Solve for y. {:[6=2(y+2)],[y=]:}

Solve for y: Solve the first equation for y. The first equation is 6 = 2(y + 2). To solve for y, we first need to divide both sides of the equation by 2 to isolate the term with y. 6 ÷ 2 = 2(y + 2) ÷ 2 3 = y + 2Subtract to find y: Subtract 2 from both sides of the equation. To find the value of y, we subtract 2 from both sides of the equation. 3 - 2 = y + 2 - 2 1 = yCheck solution: Check the solution in the second equation. The second equation is simply y =, which indicates that y can be any value. Since we have found y = 1 from the first equation, it satisfies the second equation.

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Solve for $y$ . $\newline$ $\begin{array}{l} 6=2(y+2) \\ y=\square \end{array}$

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Grade 6

Multiply using the distributive property

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Q. Solve for

y

\newline

\begin{array}{l} 6=2(y+2) \\ y=\square \end{array}

Solve for y: Solve the first equation for y. $\newline$ The first equation is $6 = 2(y + 2)$ . To solve for y, we first need to divide both sides of the equation by $2$ to isolate the term with $y$ . $\newline$ $6 \div 2 = 2(y + 2) \div 2$ $\newline$ $3 = y + 2$
Subtract to find $y$ : Subtract $2$ from both sides of the equation. $\newline$ To find the value of $y$ , we subtract $2$ from both sides of the equation. $\newline$ $3 - 2 = y + 2 - 2$ $\newline$ $1 = y$
Check solution: Check the solution in the second equation. $\newline$ The second equation is simply $y =$ , which indicates that $y$ can be any value. Since we have found $y = 1$ from the first equation, it satisfies the second equation.