Example

Solving 7x+5=6x+27x + 5 = 6x + 2 by Collecting Variables and Constants

To solve 7x+5=6x+27x + 5 = 6x + 2, observe that variable terms (7x7x and 6x6x) and constant terms (55 and 22) appear on both sides of the equation. Because 77 is greater than 66, designate the left side as the variable side and the right side as the constant side.

Step 1 — Remove the variable term from the constant side: Since 6x6x is on the constant side, subtract 6x6x from both sides using the Subtraction Property of Equality:

7x6x+5=6x6x+27x - 6x + 5 = 6x - 6x + 2

Combine like terms: x+5=2x + 5 = 2. Now the variable appears only on the left.

Step 2 — Remove the constant from the variable side: The constant 55 is on the variable side, so subtract 55 from both sides:

x+55=25x + 5 - 5 = 2 - 5

x=3x = -3

Because the coefficient of xx is already 11, no further division is needed.

Step 3 — Check by substitution: Replace xx with 3-3 in the original equation:

7(3)+5=?6(3)+27(-3) + 5 \stackrel{?}{=} 6(-3) + 2

21+5=?18+2-21 + 5 \stackrel{?}{=} -18 + 2

16=16-16 = -16 \checkmark

Because both sides are equal, x=3x = -3 is confirmed as the correct solution. Unlike equations where only variables or only constants need to be collected, this example required two collecting steps — first gathering all variable terms onto the left side and then gathering all constant terms onto the right side — before the solution could be read directly.

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Updated 2026-04-21

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