14. Write with a positive exponent 6^-3?

Which statement best reflects the solution(s) of the equation?

Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0$

Now, factoring the term:

$x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2$

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ the denominator becomes zero i.e

$\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}$

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2 in the equation,

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$

$\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2$

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

#learnwithBrainly

5 0

3 years ago

A square is shown below. Which expression can be used to find the area, in square units, of the shaded triangle in the square?

AlexFokin [52]

The area of a square would me multiplying the width times the length and then multiplying by one half would be like splitting the square in half so your answer would be that last one right there. 6*6= area then multiply by 1/2 = half of area Hope this helps! :D

6 0

2 years ago

You can make chili powder from fresh chilis. If 1 chili makes 1 tablespoon of chili powder, how many chilis should you buy?

Tresset [83]

It depends on how much chili powder you want to make. Probably just like one or two

6 0

3 years ago

Help me please ;-;!!!!! Use a coordinate grid to create a map of a town with at least five different locations, such as a house,

Sholpan [36]

A. Our diagram have six location: my house (2,1), a movie theater (5,3), the school (6,6), the police station (3,-2), the airport (-3,3), and the mall (-2,-3)

B. Distance between my house and the movie theater:
To calculate the distance we are going to use the points (2,1) and (5,3) to create a right triangle with tow legs of measures 3 and 2; the hypotenuse of the right triangle will be the distance between my house and the movie theater.
Using the Pythagorean theorem:
$d^2=3^2+2^2$
$d^2=9+4$
$d^2=13$
$d= \sqrt{13}$
We can conclude that the distance between my house and the movie theater is $\sqrt{13}$

Our second distance is the distance between the police station and the airport:
This time we are using the points (3,-2) and (-3,3) to create a right triangle with legs of measures 6 and 5; the hypotenuse of our triangle will be the the distance between our points:
$d^2=6^2+5^2$
$d^2=36+25$
$d^2=61$
$d= \sqrt{61}$
We can conclude that the distance between the police station and the airport is $\sqrt{61}$

C. Distance form the park to the Fire Station:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance from the park to the Fire Station is $\sqrt{10}$

Distance from the stadium to the mall:
$d^2=2^2+1^2$
$d^2=4+1$
$d^2=5$
$d= \sqrt{5}$
We can conclude that the distance from the stadium to the mall is $\sqrt{5}$

D. What is the distance between your house and the mall?
Answer:
$d^2=4^2+4^2$
$d^2=16+16$
$d^2=32$
$d= \sqrt{32}$
$d=4 \sqrt{2}$
We can conclude that the distance between my house and the mall is $4 \sqrt{2}$

What is the distance between the movie theater and the school?
answer:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance between the movie theater and the school is $\sqrt{10}$