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Solve for yy.\newline3(y+4)=153(y + 4) = 15\newliney=y = ____\_\_\_\_

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Q. Solve for yy.\newline3(y+4)=153(y + 4) = 15\newliney=y = ____\_\_\_\_
  1. Distribute 33 to terms: Distribute the 33 to both terms inside the parentheses.\newlineWe need to apply the distributive property, which states that a(b+c)=ab+aca(b + c) = ab + ac. In this case, we distribute 33 to both yy and 44.\newline3(y+4)=3y+343(y + 4) = 3\cdot y + 3\cdot 4\newlineThis simplifies to:\newline3y+12=153y + 12 = 15
  2. Subtract 1212 to isolate: Subtract 1212 from both sides of the equation to isolate the term with yy. To solve for yy, we need to get yy on one side of the equation by itself. We do this by subtracting 1212 from both sides. 3y+1212=15123y + 12 - 12 = 15 - 12 This simplifies to: 3y=33y = 3
  3. Divide by 33 to solve: Divide both sides of the equation by 33 to solve for yy. Now, we divide both sides by 33 to get yy by itself. 3y3=33\frac{3y}{3} = \frac{3}{3} This simplifies to: y=1y = 1