Solve for k. {:[(6)/(2k-6)=(1)/(3)],[k=]:}

Write down the equation: Write down the first equation from the system. We have the equation (6)/(2k-6)=(1)/(3). We need to solve for k.Cross-multiply to eliminate fractions: Cross-multiply to eliminate the fractions. 6 * 3 = (2k - 6) * 1 This gives us 18 = 2k - 6.Isolate the term with k: Add 6 to both sides of the equation to isolate the term with k on one side. 18 + 6 = 2k - 6 + 6 This simplifies to 24 = 2k.Solve for k: Divide both sides of the equation by 2 to solve for k. 24 / 2 = 2k / 2 This simplifies to k = 12.

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Solve for
k.

{:[(6)/(2k-6)=(1)/(3)],[k=]:}

Solve for $k$ . $\newline$ $\begin{array}{l} \frac{6}{2 k-6}=\frac{1}{3} \\ k=\square \end{array}$

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Math Problems

Grade 6

Multiply using the distributive property

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Q. Solve for

k

\newline

\begin{array}{l} \frac{6}{2 k-6}=\frac{1}{3} \\ k=\square \end{array}

Write down the equation: Write down the first equation from the system. $\newline$ We have the equation $\frac{6}{2k-6}=\frac{1}{3}$ . We need to solve for $k$ .
Cross-multiply to eliminate fractions: Cross-multiply to eliminate the fractions. $\newline$ $6 \times 3 = (2k - 6) \times 1$ $\newline$ This gives us $18 = 2k - 6$ .
Isolate the term with $k$ : Add $6$ to both sides of the equation to isolate the term with $k$ on one side. $\newline$ $18 + 6 = 2k - 6 + 6$ $\newline$ This simplifies to $24 = 2k$ .
Solve for $k$ : Divide both sides of the equation by $2$ to solve for $k$ . $\frac{24}{2} = \frac{2k}{2}$ This simplifies to $k = 12$ .