Question

equations-linear equations: problem 8 (1 point) solve for (k). (\frac{2}{-6}k+\frac{2}{9}=-9 - \frac{-6}{5}k) answer: (k=) preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. ask for help

Explanation:

Step1: Simplify the left - hand side fraction

Simplify $\frac{2}{-6}k$ to $-\frac{1}{3}k$. The equation becomes $-\frac{1}{3}k+\frac{2}{9}=-9 + \frac{6}{5}k$.

Step2: Move terms with $k$ to one side

Add $\frac{1}{3}k$ to both sides: $\frac{2}{9}=-9+\frac{6}{5}k+\frac{1}{3}k$.

Step3: Find a common denominator for $k$ terms

The common denominator of 5 and 3 is 15. So $\frac{6}{5}k+\frac{1}{3}k=\frac{18}{15}k+\frac{5}{15}k=\frac{18 + 5}{15}k=\frac{23}{15}k$. The equation is $\frac{2}{9}=-9+\frac{23}{15}k$.

Step4: Move the constant to the left - hand side

Add 9 to both sides: $\frac{2}{9}+9=\frac{23}{15}k$. Since $9=\frac{81}{9}$, then $\frac{2}{9}+\frac{81}{9}=\frac{83}{9}=\frac{23}{15}k$.

Step5: Solve for $k$

Multiply both sides by $\frac{15}{23}$. So $k=\frac{83}{9}\times\frac{15}{23}=\frac{83\times15}{9\times23}=\frac{83\times5}{3\times23}=\frac{415}{69}$.

Answer:

$\frac{415}{69}$