Linear Equations in One Variable: Class 8 Maths Guide

What is a Linear Equation in One Variable?

Let's break this down. First, remember what an algebraic expression is? It's a combination of numbers, variables (like x, y, t), and operations (+, -, ×, ÷). For example, 5x - 3 is an expression.

Now, an equation is a statement that two expressions are equal. It always has an 'equals' sign (=). For example, 5x - 3 = 12 is an equation. It's like a balance scale where the Left Hand Side (LHS) must equal the Right Hand Side (RHS).

So, what makes an equation a Linear Equation in One Variable?

1. Linear: This means the highest power of the variable in the equation is 1. You will see x, but not x², x³, etc.
2. One Variable: The equation contains only a single type of variable. For instance, it will only have 'x' or only 'y', but not both in the same equation.

So, an equation like 2y + 7 = 15 is a linear equation in one variable (the variable is 'y'). But x² + 4 = 8 is not linear (power is 2), and x + y = 10 is not in one variable (it has x and y).

How to Solve a Linear Equation

1. Step 1: Simplify Both Sides — If there are any brackets (parentheses), use the distributive property to remove them. Combine any like terms on the Left Hand Side (LHS) and on the Right Hand Side (RHS) separately. For example, simplify 3(x + 2) + 2x = 21 to 3x + 6 + 2x = 21, and then to 5x + 6 = 21.
2. Step 2: Move Variables to One Side — Use the method of 'transposing' to get all the variable terms on one side of the equation (usually the LHS) and all the constant terms (plain numbers) on the other side. When you move a term from one side to the other, you must change its sign. A + becomes a -, and a - becomes a +.
3. Step 3: Isolate the Variable — After combining like terms again, you will be left with something like ax = b. To find the value of x, divide both sides of the equation by the coefficient 'a'. This will leave 'x' by itself.
4. Step 4: Check Your Answer — This is a very important step! Substitute the value you found for the variable back into the original equation. If the LHS equals the RHS, your solution is correct. This helps you catch any mistakes you might have made.

Worked Examples: Solving Step-by-Step

• Example 1: Solve 2x - 3 = 7 Step 1: The equation is already simple. Step 2: We need to move the constant -3 to the RHS. Transposing -3 to the RHS makes it +3. 2x = 7 + 3 2x = 10 Step 3: Isolate x by dividing both sides by its coefficient, which is 2. x = 10 / 2 x = 5 Step 4 (Check): Substitute x = 5 into the original equation: 2(5) - 3 = 10 - 3 = 7. Since LHS (7) = RHS (7), our answer is correct.
• Example 2: Solve 5t - 3 = 3t - 5 Step 1: The equation is simple. Step 2: Move variable terms to LHS and constants to RHS. Transpose 3t to LHS (becomes -3t) and -3 to RHS (becomes +3). 5t - 3t = -5 + 3 Step 3: Combine like terms on both sides. 2t = -2 Now, isolate t by dividing both sides by 2. t = -2 / 2 t = -1 Step 4 (Check): LHS: 5(-1) - 3 = -5 - 3 = -8. RHS: 3(-1) - 5 = -3 - 5 = -8. Since LHS = RHS, the answer is correct.
• Example 3: Solve x/3 + 1 = 7/15 Step 1: The equation is simple. Step 2: Transpose the constant +1 to the RHS. It becomes -1. x/3 = 7/15 - 1 To subtract, find a common denominator: x/3 = 7/15 - 15/15 x/3 = (7 - 15) / 15 x/3 = -8/15 Step 3: To isolate x, we need to remove the /3. Multiply both sides by 3. x = (-8/15) * 3 x = -24/15 Simplify the fraction by dividing numerator and denominator by 3: x = -8/5.

Exam Tip: Watch Out for These Common Mistakes!

Solving linear equations is straightforward, but small mistakes can lead to the wrong answer. Be extra careful with these points:

• Sign Errors During Transposition: This is the most common error! Always remember to change the sign of a term when you move it across the equals sign. A positive term becomes negative, and a negative term becomes positive.
• Incorrect Distribution: When removing brackets like 3(x - 4), make sure to multiply the outside number with every term inside. It should be 3x - 34 = 3x - 12, not 3x - 4.
• Forgetting to Operate on Both Sides: The golden rule of equations is: whatever you do to one side, you must do to the other. If you divide the LHS by 2, you must also divide the entire RHS by 2.

Practice Questions with Solutions

• Q: Solve for y: 8y - 4 = 12 A: Step 1: Transpose the constant term -4 to the right side. It becomes +4. 8y = 12 + 4 8y = 16 Step 2: Isolate y by dividing both sides by the coefficient of y, which is 8. y = 16 / 8 y = 2 Final answer: y = 2
• Q: Find the value of x: x/5 + 11 = 1/15 A: Step 1: Transpose the constant +11 to the right side. It becomes -11. x/5 = 1/15 - 11 Step 2: Find a common denominator to subtract the numbers on the RHS. x/5 = 1/15 - (11 15)/15 x/5 = 1/15 - 165/15 x/5 = -164/15 Step 3: Isolate x by multiplying both sides by 5. x = (-164/15) 5 x = -164/3 Final answer: x = -164/3
• Q: Solve the equation: 3(t - 3) = 5(2t + 1) A: Step 1: Remove the brackets on both sides by using the distributive property. 3t - 33 = 52t + 51 3t - 9 = 10t + 5 Step 2: Transpose variable terms to one side and constants to the other. Let's move 3t to RHS and 5 to LHS. -9 - 5 = 10t - 3t -14 = 7t Step 3: Isolate t by dividing both sides by 7. -14 / 7 = t -2 = t Final answer: t = -2
• Q: A number is such that it is as much greater than 84 as it is less than 108. Find the number. A: Step 1: Let the unknown number be x. "Greater than 84" can be written as x - 84. "Less than 108" can be written as 108 - x. Step 2: According to the problem, these two quantities are equal. So we can form the equation: x - 84 = 108 - x Step 3: Solve the equation. Transpose -x to LHS and -84 to RHS. x + x = 108 + 84 2x = 192 Step 4: Isolate x by dividing both sides by 2. x = 192 / 2 x = 96 Final answer: The number is 96.

Frequently Asked Questions