Equations with Variables on Both Sides
In simpler linear equations, the variable appears on only one side — for example, 3x + 5 = 20. But many equations have the variable on both sides of the equals sign, such as 5x + 3 = 2x + 15.
Solving such equations requires collecting all variable terms on one side and all constant terms on the other side.
This topic from Linear Equations in Grade 8 teaches you systematic methods for solving equations where the variable appears on both the LHS and RHS.
In Classes 6 and 7, you solved equations like 3x + 5 = 20 where the variable x appears only on the left side. The solution is straightforward: isolate x by performing inverse operations.
In Grade 8, you encounter more complex equations where x (or any variable) appears on both sides. These arise naturally in many real-world situations:
- Comparing two quantities that both depend on an unknown.
- Problems where the same variable appears in different expressions.
- Equations formed from geometry, age problems, and number puzzles.
The key technique is to collect all variable terms on one side and all constant terms on the other side. This reduces the equation to the simple form ax = b, which you already know how to solve.
What is Equations with Variables on Both Sides - Grade 8 Maths (Linear Equations)?
Definition: An equation with variables on both sides is a linear equation where the unknown (variable) appears in expressions on both the left-hand side (LHS) and the right-hand side (RHS).
General form:
ax + b = cx + d
Where:
- a and c are the coefficients of the variable on LHS and RHS.
- b and d are the constants on LHS and RHS.
- The goal is to find the value of x that makes both sides equal.
Examples:
- 5x + 3 = 2x + 15 (variable x on both sides)
- 7y - 4 = 3y + 12 (variable y on both sides)
- 2(x + 3) = 5(x - 1) (variable x on both sides after expanding)
How to recognise equations with variables on both sides:
- Look for the variable (usually x or y) in expressions on both the LHS and RHS of the equals sign.
- After expanding all brackets, check if the variable appears on both sides.
- Examples: 5x + 3 = 2x + 15, 4(x−1) = 3(x+2), (2x+1)/3 = (x−5)/2.
Strategy overview:
- Simplify both sides independently (expand, combine like terms).
- Move variables to one side, constants to the other.
- Solve the simplified equation.
- Verify the solution.
Methods
Method: Transposition
Steps to solve ax + b = cx + d:
- Expand brackets (if any) on both sides.
- Transpose variable terms — move all terms with the variable to the LHS (or RHS).
- Transpose constant terms — move all constant terms to the other side.
- Simplify both sides by combining like terms.
- Divide both sides by the coefficient of the variable.
- Verify by substituting the answer back into the original equation.
Rule: When transposing (moving a term to the other side), change its sign:
- + becomes -
- - becomes +
- multiplication becomes division
- division becomes multiplication
Alternative: Balancing method
- Subtract cx from both sides to remove the variable from the RHS.
- Subtract b from both sides to remove the constant from the LHS.
- Divide both sides by the remaining coefficient.
Solved Examples
Example 1: Example 1: Simple equation — variables on both sides
Problem: Solve: 5x + 3 = 2x + 15
Solution:
Steps:
- Transpose 2x from RHS to LHS: 5x - 2x + 3 = 15
- Transpose 3 from LHS to RHS: 5x - 2x = 15 - 3
- Simplify: 3x = 12
- Divide by 3: x = 4
Verify: LHS = 5(4) + 3 = 23. RHS = 2(4) + 15 = 23. LHS = RHS.
Answer: x = 4.
Example 2: Example 2: Negative coefficients
Problem: Solve: 7y - 4 = 3y + 12
Solution:
Steps:
- Transpose 3y to LHS: 7y - 3y - 4 = 12
- Transpose -4 to RHS: 7y - 3y = 12 + 4
- Simplify: 4y = 16
- Divide by 4: y = 4
Verify: LHS = 7(4) - 4 = 24. RHS = 3(4) + 12 = 24. LHS = RHS.
Answer: y = 4.
Example 3: Example 3: Equation with brackets
Problem: Solve: 2(x + 3) = 5(x - 1)
Solution:
Steps:
- Expand brackets: 2x + 6 = 5x - 5
- Transpose 5x to LHS and 6 to RHS: 2x - 5x = -5 - 6
- Simplify: -3x = -11
- Divide by -3: x = 11/3
Verify: LHS = 2(11/3 + 3) = 2(20/3) = 40/3. RHS = 5(11/3 - 1) = 5(8/3) = 40/3. LHS = RHS.
Answer: x = 11/3.
Example 4: Example 4: Variable with larger coefficient on RHS
Problem: Solve: 3x + 8 = 7x - 12
Solution:
Tip: Collect variables on the side with the larger coefficient (RHS has 7x).
- Transpose 3x to RHS: 8 = 7x - 3x - 12
- Transpose -12 to LHS: 8 + 12 = 7x - 3x
- Simplify: 20 = 4x
- Divide by 4: x = 5
Verify: LHS = 3(5) + 8 = 23. RHS = 7(5) - 12 = 23. LHS = RHS.
Answer: x = 5.
Example 5: Example 5: Multiple brackets on both sides
Problem: Solve: 3(2x - 1) = 2(3x + 4) - 17
Solution:
Steps:
- Expand: 6x - 3 = 6x + 8 - 17
- Simplify RHS: 6x - 3 = 6x - 9
- Transpose 6x from RHS: 6x - 6x - 3 = -9
- Simplify: -3 = -9
Result: -3 = -9 is a false statement.
Answer: This equation has no solution. No value of x can satisfy it.
Example 6: Example 6: Identity equation
Problem: Solve: 4(x + 2) - 3 = 4x + 5
Solution:
Steps:
- Expand: 4x + 8 - 3 = 4x + 5
- Simplify LHS: 4x + 5 = 4x + 5
- Transpose 4x: 5 = 5
Result: 5 = 5 is always true.
Answer: This is an identity. It is true for all values of x.
Example 7: Example 7: Equation with fractions on both sides
Problem: Solve: (2x + 1)/3 = (3x - 2)/5
Solution:
Steps:
- Cross-multiply: 5(2x + 1) = 3(3x - 2)
- Expand: 10x + 5 = 9x - 6
- Transpose 9x to LHS: 10x - 9x + 5 = -6
- Transpose 5 to RHS: x = -6 - 5
- x = -11
Verify: LHS = (2(-11) + 1)/3 = -21/3 = -7. RHS = (3(-11) - 2)/5 = -35/5 = -7. LHS = RHS.
Answer: x = -11.
Example 8: Example 8: Word problem — ages
Problem: A father is 3 times as old as his son. After 12 years, the father will be twice as old as the son. Find their ages.
Solution:
Setup:
- Let son's age = x. Father's age = 3x.
- After 12 years: son = x + 12, father = 3x + 12.
- Condition: 3x + 12 = 2(x + 12)
Solve:
- Expand: 3x + 12 = 2x + 24
- Transpose: 3x - 2x = 24 - 12
- x = 12
Verify: Son = 12, Father = 36. After 12 years: Son = 24, Father = 48. Is 48 = 2 x 24? Yes.
Answer: Son is 12 years, Father is 36 years.
Example 9: Example 9: Word problem — consecutive odd numbers
Problem: The sum of two consecutive odd numbers is 36. Find them.
Solution:
Setup:
- Let the two consecutive odd numbers be x and x + 2.
- x + (x + 2) = 36
Solve:
- Simplify: 2x + 2 = 36
- 2x = 34
- x = 17
Numbers: 17 and 19.
Verify: 17 + 19 = 36.
Answer: The two consecutive odd numbers are 17 and 19.
Example 10: Example 10: Multi-step with distribution
Problem: Solve: 5(x - 2) - 3(x + 1) = 2(x - 4)
Solution:
Steps:
- Expand all brackets: 5x - 10 - 3x - 3 = 2x - 8
- Simplify LHS: 2x - 13 = 2x - 8
- Transpose 2x: -13 = -8
Result: -13 = -8 is a false statement.
Answer: This equation has no solution.
Real-World Applications
Real-world applications:
- Age problems: Father is 3 times the son's age now. After n years, father is twice the son's age. The condition 3x + n = 2(x + n) has variables on both sides. These are among the most common word problems in NCERT textbooks.
- Distance problems: Two cars start from different points. Car A travels at 60 km/h and is 100 km ahead. Car B travels at 80 km/h. When does B catch A? Setting 60t + 100 = 80t gives variables on both sides.
- Pricing and comparison: Two phone plans — Plan A charges Rs 200 fixed + Rs 1 per minute. Plan B charges Rs 100 fixed + Rs 2 per minute. When are they equal? 200 + 1x = 100 + 2x gives x on both sides.
- Geometry: When perimeters or areas of different shapes are set equal, variables often appear on both sides. For example: a rectangle with length 2x and width 5 has the same perimeter as a square of side (x + 3). Setting 2(2x + 5) = 4(x + 3) gives variables on both sides.
- Physics: Balancing forces, equating speeds from different reference frames, and equilibrium problems often produce equations with variables on both sides.
- Business: A company's revenue (selling price × quantity) equals its cost (fixed cost + variable cost × quantity) at the break-even point. Both expressions involve the quantity variable.
- Chemistry: Balancing chemical equations and calculating dilution concentrations involve setting two expressions equal, often with the unknown on both sides.
Key Points to Remember
- An equation with variables on both sides has the form ax + b = cx + d where a and c are both non-zero.
- Solve by collecting variable terms on one side and constant terms on the other using transposition.
- When transposing, change the sign: + becomes −, and − becomes +. Multiplication becomes division, and vice versa.
- Always expand brackets using the distributive property before transposing terms.
- Collect variables on the side with the larger coefficient to avoid working with negative coefficients.
- If fractions are present, cross-multiply (for proportions) or use the LCM method to eliminate fractions first.
- If all variable terms cancel and you get a true statement (like 5 = 5 or 0 = 0), the equation is an identity — it has infinitely many solutions.
- If all variable terms cancel and you get a false statement (like 3 = 7 or −2 = 5), the equation has no solution — it is a contradiction.
- Always verify your answer by substituting it back into the original equation and checking LHS = RHS.
- Word problems (ages, distances, pricing) often lead to equations with variables on both sides.
- A standard linear equation in one variable has exactly one solution. The exceptions are identities (infinitely many) and contradictions (none).
- The solution of ax + b = cx + d is x = (d − b)/(a − c), provided a ≠ c. If a = c and b ≠ d, there is no solution. If a = c and b = d, it is an identity.
Practice Problems
- Solve: 8x - 5 = 3x + 20.
- Solve: 4(x + 3) = 3(x + 5) + 2.
- Solve: 6y - 7 = 2y + 13.
- Solve: 2(3x + 1) = 3(2x - 5) + 17. Is this an identity?
- Solve: (5x - 3)/4 = (3x + 1)/2.
- A number added to its double gives the same result as 5 times the number minus 12. Find the number.
- Meena's age is 5 more than twice Ram's age. In 4 years, Meena's age will be 3 times Ram's present age. Find their ages.
- Solve: 3(x - 4) + 2(x + 1) = 4(x - 1) + x - 6. What type of equation is this?
Frequently Asked Questions
Q1. What does it mean to have variables on both sides?
It means the unknown (like x) appears in expressions on both the left-hand side and right-hand side of the equation. Example: 5x + 3 = 2x + 15 has x on both sides.
Q2. How do you solve equations with variables on both sides?
Move all variable terms to one side and all constant terms to the other side using transposition. Then simplify and divide by the coefficient of the variable.
Q3. What if the variable cancels out?
If you get a <strong>true statement</strong> (like 0 = 0), the equation is an identity — true for all values. If you get a <strong>false statement</strong> (like 5 = 3), the equation has no solution.
Q4. Should I always collect variables on the LHS?
No. You can collect variables on either side. It is easier to collect them on the side where the coefficient is <strong>larger</strong> to avoid negative coefficients.
Q5. What if there are brackets on both sides?
Expand all brackets first using the distributive property, then collect variable terms on one side and constants on the other.
Q6. What if there are fractions on both sides?
If the equation is a/b = c/d, cross-multiply. If there are multiple fractions, multiply every term by the LCM of all denominators to eliminate fractions.
Q7. How do you verify the solution?
Substitute your answer back into the <strong>original</strong> equation. Calculate the LHS and RHS separately. If LHS = RHS, your solution is correct.
Q8. Can a linear equation have more than one solution?
A standard linear equation has <strong>exactly one</strong> solution. The only exception is an identity (like 2x + 3 = 2x + 3), which has infinitely many solutions.
Q9. What is the difference between an equation and an identity?
An <strong>equation</strong> is true only for specific values of the variable (e.g., 2x = 6 is true only for x = 3). An <strong>identity</strong> is true for ALL values of the variable (e.g., 2(x+1) = 2x + 2).
Q10. How do word problems lead to equations with variables on both sides?
When a problem compares two quantities that both depend on the unknown (e.g., comparing ages after some years, or two plans with different rates), setting them equal gives variables on both sides.
