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3 years ago

What is x for this problem? i really need help. will offer a lot of points

Mathematics

2 answers:

Umnica [9.8K]3 years ago

8 0

X should equal 52. Sorry if I am wrong.

Valentin [98]3 years ago

3 0

Since we are trying to isolate $x$ , we are going to do the reverse operations of what is "connected to" $x$ .

$\dfrac{x + 8}{-4} + 9 = -6$

Set up

$\dfrac{x + 8}{-4} = -15$

Subtract 9 from both sides

$x + 8 = 60$

Multiply both sides by -4

$x = 52$

Subtract 8 from both sides

The answer to this problem is x = 52.

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I don’t really understand this?

Serga [27]

You replace the p in the given expression by -2 and the m by 3 and simplify

So its (-2)^ / 4 - 3

= 4/4 - 3

= 1 - 3 (note the division 4/4 is done first according to the order of operations rule)

= -2 answer

4 0

3 years ago

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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

Wazi has sixty three 5 $ bills he takes the 5 $ bills to bank to exchange for 20$ how many 20$ bills does Wes I get

dmitriy555 [2]

15

63 x 5 = 315
315/20 = 15.75
hence, 15 $20 bills

4 0

3 years ago

Please help me

nevsk [136]

25.20/420= .06. 6% sales tax

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3 years ago

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The population of rabbits on an island is growing exponentially. In the year 2001,

Alexxandr [17]

I think the answer is 666.

2050-940=1110

1110/5=222

222x3=666

6 0

2 years ago

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