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3 years ago

Need held with this question

Mathematics

2 answers:

Alex3 years ago

5 0

What you should do in this case is to follow the following steps:
1) Take out common factor (5 / 3u)
2) Rewrite the expression
3) Take out common factor (1/5)
4) Rewrite the expression
Answer:
The simplified expression is:
(6u-5) / (21u-5)
See attached image.

Lapatulllka [165]3 years ago

3 0

The answer to your question is 6u-5/21u-5

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The sum of three consecutive odd integers is 18

DIA [1.3K]

Answer: If x =7, then x + 2 = 9 and x + 4 = 11. These are your three consecutive odd integers.

Step-by-step explanation:

4 0

3 years ago

Dos anyone know how much a is 2(a+7)-7=9

Oduvanchick [21]

Answer:

a =1

Step-by-step explanation:

2(a+7)-7=9

Add 7 to each side

2(a+7)-7+7=9+7

2(a+7)=16

Divide by 2

a+7 = 16/2

a +7 = 8

Subtract 7

a+7-7 = 8-7

a = 1

3 0

4 years ago

Read 2 more answers

-3|9x-7|=2 Find the answer for x

labwork [276]

<h2>Solving Equations with Absolute Expressions</h2><h3>Answer:</h3>

No Solutions

<h3>Step-by-step explanation:</h3>

Given:

$-3|9x -7| = 2$

Rewriting the given equation:

$-3|9x -7| = 2 \\ |9x -7| = -\frac{2}{3}$

We have to realize that the right side of the equation, $|9x -7|$ , will always be positive no matter what real values of $x$ (because we're taking the absolute value of the expression) and we are equating it to a negative constant number, $-\frac{2}{3}\\$ . Something that is always positive will never be negative so there's no value for $x$ that satisfies the solution.

$\rule{6.5cm}{0.5pt}$

You may not read the following passage that I have written.

$\rule{6.5cm}{0.5pt}$

Solving by positive of the expression:

$9x -7 = -\frac{2}{3} \\ 9x = -\frac{2}{3} +7 \\ 9x = -\frac{2}{3} +\frac{21}{3} \\ 9x = \frac{19}{3} \\ 9x \times \frac{1}{9} = \frac{19}{3} \times \frac{1}{9} \\ x = \frac{19}{27}$

Solving by the negative of the expression:

$-(9x -7)= -\frac{2}{3} \\ 9x -7 = \frac{2}{3} \\ 9x = \frac{2}{3} +7 \\ 9x = \frac{2}{3} +\frac{21}{3} \\ 9x = \frac{23}{3} \\ 9x \times \frac{1}{9} = \frac{23}{3} \times \frac{1}{9} \\ x = \frac{23}{27}$

Checking: $x = \frac{19}{27}\\$

$-3|9(\frac{19}{27}) -7| \stackrel{?}{=} 2 \\ -3|\frac{19}{3} -7| \stackrel{?}{=} 2 \\ -3|\frac{19}{3} -\frac{21}{3}| \stackrel{?}{=} 2 \\ -3|-\frac{2}{3}| \stackrel{?}{=} \\ -3(\frac{2}{3}) \stackrel{?}{=} 2 \\ -2 \stackrel{?}{=} 2 \\ -2 \neq 2$

$x = \frac{19}{27}\\$ is an extraneous solution.

Checking: $x = \frac{23}{27}\\$

$-3|9(\frac{23}{27}) -7| \stackrel{?}{=} 2 \\ -3|\frac{23}{3} -7| \stackrel{?}{=} 2 \\ -3|\frac{23}{3} -\frac{21}{3}| \stackrel{?}{=} 2 \\ -3|\frac{2}{3}| \stackrel{?}{=} \\ -3(\frac{2}{3}) \stackrel{?}{=} 2 \\ -2 \stackrel{?}{=} 2 \\ -2 \neq 2$