Eighth Grade Mathematics: Solving Equations with Variables on Both Sides

In many real-world modeling scenarios, two competing options or processes change simultaneously: comparing two subscription plans with different base fees and hourly rates, or finding when two moving vehicles at different starting points will meet. In algebra, these scenarios translate into equations where the unknown variable appears on both sides of the equal sign. In this chapter, you will learn how to strategically collect variable terms onto one side, balance equations efficiently, and solve equations with variables and grouping symbols on both sides.

The Strategy of Variable Elimination

When an equation contains variable terms on both sides of the equal sign, such as 7x + 4 = 3x + 24, you cannot solve for x until all variable terms are consolidated on a single side.

The Balancing Scale Analogy:
Left Scale Pan:   7 unknown boxes (x) and 4 single weights
Right Scale Pan:  3 unknown boxes (x) and 24 single weights

Action: Remove 3 unknown boxes (x) from BOTH pans simultaneously!
Result: The scale remains perfectly balanced:
        4 unknown boxes (x) and 4 weights equal 24 weights.

Algebraically, removing variable terms is accomplished using the Addition or Subtraction Property of Equality.

Solve: 7x + 4 = 3x + 24

Step 1: Subtract 3x from both sides to eliminate the variable on the right:
        7x - 3x + 4 = 3x - 3x + 24
        4x + 4 = 24

Step 2: Subtract 4 from both sides to isolate the variable term:
        4x = 20

Step 3: Divide both sides by 4:
        x = 5

Strategic Tip: Keeping the Coefficient Positive

While you can subtract either variable term, it is usually advantageous to eliminate the variable term with the smaller numerical coefficient. This ensures that the remaining variable term has a positive coefficient, avoiding negative signs during division.

Comparing Approaches for: 2x - 8 = 6x + 12

Approach A (Subtract 2x):            Approach B (Subtract 6x):
2x - 2x - 8 = 6x - 2x + 12           2x - 6x - 8 = 6x - 6x + 12
-8 = 4x + 12                         -4x - 8 = 12
-20 = 4x                             -4x = 20
x = -5                               x = -5
(Keeps variable coefficient +4!)     (Produces negative coefficient -4)

Both methods are mathematically sound and yield the identical solution x = -5, but Approach A minimizes potential sign errors.

Equations with Parentheses on Both Sides

When both sides of an equation contain grouping symbols, the distributive property must be applied across both sides before moving variable terms.

Solve: 4*(2x - 3) = 2*(x + 9)

Step 1: Distribute on both sides:
        8x - 12 = 2x + 18

Step 2: Subtract 2x from both sides:
        6x - 12 = 18

Step 3: Add 12 to both sides:
        6x = 30

Step 4: Divide by 6:
        x = 5

By methodically simplifying each side independently before attempting to cross the equal sign, complex multi-parentheses problems become straightforward two-step equations.

Chapter Exercises

Exercise 1: Solve the linear equation 9x - 14 = 4x + 21. Show all balancing steps and verify your answer through substitution into the original equation.

Exercise 2: Solve the multi-step equation 5(x - 3) + 2 = 2(3x - 7) - 4x. Fully simplify both sides before collecting variable terms.

Exercise 3: Two fitness centers offer different membership pricing. Gym Alpha charges a 50 dollar enrollment fee plus 25 dollars per month. Gym Beta charges a 20 dollar enrollment fee plus 30 dollars per month. Write an equation setting the total costs equal, solve for the number of months when the costs are identical, and state that total cost.

Solutions and Explanations

Solution 1: To solve 9x - 14 = 4x + 21, first eliminate the smaller variable term 4x from the right side by subtracting 4x from both sides of the equation: 9x - 4x - 14 = 4x - 4x + 21. This simplifies to 5x - 14 = 21. Next, isolate the variable term 5x by adding 14 to both sides: 5x = 21 + 14, which gives 5x = 35. Finally, divide both sides by 5: x = 35 / 5 = 7. To check, substitute x = 7 into the original equation: Left side is 9(7) - 14 = 63 - 14 = 49. Right side is 4(7) + 21 = 28 + 21 = 49. Because both sides evaluate to 49, the solution x = 7 is correct.

Solution 2: To solve 5(x - 3) + 2 = 2(3x - 7) - 4x, begin by distributing on both sides. On the left side, distribute 5 into (x - 3): 5x - 15 + 2 = 5x - 13. On the right side, distribute 2 into (3x - 7): 6x - 14 - 4x. Combine like terms on the right side: (6x - 4x) - 14 = 2x - 14. The equation is now 5x - 13 = 2x - 14. Next, subtract 2x from both sides to collect variables on the left: 3x - 13 = -14. Now add 13 to both sides to isolate the variable term: 3x = -14 + 13 = -1. Finally, divide both sides by 3: x = -1/3.

Solution 3: Let m represent the number of months of gym membership. The total cost for Gym Alpha is modeled by the expression 25m + 50. The total cost for Gym Beta is modeled by the expression 30m + 20. Setting the two costs equal gives the equation 30m + 20 = 25m + 50. To solve for m, subtract 25m from both sides: 5m + 20 = 50. Next, subtract 20 from both sides: 5m = 30. Dividing both sides by 5 gives m = 6 months. To find the total cost at 6 months, substitute m = 6 into either cost expression: for Gym Alpha, 25(6) + 50 = 150 + 50 = 200 dollars; for Gym Beta, 30(6) + 20 = 180 + 20 = 200 dollars. Thus, at 6 months, both fitness centers cost exactly 200 dollars.