Example

Solving 3(2y−1)−5y=2(y+1)−2(y+3)3(2y - 1) - 5y = 2(y + 1) - 2(y + 3)

To solve the equation 3(2y−1)−5y=2(y+1)−2(y+3)3(2y - 1) - 5y = 2(y + 1) - 2(y + 3), apply a systematic strategy: simplify both sides first, then isolate the variable.

  1. Simplify both sides: Distribute on both sides of the equation: 6y−3−5y=2y+2−2y−66y - 3 - 5y = 2y + 2 - 2y - 6 Next, use the Commutative Property of Addition to rearrange terms so that like terms are adjacent, then combine like terms to simplify each side. On the left side, 6y−5y6y - 5y becomes yy. On the right side, 2y−2y2y - 2y becomes 00, and 2−62 - 6 becomes −4-4. The equation simplifies to: y−3=−4y - 3 = -4
  2. Isolate the variable: To undo the subtraction of 33, apply the Addition Property of Equality by adding 33 to both sides: y−3+3=−4+3y - 3 + 3 = -4 + 3 y=−1y = -1
  3. Check the solution: Substitute −1-1 for yy in the original equation to verify: 3(2(−1)−1)−5(−1)=2(−1+1)−2(−1+3)3(2(-1) - 1) - 5(-1) = 2(-1 + 1) - 2(-1 + 3) 3(−2−1)+5=2(0)−2(2)3(-2 - 1) + 5 = 2(0) - 2(2) 3(−3)+5=−43(-3) + 5 = -4 −9+5=−4-9 + 5 = -4 −4=−4-4 = -4 Because both sides are equal, the solution y=−1y = -1 is confirmed.
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Updated 2026-04-21

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