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Kobotan [32]
2 years ago
9

3(x – 2) = 3x + 4 solve for X

Mathematics
2 answers:
olya-2409 [2.1K]2 years ago
6 0

Answer: There is no solution.

Step-by-step explanation:

Simplify both sides of the equation

subtract 3x from both sides

add 6 to both sides

0=10

The answer is - there are no solutions

Molodets [167]2 years ago
4 0

Answer:

The answer is: Use a calculator.

Step-by-step explanation

With the calculator you are able to do ANY problem.

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Xis less than or equal to 3

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The length of a rectangle is 4 units shorter than half the width. If the length of the rectangle is 18 units, which equation can
never [62]

Answer:

The width is 11 units.

Step-by-step explanation:

First, you have to add 4 to 18 which is 22. Then divide by 2.

8 0
3 years ago
Carly solved a quadratic equation by completing the square, but her work has errors. Identify the first error in Carly's work.
Savatey [412]

Answer:

Her first error was adding the wrong number to both sides. Instead of 4², it should be 2²

Step-by-step explanation:

When completing the square, the number that needs to be added to both sides is (b/2)²

In this problem, b is 4. So, Carly should have added 2² to both sides, but she added 4² instead.

So, her first error was adding 4² to both sides instead of 2²

8 0
3 years ago
What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove
wolverine [178]

Answer:

The solutions are:

b=0,\:b=4

Step-by-step explanation:

Considering the expression

  • \frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

Solving the expression

\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2

5b^3-10=6b^3-4b^2-10

\mathrm{Switch\:sides}

6b^3-4b^2-10=5b^3-10

6b^3-4b^2-10+10=5b^3-10+10

6b^3-4b^2=5b^3

\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}

6b^3-4b^2-5b^3=5b^3-5b^3

b^3-4b^2=0

Using\:the\:Zero\:Factor\:Principle: if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

So,

b=0,b-4=0

b=0,b=4

Therefore, the solutions are:

b=0,\:b=4

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