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 both \(y\) and 3, so \$3y + 9\(, then rewrite the equation as \)3y + 9 + 1 - y = 3y + 13$.
3. Combine like terms on the left side: \$3y - y\( becomes \)2y\(, and \)9 + 1\( becomes \)10\(, so the equation simplifies to \)2y + 10 = 3y + 13$.
4. Use the addition and subtraction properties of equality to isolate the variable \(y\): subtract \$2y\( from both sides to get \)10 = y + 13\(, then subtract 13 from both sides to get \)10 - 13 = y$.
5. Simplify the right side to express \(y\) explicitly: \(y = 10 - 13\).