Solve the given linear equation using addition and subtraction properties of equality.
$3(y+3)+(1-y)=3y+14$

• A. $y=-4$
• B. $y=4$
• C. $y=0$
• D. $y=13$

Answer: (A)

Verified step by step guidance

1. Start with the given equation: \(3\left(y+3\right) + \left(1 - y\right) = 3y + 14\).
2. Apply the distributive property to remove parentheses: multiply 3 by each term inside the first parentheses to get \$3y + 9\(, and rewrite the second parentheses as \)1 - y\(. So the left side becomes \)3y + 9 + 1 - y$.
3. Combine like terms on the left side: combine \$3y\( and \)-y\( to get \)2y\(, and combine \(9\) and \(1\) to get \(10\). Now the equation is \)2y + 10 = 3y + 14$.
4. Use the subtraction property of equality to get all variable terms on one side: subtract \$2y\( from both sides to isolate variable terms on the right side, resulting in \)10 = 3y - 2y + 14\(, which simplifies to \)10 = y + 14$.
5. Use the subtraction property of equality again to isolate \(y\): subtract \(14\) from both sides to get \$10 - 14 = y\(, which simplifies to \)y = 10 - 14$.