Equations in One Variable
Class-8-Mathematics-Chapter-12-Maharashtra Board
Notes
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Topics to be learn :
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Recall :
The main goal of solving an equation is to find the value of the variable (like x) that makes the equation true. This value is called the solution.
To solve an equation, we must keep both sides equal. So, whatever we do to one side, we must do the same to the other side.
We can use these operations:
- Addition – Add the same number to both sides
- Subtraction – Subtract the same number from both sides
- Multiplication – Multiply both sides by the same number
- Division – Divide both sides by the same number (but not by zero)
How to solve simple equations:
We try to get the variable alone on one side. To do this, we use the opposite (inverse) operation:
- If a number is added, we subtract it
- If a number is subtracted, we add it
- If a number is multiplied, we divide it
- If a number is divided, we multiply it
Example (Addition):
x + 4 = 9
→ x + 4 − 4 = 9 − 4
→ x = 5
Example (Subtraction):
x − 2 = 7
→ x − 2 + 2 = 7 + 2
→ x = 9
Example (Multiplication):
4x = 24
→ 4x ÷ 4 = 24 ÷ 4
→ x = 6
Example (Division):
x ÷ 3 = 4
→ (x ÷ 3) × 3 = 4 × 3
→ x = 12
Solution of equations in one variable :
While solving an equation, sometimes we have to perform many operations on it.
Multi-Step Solving and Parentheses :
When equations are a little longer or have brackets, follow these steps:
- Remove Parentheses : Multiply the number outside the bracket with each term inside. Example: 2(x − 3) = 2x − 6
- Clear Fractions (if any) : Multiply all terms by the same number to remove fractions.
- Collect Like Terms : Bring all variables (like x) to one side and numbers to the other side.
- Make the Variable Alone : Divide by the number in front of the variable to get the answer.
Cross-Multiplication (for Fractions): When two fractions are equal, we can use cross-multiplication.
If \(\frac{A}{B}=\frac{C}{D}\) Then A × D = B × C
Example:
\(\frac{x-7}{x-2}=\frac{5}{4}\)
Step 1: Cross multiply
4(x − 7) = 5(x − 2)
Step 2: Open brackets
4x − 28 = 5x − 10
Step 3: Bring like terms together
4x − 5x = −10 + 28
Step 4: Simplify
−x = 18
Step 5: Find x
x = −18
Word Problems :
Word problems are questions written in sentences. We change them into maths using variables (like x).
- First, choose a variable (usually x) for the unknown value.
- Then write other values using x.
Changing Words into Algebra :
Let x = my age
- My age → x
- 3 times my age → 3x
- 10 more than 4 times my age → 4x + 10
- 4 less than my age → x – 4
- Half of my age + 5 → x/2 + 5
- 32 more than my age → x + 32
Applications :
Word problems are used to solve for unknowns across several domains,
Examples :
- Age Problems: Determining current or future ages based on ratios (e.g., Mother and son age ratios after 8 years).
- Geometry: Finding dimensions of rectangles using perimeter formulas (P = 2(l + b)).
- Fraction Problems: Solving for numerators and denominators when their values are altered by specific constants.
- Ratio and Proportion: Calculating weights of components in alloys (e.g., Copper and Zinc ratio in brass).
- Natural/Whole Numbers: Identifying three consecutive numbers based on their sum.
- Commercial Math: Solving for ticket sales (drama theatre), profit/loss (bicycle resale), and averages (cricket runs).
Example :
(1) A cricket player scored 180 runs in his first match and 257 runs in his second match. How many runs must he score in his third match so that his average for the three matches is 230?
Solution:
Step 1: Let the number of runs the player should score in the third match be x.
Step 2: The formula for the average of three matches is the sum of the runs divided by 3.
Step 3: Set up the equation based on the required average of 230:
\(\frac{180+257+x}{3}\) = 230.
Step 4: Simplify the sum of the first two matches: \(\frac{437+x}{3}\) = 230.
Step 5: Multiply both sides of the equation by 3 to clear the fraction:
437 + x = 230 × 3, which results in 437 + x = 690.
Step 6: Subtract 437 from both sides to find x:
x = 690 – 437 = 253
Step 7: The player should score 253 runs in the third match
(2) Joseph’s weight is two times the weight of his younger brother. Find Joseph’s weight if sum of their weights is 63 kg
Solution:
Let the weight of Joseph’s younger brother be x kg
∴ Joseph’s weight is two times the weight of his younger brother = 2x
∴ from the given information x + 2x = 63
∴ 3x = 63
∴ x = 21
∴ Joseph’s weight = 2x = 2 × 21 = 42 kg
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