Solving by Substitution MCQs Quiz | Class 10

Understanding the Substitution Method for Linear Equations

The substitution method is a powerful algebraic technique used to solve systems of linear equations. It involves solving one of the equations for one variable in terms of the other, and then substituting that expression into the second equation. This reduces the system to a single linear equation with one variable, which is then straightforward to solve.

Key Steps of the Substitution Method:

1. Isolate a Variable: Choose one of the two given linear equations and express one variable (e.g., x) in terms of the other variable (e.g., y). This means you rearrange the equation to get `x = (an expression involving y)` or `y = (an expression involving x)`.
2. Substitute: Take the expression obtained in step 1 and substitute it into the other equation. This will result in a new equation that has only one variable.
3. Solve for the First Variable: Solve the single-variable equation obtained in step 2. This will give you the numerical value of one of the variables.
4. Substitute Back: Substitute the numerical value found in step 3 back into the expression from step 1 (where one variable was expressed in terms of the other). This will give you the numerical value of the second variable.
5. Check the Solution: (Optional but recommended) Substitute both numerical values back into both original equations to ensure they satisfy both equations.

Example Walkthrough:

Solve the system:

• `x + y = 7`
• `x – y = 3`

Step 1: Isolate a variable.
From equation (1), we can express `x` in terms of `y`: `x = 7 – y`.

Step 2: Substitute.
Substitute `(7 – y)` for `x` in equation (2):
`(7 – y) – y = 3`

Step 3: Solve for the first variable.
`7 – 2y = 3`
`-2y = 3 – 7`
`-2y = -4`
`y = 2`

Step 4: Substitute back.
Substitute `y = 2` into `x = 7 – y`:
`x = 7 – 2`
`x = 5`

Solution: The solution is `x = 5, y = 2`.

Applications of the Substitution Method:

The substitution method is widely used to solve real-world problems that can be modeled using systems of linear equations. Some common applications include:

• Age Problems: Determining the current ages of individuals based on given conditions relating their ages.
• Cost and Price Problems: Finding the individual cost or price of multiple items when given total costs and quantities.
• Number Problems: Solving for unknown numbers based on relationships like their sum, difference, or products.
• Mixture Problems: Determining the quantities of different components in a mixture.
• Distance, Speed, and Time Problems: Analyzing scenarios involving travel.

Quick Revision Points:

• Goal: To eliminate one variable by expressing it in terms of the other.
• Choice: Select the equation and variable that is easiest to isolate (e.g., a variable with a coefficient of 1 or -1).
• Substitution: Always substitute the expression into the other equation.
• Verification: Always check your answers by plugging them back into both original equations.

Practice Questions (No Options):

1. Solve for x and y:
   • `x = 2y + 3`
   • `4x – 5y = 12`
2. Solve for m and n:
   • `m – 3n = 1`
   • `2m + n = 12`
3. The sum of two numbers is 20, and their difference is 4. Find the numbers.
4. The cost of 2 tables and 3 chairs is Rs 1800. If the cost of a table is Rs 150 more than the cost of a chair, find the cost of each table and chair.
5. Solve the system:
   • `2x + y = 10`
   • `3x – 2y = 1`