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3 years ago

What is the solution to this equation? –5(s – 30) = –10

Mathematics

2 answers:

Margarita [4]3 years ago

7 0

Distribute first:

-5s + 150 = -10

Subtract 150:

-5s = -160

Divide by -5:

s = 32

KonstantinChe [14]3 years ago

4 0

Answer:

There is nothing more that can be done with this equation to solve it. If you mean to solve for s, your answer would be 32.

How? You divide both sides by the numeric factor on the left and solve.

If you mean something else, let me know in the comments and I'll help! I just don't fully understand since the question is already solved.

Hope I helped! ~Chaos

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3 years ago

-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.

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

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

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

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

$x = \frac{23}{27}\\$ is an extraneous solution.

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