Reducing Equations to Simpler Form
Many linear equations appear complicated at first — they may contain fractions, decimals, brackets within brackets, or variables in the denominator. But these can all be reduced to the simple form ax + b = 0.
The process of reducing means transforming a complex-looking equation into a simpler equivalent equation that is easier to solve.
This topic from Linear Equations in Grade 8 covers techniques for simplifying equations involving fractions, cross-multiplication, LCM multiplication, and nested brackets before solving.
Consider the equation (2x + 1)/3 − (x − 2)/5 = 7. At first glance, this looks difficult. But by multiplying both sides by 15 (the LCM of 3 and 5), all fractions disappear, and you get a simple equation: 5(2x + 1) − 3(x − 2) = 105. Expanding gives 10x + 5 − 3x + 6 = 105, which simplifies to 7x + 11 = 105, and then x = 94/7.
The process of reducing does not change the solution — it only changes the appearance of the equation. The reduced equation is equivalent to the original (same solution), but much easier to work with.
In Grade 8, you learn four key reduction techniques:
- LCM method — for equations with multiple fractions.
- Cross-multiplication — for equations with one fraction on each side.
- Bracket expansion — for nested brackets.
- Decimal clearing — for equations with decimals.
What is Reducing Equations to Simpler Form - Grade 8 Maths (Linear Equations)?
Definition: Reducing an equation to simpler form means transforming it into an equivalent equation that is easier to solve, typically by removing fractions, decimals, or nested brackets.
Common forms that need reduction:
- Equations with fractions: (x + 1)/3 + (x - 2)/5 = 7
- Equations with variable in denominator: 5/(x + 1) = 3/7
- Equations with nested brackets: 3[2(x + 1) - 4] = 18
- Equations with decimals: 0.3x + 1.5 = 0.7x - 0.9
Goal: Reduce any of the above to the form ax + b = 0 or ax = c, then solve.
Why reduce first?
- Fractions are harder to work with than whole numbers. Reducing removes them.
- Nested brackets make equations hard to read. Expansion simplifies them.
- Decimals can lead to rounding errors. Converting to whole numbers avoids this.
- A reduced equation in the form ax = c is trivially solved by division.
Equivalence:
- Two equations are equivalent if they have exactly the same solution.
- Multiplying both sides by a non-zero number gives an equivalent equation.
- Adding or subtracting the same quantity from both sides gives an equivalent equation.
- The reduced equation and the original equation are always equivalent (assuming no division by zero).
Methods
Techniques for reducing equations:
Technique 1: LCM Method (for equations with fractions)
- Find the LCM of all denominators.
- Multiply every term on both sides by the LCM.
- This eliminates all fractions.
- Solve the resulting simple equation.
Technique 2: Cross-Multiplication (for proportions)
- If the equation is in the form a/b = c/d, cross-multiply: a x d = b x c.
- This removes both fractions at once.
Technique 3: Removing Nested Brackets
- Work from the innermost bracket outward.
- Expand each bracket using the distributive property.
- Combine like terms at each step.
Technique 4: Removing Decimals
- Multiply both sides by a power of 10 (10, 100, 1000...) to eliminate all decimal places.
- Solve the resulting whole-number equation.
Technique 5: Variable in the Denominator
- If the variable is in a denominator, cross-multiply to remove it.
- Or multiply both sides by the expression containing the variable.
- Important: After solving, verify that the solution does not make any denominator zero.
Solved Examples
Example 1: Example 1: LCM method — two fractions
Problem: Solve: x/2 + x/3 = 5
Solution:
Given: x/2 + x/3 = 5
Steps:
- LCM of 2 and 3 = 6
- Multiply every term by 6: 6(x/2) + 6(x/3) = 6(5)
- Simplify: 3x + 2x = 30
- Combine: 5x = 30
- Divide: x = 6
Verify: 6/2 + 6/3 = 3 + 2 = 5. Correct.
Answer: x = 6.
Example 2: Example 2: LCM method — three fractions
Problem: Solve: x/3 + x/4 + x/5 = 47
Solution:
Given: x/3 + x/4 + x/5 = 47
Steps:
- LCM of 3, 4, 5 = 60
- Multiply every term by 60: 20x + 15x + 12x = 2820
- Combine: 47x = 2820
- Divide: x = 60
Verify: 60/3 + 60/4 + 60/5 = 20 + 15 + 12 = 47. Correct.
Answer: x = 60.
Example 3: Example 3: Cross-multiplication
Problem: Solve: (2x + 3)/5 = (3x - 1)/7
Solution:
Given: (2x + 3)/5 = (3x - 1)/7
Steps:
- Cross-multiply: 7(2x + 3) = 5(3x - 1)
- Expand: 14x + 21 = 15x - 5
- Transpose: 14x - 15x = -5 - 21
- Simplify: -x = -26
- x = 26
Verify: LHS = (52 + 3)/5 = 55/5 = 11. RHS = (78 - 1)/7 = 77/7 = 11. LHS = RHS.
Answer: x = 26.
Example 4: Example 4: Variable in the denominator
Problem: Solve: 5/(x - 2) = 3/4
Solution:
Given: 5/(x - 2) = 3/4
Steps:
- Cross-multiply: 5 x 4 = 3(x - 2)
- Simplify: 20 = 3x - 6
- Transpose: 3x = 20 + 6 = 26
- x = 26/3
Check denominator: x - 2 = 26/3 - 2 = 20/3, which is not zero. Valid.
Verify: LHS = 5/(20/3) = 5 x 3/20 = 15/20 = 3/4. RHS = 3/4. LHS = RHS.
Answer: x = 26/3.
Example 5: Example 5: Nested brackets
Problem: Solve: 2[3(x + 1) - 5] = 16
Solution:
Given: 2[3(x + 1) - 5] = 16
Steps (inside out):
- Expand inner bracket: 3(x + 1) = 3x + 3
- Replace: 2[3x + 3 - 5] = 16
- Simplify inside: 2[3x - 2] = 16
- Expand outer bracket: 6x - 4 = 16
- Transpose: 6x = 20
- x = 20/6 = 10/3
Verify: Inner: 3(10/3 + 1) - 5 = 3(13/3) - 5 = 13 - 5 = 8. Outer: 2(8) = 16. Correct.
Answer: x = 10/3.
Example 6: Example 6: Removing decimals
Problem: Solve: 0.3x + 1.5 = 0.7x - 0.9
Solution:
Given: 0.3x + 1.5 = 0.7x - 0.9
Steps:
- Multiply every term by 10: 3x + 15 = 7x - 9
- Transpose: 3x - 7x = -9 - 15
- Simplify: -4x = -24
- x = 6
Verify: LHS = 0.3(6) + 1.5 = 1.8 + 1.5 = 3.3. RHS = 0.7(6) - 0.9 = 4.2 - 0.9 = 3.3. LHS = RHS.
Answer: x = 6.
Example 7: Example 7: Complex fraction equation
Problem: Solve: (x + 1)/2 - (x - 2)/3 = 1
Solution:
Given: (x + 1)/2 - (x - 2)/3 = 1
Steps:
- LCM of 2 and 3 = 6
- Multiply every term by 6: 3(x + 1) - 2(x - 2) = 6
- Expand: 3x + 3 - 2x + 4 = 6
- Simplify: x + 7 = 6
- x = 6 - 7 = -1
Verify: LHS = (-1 + 1)/2 - (-1 - 2)/3 = 0/2 - (-3)/3 = 0 + 1 = 1. RHS = 1. Correct.
Answer: x = -1.
Example 8: Example 8: Sum of fractions with variable in denominator
Problem: Solve: 3/(x + 1) + 2/(x + 1) = 1
Solution:
Given: 3/(x + 1) + 2/(x + 1) = 1
Steps:
- Same denominator, so add numerators: 5/(x + 1) = 1
- Cross-multiply: 5 = x + 1
- x = 4
Check denominator: x + 1 = 5, not zero. Valid.
Verify: 3/5 + 2/5 = 5/5 = 1. Correct.
Answer: x = 4.
Example 9: Example 9: Fraction equation with constants
Problem: Solve: (3x - 2)/4 + (2x + 3)/3 = 5
Solution:
Given: (3x - 2)/4 + (2x + 3)/3 = 5
Steps:
- LCM of 4 and 3 = 12
- Multiply every term by 12: 3(3x - 2) + 4(2x + 3) = 60
- Expand: 9x - 6 + 8x + 12 = 60
- Simplify: 17x + 6 = 60
- 17x = 54
- x = 54/17
Verify: LHS = (3(54/17) - 2)/4 + (2(54/17) + 3)/3 = (162/17 - 34/17)/4 + (108/17 + 51/17)/3 = (128/17)/4 + (159/17)/3 = 128/68 + 159/51 = 32/17 + 53/17 = 85/17 = 5. Correct.
Answer: x = 54/17.
Example 10: Example 10: Word problem requiring reduction
Problem: The numerator of a fraction is 3 less than the denominator. If 1 is added to both the numerator and denominator, the fraction becomes 3/4. Find the fraction.
Solution:
Setup:
- Let denominator = x. Then numerator = x - 3.
- Fraction = (x - 3)/x
- Condition: (x - 3 + 1)/(x + 1) = 3/4
- Simplify: (x - 2)/(x + 1) = 3/4
Solve by cross-multiplication:
- 4(x - 2) = 3(x + 1)
- 4x - 8 = 3x + 3
- 4x - 3x = 3 + 8
- x = 11
The fraction: (11 - 3)/11 = 8/11
Verify: Add 1 to both: (8 + 1)/(11 + 1) = 9/12 = 3/4. Correct.
Answer: The fraction is 8/11.
Real-World Applications
Real-world applications:
- Splitting bills: When a bill is shared unequally among friends with different contributions as fractions, the equation involves fractions that need reduction. If Amit pays 1/3 of the bill, Bina pays 1/4, and the remaining Rs 500 is paid by Charu, you set up: 1/3 × T + 1/4 × T + 500 = T. This has fractions that need the LCM method.
- Speed problems: Time = distance/speed involves fractions. When two journeys at different speeds are compared, the equation requires reduction. For example: x/60 + x/80 = 7 (total time of 7 hours for a journey at two speeds).
- Mixing solutions: When solutions of different concentrations are mixed in certain ratios, fraction equations arise. If 500 ml of 20% salt solution is mixed with x ml of 50% solution to get 30% solution: (0.2 × 500 + 0.5x)/(500 + x) = 0.3. Cross-multiplication reduces this.
- Proportional reasoning: Recipes, map scales, and gear ratios all involve proportions that reduce to cross-multiplication. If 3/4 cup of sugar is needed for 12 cookies, how much for 20 cookies?
- Finance: EMI calculations, interest rate comparisons, and tax computations involve equations with fractions and decimals. Monthly interest calculations often give equations like P × r/1200 = 500.
- Work problems: If person A can do a job in 6 hours and person B in 8 hours, together they complete 1/6 + 1/8 of the work per hour. Setting 1/6 + 1/8 = 1/t and solving for t requires the LCM method.
- Age problems with fractions: A person's age is 2/3 of another's age, plus 5 years. Such conditions create equations with fractions on both sides.
Key Points to Remember
- Reducing an equation means transforming it into a simpler equivalent form that has the same solution.
- For fractions: multiply all terms on both sides by the LCM of all denominators. This eliminates every fraction at once.
- For proportions (a/b = c/d): use cross-multiplication to get ad = bc. This works only when there is one fraction on each side.
- For nested brackets: expand from the innermost bracket outward, step by step.
- For decimals: multiply both sides by 10, 100, or 1000 (based on the maximum number of decimal places) to convert to whole numbers.
- For variable in the denominator: cross-multiply or multiply both sides by the expression containing the variable. Always check that the solution does not make any denominator zero.
- When using LCM, multiply every single term on both sides — a common mistake is forgetting to multiply a standalone constant term.
- Cross-multiplication works only when the equation is in the form one fraction = one fraction. If there are multiple fractions on one side, use LCM instead.
- Always verify your answer in the original equation, not the reduced form. Verification catches errors made during reduction.
- The reduced equation should be in the form ax + b = 0 or ax = c, which is trivially solved by x = −b/a or x = c/a.
- An extraneous solution is a value that satisfies the reduced equation but not the original. This can occur when variables appear in denominators.
- Multiple techniques can be combined: first remove decimals, then expand brackets, then use LCM for remaining fractions.
Practice Problems
- Solve: x/4 + x/6 = 5.
- Solve: (3x + 5)/7 = (x + 3)/3.
- Solve: 7/(2x - 1) = 3/5.
- Solve: (x + 1)/3 - (x - 1)/4 = 2.
- Solve: 0.5x - 0.3 = 0.2x + 1.2.
- Solve: 3[2(x - 1) + 5] = 27.
- The denominator of a fraction is 5 more than the numerator. If 3 is added to both, the fraction becomes 4/5. Find the fraction.
- Solve: (2x - 1)/3 + (3x + 2)/4 = (x + 3)/2.
Frequently Asked Questions
Q1. What does it mean to reduce an equation?
Reducing an equation means converting it into a simpler form — removing fractions, decimals, or brackets — so that it becomes easy to solve. The reduced equation has the same solution as the original.
Q2. When do you use the LCM method?
Use the LCM method when the equation has <strong>multiple fractions with different denominators</strong>. Multiply every term by the LCM of all denominators to eliminate fractions at once.
Q3. When do you use cross-multiplication?
Use cross-multiplication when the equation is in the form <strong>one fraction equals another fraction</strong> (a/b = c/d). Cross-multiply to get ad = bc.
Q4. Can you cross-multiply when there are three fractions?
No. Cross-multiplication works only for <strong>a/b = c/d</strong> (one fraction on each side). If there are three or more fractions, use the LCM method instead.
Q5. How do you handle nested brackets?
Expand from the <strong>innermost</strong> bracket outward. First expand the inner brackets, simplify, then expand the outer brackets.
Q6. What if the variable is in the denominator?
Cross-multiply or multiply both sides by the expression containing the variable. After solving, <strong>verify</strong> that your answer does not make any denominator zero.
Q7. How do you remove decimals from an equation?
Multiply both sides by a power of 10. If the most decimal places is 1, multiply by 10. If 2, multiply by 100. This converts all decimals to whole numbers.
Q8. Why is verification important for reduced equations?
Reducing can introduce extraneous solutions (especially when variables are in denominators). Verification ensures the solution works in the <strong>original</strong> equation, not just the reduced form.
Q9. What is an extraneous solution?
An extraneous solution is a value that satisfies the reduced equation but NOT the original equation. This happens when the solution makes a denominator zero in the original equation.
Q10. Can every equation be reduced to ax + b = 0?
Every linear equation in one variable can be reduced to the form ax + b = 0 (or ax = c). This is the <strong>standard form</strong> of a linear equation, and x = -b/a (or x = c/a) gives the solution.
