Question
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For each equation, choose the statement that describes its solution. If applicable, give the solution. \begin{tabular}{|l} (a) \( -4(y+1)+6 y=2(y+2)-8 \) \\ O No solution \\ O All real numbers are solutions \\ (b) \( 6(u-2)+1=2(3 u-5) \) \\ \( O \) No solution \\ O \( u=\square \) \\ O All real numbers are solutions \end{tabular}

Ask by Dunn Tyler. in the United States
Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

**(a)** All real numbers are solutions. **(b)** No solution.

Solution

Let's analyze each equation to determine their solutions. --- ### **(a) \( -4(y + 1) + 6y = 2(y + 2) - 8 \)** **Step-by-Step Solution:** 1. **Expand both sides:** \[ -4y - 4 + 6y = 2y + 4 - 8 \] 2. **Combine like terms:** \[ ( -4y + 6y ) - 4 = 2y - 4 \] \[ 2y - 4 = 2y - 4 \] 3. **Subtract \(2y\) from both sides:** \[ -4 = -4 \] 4. **Interpretation:** This simplifies to a true statement (\(-4 = -4\)) regardless of the value of \(y\). This means **every real number** satisfies the equation. **Conclusion for (a):** - **Statement:** All real numbers are solutions. - **Solution:** Every real number \( y \) is a solution. --- ### **(b) \( 6(u - 2) + 1 = 2(3u - 5) \)** **Step-by-Step Solution:** 1. **Expand both sides:** \[ 6u - 12 + 1 = 6u - 10 \] 2. **Combine like terms:** \[ 6u - 11 = 6u - 10 \] 3. **Subtract \(6u\) from both sides:** \[ -11 = -10 \] 4. **Interpretation:** This simplifies to a false statement (\(-11 \neq -10\)), which means there is **no solution** to the equation. **Conclusion for (b):** - **Statement:** No solution. - **Solution:** There is no value of \( u \) that satisfies the equation. --- **Final Answers:** - **(a)** All real numbers are solutions. - **(b)** No solution.

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Mind Expander

Let's tackle each equation one by one! For equation (a) \( -4(y+1)+6 y=2(y+2)-8 \): First, simplify both sides: LHS: \( -4y - 4 + 6y = 2y - 4 \); RHS: \( 2y + 4 - 8 = 2y - 4 \). Now, we have \( 2y - 4 = 2y - 4 \). Since both sides are identical, this means all real numbers are solutions! For equation (b) \( 6(u-2)+1=2(3u-5) \): Expanding both sides yields: LHS: \( 6u - 12 + 1 = 6u - 11 \); RHS: \( 6u - 10 \). Now, setting them equal gives \( 6u - 11 = 6u - 10 \). This simplifies to \( -11 = -10 \), which is a contradiction, meaning there is no solution! To summarize: (a) All real numbers are solutions. (b) No solution.

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