Question

question #4
eight quarts of an 80% alcohol solution was mixed with two quarts of a 50% alcohol solution. determine the concentration of the new mixture. round your answer to the nearest hundredth.
74%
50%
80%
25%
question #5
solve the equation. remember to check for extraneous solutions.
$\frac{4}{x + 6}+\frac{2x}{x + 6}=\frac{6}{x^{2}+6x}$
x = - 3
x = - 1 and x = 3
x = 1 and x = - 3
x = 1

Explanation:

Question #4

Step1: Calculate total alcohol in first solution

$8\times0.8 = 6.4$ (quarts of alcohol in 8 - quart 80% solution)

Step2: Calculate total alcohol in second solution

$2\times0.5=1$ (quarts of alcohol in 2 - quart 50% solution)

Step3: Calculate total volume of mixture

$8 + 2=10$ (quarts)

Step4: Calculate total alcohol in mixture

$6.4+1 = 7.4$ (quarts)

Step5: Calculate concentration of new mixture

$\frac{7.4}{10}\times100\%=74\%$

Step1: Combine left - hand side fractions

$\frac{4 + 2x}{x + 6}=\frac{6}{x^{2}+6x}$

Step2: Factor the denominator on the right

$\frac{4 + 2x}{x + 6}=\frac{6}{x(x + 6)}$

Step3: Cross - multiply (assuming $x

eq - 6$ and $x
eq0$)
$(4 + 2x)\times x=6$

Step4: Expand the left - hand side

$4x+2x^{2}=6$

Step5: Rearrange to standard quadratic form

$2x^{2}+4x - 6 = 0$

Step6: Divide through by 2

$x^{2}+2x - 3=0$

Step7: Factor the quadratic equation

$(x + 3)(x - 1)=0$

Step8: Solve for x

$x=-3$ or $x = 1$

Step9: Check for extraneous solutions

When $x=-3$:
Left - hand side: $\frac{4+2\times(-3)}{-3 + 6}=\frac{4-6}{3}=-\frac{2}{3}$
Right - hand side: $\frac{6}{(-3)^{2}+6\times(-3)}=\frac{6}{9 - 18}=-\frac{2}{3}$
When $x = 1$:
Left - hand side: $\frac{4+2\times1}{1 + 6}=\frac{6}{7}$
Right - hand side: $\frac{6}{1^{2}+6\times1}=\frac{6}{7}$
So $x = 1$ and $x=-3$ are valid solutions.

Answer:

74%

Question #5