Question

question #9 solve the equation. remember to check for extraneous solutions. $\frac{5k - 5}{k}=\frac{3k + 18}{k}$ $k=\frac{13}{2}$ $k=\frac{13}{8}$ $k=\frac{23}{8}$ $k=\frac{23}{2}$ question #10 in order to create a 56% solution, two different saline solutions will be mixed together. how many milliliters of a 60% saline solution must be mixed with 2 milliliters of a 40% saline solution to create the 56% solution? round your answer to the nearest hundredth 6 ml 3 ml 8 ml 14 ml

Explanation:

Step1: Cross - multiply the equation

Since $\frac{5k - 5}{k}=\frac{3k + 18}{k}$, cross - multiplying gives $k(5k - 5)=k(3k + 18)$. But since $k
eq0$ (as it is in the denominator), we can cancel out $k$ on both sides and get $5k-5 = 3k + 18$.

Step2: Isolate the variable $k$

Subtract $3k$ from both sides: $5k-3k-5=3k - 3k+18$, which simplifies to $2k-5 = 18$. Then add 5 to both sides: $2k-5 + 5=18 + 5$, so $2k=23$. Divide both sides by 2: $k=\frac{23}{2}$.

Step3: Check for extraneous solutions

The original equation has a denominator of $k$. When $k = \frac{23}{2}$, the denominators in the original rational equation are non - zero. So $k=\frac{23}{2}$ is a valid solution.

Step4: Solve the mixture problem

Let $x$ be the volume (in milliliters) of the 60% saline solution. The amount of salt in the 40% solution is $0.4\times2$ milliliters, the amount of salt in the 60% solution is $0.6x$ milliliters, and the amount of salt in the final 56% solution is $0.56(x + 2)$ milliliters.
Set up the equation based on the conservation of the amount of salt: $0.4\times2+0.6x=0.56(x + 2)$.
Expand the right - hand side: $0.8+0.6x=0.56x+1.12$.
Subtract $0.56x$ from both sides: $0.6x-0.56x+0.8=0.56x-0.56x + 1.12$, so $0.04x+0.8=1.12$.
Subtract 0.8 from both sides: $0.04x+0.8 - 0.8=1.12 - 0.8$, so $0.04x=0.32$.
Divide both sides by 0.04: $x = 8$.

Answer:

For the first equation, $k=\frac{23}{2}$
For the second problem, 8 ml