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

Set up the equation: Set up the equation from the problem. We are given the equation (6)/(2k-6) = 3. We need to solve for k.Multiply to eliminate the denominator: Multiply both sides of the equation by (2k-6) to eliminate the denominator. (6)/(2k-6) * (2k-6) = 3 * (2k-6) This simplifies to: 6 = 3(2k-6)Distribute 3 on the right side: Distribute 3 on the right side of the equation. 6 = 3 * 2k - 3 * 6 6 = 6k - 18Isolate the term with k: Add 18 to both sides of the equation to isolate the term with k on one side. 6 + 18 = 6k - 18 + 18 24 = 6kDivide to solve for k: Divide both sides of the equation by 6 to solve for k. 24 / 6 = 6k / 6 4 = k

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

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Math Problems

Grade 8

Solve multi-step equations with fractional coefficients

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Q. Solve for

k

\newline

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

Set up the equation: Set up the equation from the problem. $\newline$ We are given the equation $\frac{6}{2k-6} = 3$ . We need to solve for $k$ .
Multiply to eliminate the denominator: Multiply both sides of the equation by $(2k-6)$ to eliminate the denominator. $\newline$ $\frac{6}{2k-6} \times (2k-6) = 3 \times (2k-6)$ $\newline$ This simplifies to: $\newline$ $6 = 3(2k-6)$
Distribute $3$ on the right side: Distribute $3$ on the right side of the equation. $6 = 3 \times 2k - 3 \times 6$ $6 = 6k - 18$
Isolate the term with $k$ : Add $18$ to both sides of the equation to isolate the term with $k$ on one side. $\newline$ $6 + 18 = 6k - 18 + 18$ $\newline$ $24 = 6k$
Divide to solve for k: Divide both sides of the equation by $6$ to solve for k. $\newline$ $\frac{24}{6} = \frac{6k}{6}$ $\newline$ $4 = k$