Methods of Solving Linear Equations: Substitution & Elimination

The Substitution and Elimination methods are powerful techniques for solving systems of linear equations. A system of linear equations is a set of two or more linear equations containing the same variables. The goal is to find the values of the variables that satisfy all equations in the system.

The Substitution Method involves solving one equation for one variable and then substituting that expression into the other equation. This results in a single equation with one variable, which can then be solved. Finally, the value of that variable is substituted back into one of the original equations to find the value of the other variable.

The Elimination Method (also known as the addition or subtraction method) involves manipulating the equations (by multiplying them by constants) to make the coefficients of one of the variables opposites. Then, the equations are added together, eliminating that variable. This leaves a single equation with one variable, which can be solved. The value of that variable is then substituted back into one of the original equations to find the value of the other variable.

Formulae

There aren’t specific “formulae” in the same way as a single equation like $y = mx + b$. Instead, these methods involve a series of algebraic manipulations.

Substitution Method:

1. Solve one equation for one variable in terms of the other.
2. Substitute the expression from step 1 into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found in step 3 back into either of the original equations to find the value of the other variable.

Elimination Method:

1. Multiply one or both equations by a constant so that the coefficients of one of the variables are opposites.
2. Add the equations together to eliminate one variable.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found in step 3 back into either of the original equations to find the value of the other variable.

Examples

Example-1: Substitution Method

Solve the following system of equations:

$x + y = 5$
$x – y = 1$

1. Solve the first equation for $x$: $x = 5 – y$
2. Substitute $(5 – y)$ for $x$ in the second equation: $(5 – y) – y = 1$
3. Solve for $y$: $5 – 2y = 1 \Rightarrow -2y = -4 \Rightarrow y = 2$
4. Substitute $y = 2$ back into $x = 5 – y$: $x = 5 – 2 \Rightarrow x = 3$

Solution: $x = 3, y = 2$

Example-2: Elimination Method

Solve the following system of equations:

$2x + y = 7$
$x – y = 2$

1. The coefficients of $y$ are already opposites (+1 and -1), so add the equations together.
2. Add the equations: $(2x + y) + (x – y) = 7 + 2 \Rightarrow 3x = 9$
3. Solve for $x$: $x = 3$
4. Substitute $x = 3$ back into $x – y = 2$: $3 – y = 2 \Rightarrow y = 1$

Solution: $x = 3, y = 1$

Common mistakes by students

• Incorrect substitution: Failing to correctly substitute the expression into the other equation.
• Sign errors: Making mistakes when adding or subtracting terms, especially when dealing with negative signs.
• Multiplying incorrectly: Not multiplying all terms of an equation when using the elimination method.
• Not solving for both variables: Stopping after finding the value of one variable and forgetting to solve for the other.
• Forgetting to check the solution: Not checking their answers by substituting the values back into the original equations.

Real Life Application

Systems of equations have many real-world applications:

• Budgeting: Calculating the quantities of different items that can be purchased with a limited budget, given their prices.
• Mixture problems: Determining the amounts of different substances to combine to create a specific mixture.
• Business and Economics: Analyzing supply and demand curves, profit maximization, and cost analysis.
• Engineering: Solving systems of equations to model the behavior of structures, circuits, and other physical systems.

Fun Fact

The concept of solving systems of equations dates back to ancient civilizations! The Babylonians, for example, used methods similar to the elimination method as early as 2000 BC.

Recommended YouTube Videos for Deeper Understanding

Q.1 Consider the system of equations: $x + y = 5$ and $2x – y = 1$. Using the substitution method, what is the value of $x$?

Check Solution

Ans: B

From the first equation, we get $y = 5 – x$. Substituting this into the second equation gives $2x – (5 – x) = 1$. Simplifying, $2x – 5 + x = 1$, which leads to $3x = 6$, so $x = 2$.

Q.2 Solve for $y$ in the system of equations: $3x – 2y = 7$ and $x = y + 1$ using the substitution method.

Check Solution

Ans: D

Substitute $x = y + 1$ into the first equation: $3(y + 1) – 2y = 7$. This simplifies to $3y + 3 – 2y = 7$, hence $y + 3 = 7$, so $y = 4$.

Q.3 Using the elimination method, solve for $x$ in the system of equations: $x + 2y = 7$ and $x – y = 1$.

Check Solution

Ans: C

Subtract the second equation from the first: $(x + 2y) – (x – y) = 7 – 1$, which simplifies to $3y = 6$, so $y = 2$. Substituting back into the second equation: $x – 2 = 1$, so $x = 3$.

Q.4 Find the value of $y$ in the system $2x + y = 4$ and $4x + 3y = 10$ using the elimination method.

Check Solution

Ans: B

Multiply the first equation by -2 to get $-4x – 2y = -8$. Add this to the second equation: $(-4x – 2y) + (4x + 3y) = -8 + 10$. This simplifies to $y = 2$.

Q.5 Solve for $x$ in the system of equations: $5x – 2y = 13$ and $3x + y = 10$ using the elimination method.

Check Solution

Ans: C

Multiply the second equation by 2 to get $6x + 2y = 20$. Add this to the first equation: $(5x – 2y) + (6x + 2y) = 13 + 20$. This simplifies to $11x = 33$, so $x = 3$.

Next Topic: Equations Reducible to Linear Form