Solve the given linear equation using addition and subtraction properties of equality.
$3(y+3)+(1-y)=3y+13$

• 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 + 13$.
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$.
3. Combine like terms on the left side: \$3y + 9 + 1 - y\( simplifies to \)2y + 10$.
4. Rewrite the equation as \$2y + 10 = 3y + 13\( and then use the subtraction property of equality to get all variable terms on one side by subtracting \)2y\( from both sides: \)10 = y + 13$.
5. Use the subtraction property of equality again to isolate \(y\) by subtracting 13 from both sides: \$10 - 13 = y\(, which simplifies to \)y = -3$ (you can verify this step to confirm the exact value).