of linear equations. In this method, you add or subtract the equations to eliminate one of the variables, leading to an equation with only one variable. If the coefficients of one variable are the same but have opposite signs, you can add the equations to eliminate that variable. 

If the equations are not initially set up to directly add or subtract to eliminate a variable, you can start by multiplying one or both equations by a constant to create an equivalent system where elimination is possible. Once the coefficients of one variable are the same (either with opposite signs for addition or the same signs for subtraction), you can then subtract or add the equations to eliminate that variable.

2.0Steps to Use Elimination Method

The Elimination Method formula typically involves the following steps:

1. Adjust Coefficients: Multiply one or both equations by constants to make the coefficients of one variable (e.g., x or y) equal or opposite in sign.
2. Eliminate a Variable: Add or subtract the equations to eliminate that variable, resulting in an equation with only one variable.

• If you obtain a true statement with no variables (e.g., 0 = 0), the system has infinitely many solutions.
• If you obtain a false statement with no variables (e.g., 0 = 5), the system has no solution and is inconsistent.

3. Solve for One Variable: Solve the resulting single-variable equation to find the value of one variable.
4. Find the Other Variable: Substitute the value obtained in step 3 into one of the original equations to solve for the remaining variable.

Let's understand this process with a general example.

General Case: Consider the system of two linear equations:

ax + by = c (1)

px + qy = r (2)

To use the elimination method, follow these steps:

1. Multiply Equations to Equalize Coefficients:

• Multiply equation (1) by p:

apx + bpy = cp (3)

• Multiply equation (2) by a:

apx + aqy = ar (4)

Now, the coefficients of x in equations (3) and (4) are the same.

2. Eliminate Variable:

• Subtract equation (4) from equation (3) to eliminate x:

(a p x+b p y)-(a p x+a q y)=c p-a r

Simplifying, we get:

b p y-a q y=c p-a r

(b p-a q) y=c p-a r

$y=\frac{cp-ar}{bp-aq}$

3. Solve for the Other Variable:

• From equation (1):

a x+b y=c

Solving for x:

a x=c-b y

$\Rightarrow x=\frac{c-by}{a}$

Putting the value of y

$\Rightarrow x=\frac{c-b\left(\frac{cp-ar}{bp-aq}\right)}{a}$

$\Rightarrow \mathrm{x}=\frac{\mathrm{c}(\mathrm{bp}-\mathrm{aq})-\mathrm{b}(\mathrm{cp}-\mathrm{ar})}{\mathrm{a}(\mathrm{bp}-\mathrm{aq})}$

$\Rightarrow \mathrm{x}=\frac{\mathrm{cbp}-\mathrm{caq}-\mathrm{bcp}+\mathrm{bar}}{\mathrm{a}(\mathrm{bp}-\mathrm{aq})}$

$\Rightarrow \mathrm{x}=\frac{\mathrm{a}(\mathrm{br}-\mathrm{cq})}{\mathrm{a}(\mathrm{pb}-\mathrm{aq})}=\frac{\mathrm{br}-\mathrm{cq}}{\mathrm{bp}-\mathrm{aq}}$

By following these steps, you can find the values of x and y in the given system of linear equations.

3.0System of 3 Equations Using Elimination Method

Solving a system of three equations using the Elimination Method involves extending the technique used for two equations. Here's a step-by-step outline:

1. Choose Variables to Eliminate: Decide which variables to eliminate first. The goal is to reduce the system to equations with two variables.
2. Eliminate One Variable: Proceed by eliminating one variable at a time across the equations. You can achieve this by adding or subtracting equations to eliminate the chosen variable.
3. Reduce to Two Equations: After eliminating the first variable, you should have a simplified system of two equations with two variables left.
4. Apply Elimination Again: Use the same method to eliminate another variable to reduce the system to a single equation with one variable.
5. Solve for Variables: Once you have a single equation with one variable, solve for that variable.
6. Back-Substitute: Substitute the found values back into the other equations to find the remaining variables.

4.0Elimination Method Solved Examples

Example 1: Consider the system of equations: $\{\begin{array}{l}3x+4y=10\\ 5x-6y=-2\end{array}$

Solution: Step 1: Adjust Coefficients

To eliminate y, we need the coefficients of y to be equal (with opposite signs). We can achieve this by multiplying the first equation by 3 and the second equation by 2:

$\{\begin{array}{l}3(3x+4y)=3(10)\\ 2(5x-6y)=2(-2)\end{array}$

This results in:

$\{\begin{array}{l}9x+12y=30\\ 10x-12y=-4\end{array}$

Step 2: Eliminate One Variable

Now, add the two equations to eliminate y:

(9 x+12 y)+(10 x-12 y)=30-4

Simplifying, we get:

$19x=26\Rightarrow x=\frac{26}{19}$

Step 3: Solve for the Remaining Variable

Substitute $x=\frac{26}{19}$ back into the first equation:

$3\left(\frac{26}{19}\right)+4y=10$

Simplifying:

$\frac{78}{19}+4y=10\Rightarrow 4y=10-\frac{78}{19}\Rightarrow 4y=\frac{112}{19}\Rightarrow y=\frac{28}{19}$

Solution:

The solution to the system is:

$\left(\frac{26}{19},\frac{28}{19}\right)$

Example 2: Consider the system of equations: $\{\begin{array}{l}2x+3y=8\\ 4x-y=2\end{array}$

Solution: Step 1: Adjust Coefficients

To eliminate y, we can multiply the second equation by 3:

$\{\begin{array}{l}2x+3y=8\\ 12x-3y=6\end{array}$

Step 2: Eliminate One Variable

Add the two equations to eliminate y:

(2x + 3y) + (12x – 3y) = 8 + 6 

Simplifying, we get:

$14x=14\Rightarrow x=1$

Step 3: Solve for the Remaining Variable

Substitute x = 1 back into the first equation:

$2(1)+3y=8\Rightarrow 2+3y=8\Rightarrow 3y=6\Rightarrow y=2$

Solution:

The solution to the system is:

(1, 2) 

Example 3:  Solve 3x + 4y = 10 and 2x– 2y = 2

Solution: 3x + 4y = 10 …… (1)

Let us eliminate y so multiple equation (1) by 1 and equation (2) by 2 and then add

$\begin{array}{rl}3x+4y& =10\\ +4x-4y& =4\\ 7x& =14\end{array}$

$x=\frac{14}{7}=2$

$\Rightarrow x=2$

Putting the value of x = 2 in equation (1)

3x + 4y = 10

⇒ 3(2) + 4y = 10

⇒ 6 + 4y = 10

⇒ 4y = 10 – 6 = 4

$\Rightarrow y=1$

So, x = 2, y = 1

Example 5: Solve 3x – 5y –4 = 0 and 9x = 2y + 7

Solution:

3x –5y – 4 = 0 ⇒ 3x – 5y = 4 …………. (1)

9x = 2y + 7 ⇒ 9x – 2y = 7 …………. (2)

Let us eliminate x. So, multiply equation 1 by 3 and equation 2 by 1 and then subtract

(3x–5y = 4) ×3

(9x–2y = 7) × 1

$\begin{array}{r}9x-15y=12\\ +9x-2y=7\\ -+-\\ -13y=5\end{array}$

$\mathrm{y}=-\frac{5}{13}$

Putting value of y in (1)

3x – 5y = 4

$\Rightarrow 3\mathrm{x}-5\left(\frac{-5}{13}\right)=4$

$\Rightarrow 3x+\frac{25}{13}=4\Rightarrow 3x=4-\frac{25}{13}$

$\Rightarrow 3\mathrm{x}=\frac{52-25}{13}=\frac{27}{13}$

$\Rightarrow x=\frac{27}{13\times 3}=\frac{9}{13}$

So $x=\frac{9}{13}andy=\frac{-5}{13}$

Example 6: Solve $\frac{x}{2}+\frac{2y}{3}=-1andx-\frac{y}{3}=3$

$\frac{x}{2}+\frac{2y}{3}=-1$ ⇒ 3x +4y = –6 …..(1)

$x-\frac{y}{3}=3$ ⇒ 3x –y = 9 = …..(2) 

Let's eliminate x. Subtract equation (2) from equation (1)

$\begin{array}{r}3x+4y=-6\\ 3x-y=9\\ -+-\\ 5y=-15\end{array}$

$y=-\frac{15}{5}=-3$

y=-3

Putting value in eq (1)

3x + 4 (–3) = –6 ⇒ 3x – 12 = – 6

⇒ 3x = – 6 + 12 ⇒ 3x = $6\Rightarrow \mathrm{x}=2$

So, x = 2 and y = –3

Example 7: Solve x + 2y –z = 2, 2x – 3y +z = –1 and 5x–y –2z = –3

Solution :

 x + 2y – z = 2 ..... (1)

2x –3y + z = –1 ..... (2)

5x–y – 2z= –3 ..... (3)

Let us first eliminate z, so Add equation (1) and equation (2)

$\begin{array}{c}x+2y-z=2\\ 2x-3y+z=-1\\ 3x-y=1\end{array}$

3x – y = 1 … (4)

Now we will eliminate z from equation (2) and equation (3). So Multiply equation (2) by 2 and add equation (3)

(2x–3y + z = –1)×2 ⇒ 

(5x–y–2z = –3) ×1 ⇒  

$\begin{array}{r}4x-6y+2z=-2\\ +5x-y+2z=-3\\ 9x-7y=-5\end{array}$

9x – 7y = – 5 … (5)

Now, solve equation (4) and equation (5) by elimination method

So let us eliminate x so multiply equation 4 and subject 

(3x – y = 1) × 3

(9x–7y = –5) ×1 

\begin{aligned}

& 9 x-3 y=3 \\

& 9 x-7 y=-5 \\

& -\quad+\quad+ \\

& \hline 4 y=8

\end{aligned}

y=\frac{8}{4}

$\Rightarrow y=2$

Putting value of y in equation 4

3x – (2) = 1 ⇒ 3x –2 = 1 ⇒ 3x = 3

$\Rightarrow x=1$

Putting the value of x and y in equation (1)

x + 2 y – z = 2

⇒ (1) + 2(2) – z = 2

⇒ 1 + 4 – z = 2

⇒ –z = 2 –5

⇒ –z = –3   

$\Rightarrow \mathrm{z}=3$

So, x = 1, y = 2 and z = 3 are the solution of linear equation with 3 variables. 

5.0Elimination Method Practice problems

1. Solve 3x – 5y – 4 = 0 and 9x = 2y + 7.
2. Solve 4x + y = 17 and 2x + y = 9.
3. Solve 3x + 2y = 11 and 2x – 2y = 14.
4. Solve 2x – 3y = –2 and 3x –2y = 12.
5. Solve x + 2y + z = 12, 2x – 2y – z = –6 and x + 2y – z = 2.