Solve for y. {:[(2y-4)/(y+7)=(1)/(3)],[y=]:}

Write down the system: Write down the given system of equations. We have the system: (2y - 4) / (y + 7) = 1 / 3 y = yFocus on the first equation: Since the second equation is simply y = y, which is always true, we can ignore it and focus on the first equation to find the value of y.Eliminate the fractions: Cross-multiply to eliminate the fractions. (2y - 4) * 3 = (y + 7) * 1Multiply both sides: Perform the multiplication on both sides. 6y - 12 = y + 7Rearrange the equation: Move all terms involving y to one side of the equation. 6y - y = 7 + 12Combine like terms: Combine like terms. 5y = 19Solve for y: Divide both sides by 5 to solve for y. y = 19 / 5Final answer: Simplify the fraction to get the final answer. y = 3.8

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Solve for
y.

{:[(2y-4)/(y+7)=(1)/(3)],[y=]:}

Solve for $y$ . $\newline$ $\begin{array}{l} \frac{2 y-4}{y+7}=\frac{1}{3} \\ y=\square \end{array}$

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Math Problems

Grade 6

Multiply using the distributive property

Full solution

Q. Solve for

y

\newline

\begin{array}{l} \frac{2 y-4}{y+7}=\frac{1}{3} \\ y=\square \end{array}

Write down the system: Write down the given system of equations. $\newline$ We have the system: $\newline$ $(2y - 4) / (y + 7) = 1 / 3$ $\newline$ $y = y$
Focus on the first equation: Since the second equation is simply $y = y$ , which is always true, we can ignore it and focus on the first equation to find the value of $y$ .
Eliminate the fractions: Cross-multiply to eliminate the fractions. $\newline$ $(2y - 4) \times 3 = (y + 7) \times 1$
Multiply both sides: Perform the multiplication on both sides. $\newline$ $6y - 12 = y + 7$
Rearrange the equation: Move all terms involving $y$ to one side of the equation. $6y - y = 7 + 12$
Combine like terms: Combine like terms. $\newline$ $5y = 19$
Solve for y: Divide both sides by $5$ to solve for y. $\newline$ $y = \frac{19}{5}$
Final answer: Simplify the fraction to get the final answer. $y = 3.8$