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leveled practice in 1 and 2, solve each equation. 1. 6.4n - 10 = 4.4n +…

Question

leveled practice in 1 and 2, solve each equation. 1. 6.4n - 10 = 4.4n + 6 2. $\frac{1}{3}k + 80=\frac{1}{2}k + 120$

Explanation:

Step1: Isolate variable terms

Subtract $4.4n$ from both sides of $6.4n - 10=4.4n + 6$.
$6.4n-4.4n-10=4.4n-4.4n + 6$, which simplifies to $2n-10 = 6$.

Step2: Isolate the variable

Add 10 to both sides: $2n-10 + 10=6 + 10$, getting $2n=16$.

Step3: Solve for n

Divide both sides by 2: $\frac{2n}{2}=\frac{16}{2}$, so $n = 8$.

For the second - equation $\frac{1}{3}k+80=\frac{1}{2}k + 120$:

Step1: Get a common denominator

Rewrite $\frac{1}{3}k$ as $\frac{2}{6}k$ and $\frac{1}{2}k$ as $\frac{3}{6}k$. The equation becomes $\frac{2}{6}k+80=\frac{3}{6}k + 120$.

Step2: Isolate variable terms

Subtract $\frac{2}{6}k$ from both sides: $\frac{2}{6}k-\frac{2}{6}k+80=\frac{3}{6}k-\frac{2}{6}k + 120$, which simplifies to $80=\frac{1}{6}k + 120$.

Step3: Isolate the variable - term with k

Subtract 120 from both sides: $80 - 120=\frac{1}{6}k+120 - 120$, getting $- 40=\frac{1}{6}k$.

Step4: Solve for k

Multiply both sides by 6: $6\times(-40)=6\times\frac{1}{6}k$, so $k=-240$.

Answer:

  1. $n = 8$
  2. $k=-240$