NCERT Solutions: Linear Equations in Two Variables Ex 3.3

What is the Substitution Method?

Imagine you have two equations, each with two unknown values, 'x' and 'y'. Our goal is to find a single pair of values (x, y) that makes both equations true. The Substitution Method achieves this by cleverly converting the pair of two-variable equations into a single one-variable equation.

The main idea is to 'isolate' one variable in one of the equations. This means rewriting the equation so it says 'x = ...' or 'y = ...'. This '...' part is an expression involving the other variable. You then take this expression and 'substitute' it into the other equation in place of the variable you solved for. This leaves you with one equation that only has one variable, which you already know how to solve! Once you find the value of that first variable, finding the second one is a simple plug-and-play.

Solving by Substitution: A Step-by-Step Guide

1. Step 1: Isolate a Variable — Pick one of the two equations. Choose the one where it's easiest to get either 'x' or 'y' by itself on one side of the equals sign. (Hint: Look for a variable with a coefficient of 1 or -1).
2. Step 2: Substitute the Expression — Take the expression you got from Step 1 and substitute it into the other equation. This is the most crucial step. You will now have an equation with only one variable.
3. Step 3: Solve the New Equation — Solve the single-variable linear equation you created in Step 2 to find the value of that variable (either x or y).
4. Step 4: Find the Second Variable — Take the value you found in Step 3 and plug it back into the 'isolated' expression from Step 1. This will give you the value of the other variable.
5. Step 5: Verify Your Solution — Always check your answer! Substitute the (x, y) pair you found into both of the original equations to make sure they hold true. This confirms your solution is correct.

Worked Example: Applying the Substitution Method

• Problem: Solve the pair of linear equations: x + y = 14 --- (1) x - y = 4 --- (2) Solution: Step 1: Isolate a Variable From equation (2), it is easy to isolate 'x'. x = 4 + y --- (3) Step 2: Substitute the Expression Substitute the value of 'x' from equation (3) into equation (1). (4 + y) + y = 14 Step 3: Solve the New Equation Now, solve for 'y'. 4 + 2y = 14 2y = 14 - 4 2y = 10 y = 5 Step 4: Find the Second Variable Substitute the value y = 5 into equation (3). x = 4 + 5 x = 9 Step 5: Verify Your Solution The solution is (x, y) = (9, 5). Let's check in both original equations: Equation (1): 9 + 5 = 14 (True) Equation (2): 9 - 5 = 4 (True) Since the solution satisfies both equations, it is correct. Final Answer: The solution is x = 9 and y = 5.

Exam Tip: Watch Out for These Common Errors!

When using the substitution method, students often make a few common mistakes. Be careful to avoid these:

• Substituting into the wrong equation: A classic error is to substitute the expression back into the same equation you derived it from. This will just lead you to an identity like 5 = 5 and won't help you solve the system.
• Sign errors: Be extremely careful with negative signs when rearranging equations and substituting expressions. A small sign mistake can lead to a completely wrong answer.
• Stopping halfway: Don't forget to find the value of the second variable! A complete solution is a pair of values (x, y), not just a single value for x or y.
• Skipping verification: Always take a minute to plug your final (x, y) solution back into both original equations. This is the best way to catch any calculation errors you might have made.

Practice Questions with Solutions

• Q: Solve the following pair of linear equations by the substitution method: 7x - 15y = 2 and x + 2y = 3. A: Step 1: From the second equation, x + 2y = 3, we can easily isolate x: x = 3 - 2y. Step 2: Substitute this expression for x into the first equation: 7(3 - 2y) - 15y = 2. Step 3: Solve for y: 21 - 14y - 15y = 2 => 21 - 29y = 2 => -29y = -19 => y = 19/29. Step 4: Substitute y = 19/29 back into x = 3 - 2y: x = 3 - 2(19/29) => x = 3 - 38/29 => x = (87 - 38)/29 => x = 49/29. Final answer: x = 49/29 and y = 19/29.
• Q: Solve for s and t: s - t = 3 and s/3 + t/2 = 6. A: Step 1: From the first equation, isolate s: s = 3 + t. Step 2: Substitute this into the second equation: (3 + t)/3 + t/2 = 6. To clear the fractions, find the LCM of 3 and 2, which is 6. Multiply the entire equation by 6: 2(3 + t) + 3(t) = 36. Step 3: Solve for t: 6 + 2t + 3t = 36 => 5t = 30 => t = 6. Step 4: Substitute t = 6 back into s = 3 + t: s = 3 + 6 => s = 9. Final answer: s = 9 and t = 6.
• Q: Solve the pair of equations: 3x - y = 3 and 9x - 3y = 9. A: Step 1: From the first equation, 3x - y = 3, isolate y: y = 3x - 3. Step 2: Substitute this expression for y into the second equation: 9x - 3(3x - 3) = 9. Step 3: Solve the equation: 9x - 9x + 9 = 9 => 9 = 9. This is a true statement, but the variables have disappeared. Step 4: This result means that the two equations are dependent (one is a multiple of the other). They represent the same line. Final answer: The system has infinitely many solutions.
• Q: Solve: 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3. A: Step 1: To remove decimals, multiply both equations by 10. The system becomes 2x + 3y = 13 (1) and 4x + 5y = 23 (2). Step 2: From equation (1), isolate x: 2x = 13 - 3y => x = (13 - 3y) / 2. Step 3: Substitute this into equation (2): 4((13 - 3y) / 2) + 5y = 23 => 2(13 - 3y) + 5y = 23. Step 4: Solve for y: 26 - 6y + 5y = 23 => 26 - y = 23 => -y = -3 => y = 3. Step 5: Substitute y = 3 into x = (13 - 3y) / 2: x = (13 - 3*3) / 2 => x = (13 - 9) / 2 => x = 4 / 2 => x = 2. Final answer: x = 2 and y = 3.

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