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Linear Equations in One Variable - Applications of Linear Equations

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Translating Verbal Statements: The core of application problems is converting word phrases into mathematical expressions. Visualize an equation as a balanced weighing scale; the 'is' or 'equals' in a sentence represents the central pivot point where the left-hand side (LHS) must perfectly balance the right-hand side (RHS).

Defining Variables: Always identify the unknown quantity you need to find and represent it with a variable, usually xx. If there are two unknown quantities, try to express the second one in terms of xx to keep the equation in one variable. Imagine a labeled container xx representing the hidden value.

Age-Related Problems: These involve comparing ages at different points in time. Visualize a horizontal timeline: if the present age is xx, moving to the left represents 'years ago' (xnx - n) and moving to the right represents 'years hence' or 'future' (x+nx + n).

Number and Digit Problems: For a two-digit number, the value depends on the place value of its digits. Visualize a place-value grid with 'Tens' and 'Units' columns; if the tens digit is xx and units digit is yy, the value is 10x+y10x + y. If the digits are reversed, they swap columns to become 10y+x10y + x.

Consecutive Integers: These are numbers that follow each other in a sequence without gaps. Visualize them as steps on a ladder where each step is 1 unit higher than the previous: xx, x+1x+1, x+2x+2. For consecutive even or odd numbers, the steps are 2 units apart: xx, x+2x+2, x+4x+4.

Geometric Applications: Linear equations often solve for dimensions of shapes. Visualize a rectangle where the length is ll and width is ww; the perimeter is the total boundary length. If the length is given in terms of width, you can represent the entire perimeter as a single-variable equation like 2(x+expression in x)=Total Perimeter2(x + \text{expression in } x) = \text{Total Perimeter}.

The Transposition Method: When solving the final equation, imagine moving terms across the equal sign bridge. When a term crosses the 'bridge', its sign flips: ++ becomes -, and multiplication becomes division (and vice versa) to maintain the scale's balance.

📐Formulae

Standard form of a linear equation: ax+b=cax + b = c

Perimeter of a Rectangle: P=2(l+w)P = 2(l + w) where ll is length and ww is width

Distance Formula: D=S×TD = S \times T (Distance = Speed ×\times Time)

Digit Place Value: Number=(10×Tens Digit)+(1×Units Digit)\text{Number} = (10 \times \text{Tens Digit}) + (1 \times \text{Units Digit})

Sum of nn consecutive integers: x+(x+1)+(x+2)+...+(x+n1)=Sx + (x+1) + (x+2) + ... + (x + n-1) = S

💡Examples

Problem 1:

The perimeter of a rectangular swimming pool is 154154 m. Its length is 22 m more than twice its breadth. What are the length and the breadth of the pool?

Solution:

  1. Let the breadth of the pool be xx m.
  2. According to the problem, the length is 22 m more than twice the breadth, so: Length=2x+2\text{Length} = 2x + 2.
  3. Use the perimeter formula: P=2(length+breadth)P = 2(\text{length} + \text{breadth}).
  4. Substitute the known values: 154=2((2x+2)+x)154 = 2((2x + 2) + x).
  5. Simplify inside the parentheses: 154=2(3x+2)154 = 2(3x + 2).
  6. Divide both sides by 22: 77=3x+277 = 3x + 2.
  7. Subtract 22 from both sides: 75=3x75 = 3x.
  8. Divide by 33: x=25x = 25.
  9. Therefore, Breadth=25\text{Breadth} = 25 m and Length=2(25)+2=52\text{Length} = 2(25) + 2 = 52 m.

Explanation:

We first defined the breadth as the variable xx because the length was described in relation to it. By substituting these expressions into the standard perimeter formula, we created a solvable linear equation.

Problem 2:

Aman's age is three times his son's age. Ten years ago he was five times his son's age. Find their present ages.

Solution:

  1. Let the son's present age be xx years.
  2. Then Aman's present age is 3x3x years.
  3. Ten years ago, the son's age was x10x - 10 and Aman's age was 3x103x - 10.
  4. According to the condition (Aman was 55 times his son's age): 3x10=5(x10)3x - 10 = 5(x - 10).
  5. Expand the right side: 3x10=5x503x - 10 = 5x - 50.
  6. Transpose 3x3x to the RHS and 50-50 to the LHS: 5010=5x3x50 - 10 = 5x - 3x.
  7. 40=2x40 = 2x.
  8. Divide by 22: x=20x = 20.
  9. Son's age is 2020 years and Aman's age is 3×20=603 \times 20 = 60 years.

Explanation:

This problem uses a timeline approach. By setting the son's age as xx, we established the ages for both the present and 10 years ago. The relationship given for the past allowed us to form the equation.