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STatiana [176]
3 years ago
5

1/3+2/3m=2/3m-2/3 Does it have no solution or is it an identity?

Mathematics
1 answer:
Crazy boy [7]3 years ago
3 0
\dfrac{1}{3}+\dfrac{2}{3}m=\dfrac{2}{3}m-\dfrac{2}{3}\ \ \ \ \ |subtract\ \dfrac{2}{3}m\ from\ both\ sides\\\\\dfrac{1}{3}=-\dfrac{2}{3}\ \ \ \ FALSE\\\\Answer:\ No\ solution
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Find all solutions of given equation<br> 2cosx-1=0
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There are an infinite number of solutions, so I don't plan to list them all.
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You said that      <u>2cos(x) - 1  =  0</u>

Add  1  to each side:     2cos(x) = 1

Divide each side by  2:    cos(x) = 1/2

The angles whose cosine is 1/2 are  60 degrees, 300 degrees,
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Contact [7]

Example 2: x=10

Example 3: x=-1

Example 4: x=35

Example 5: a-b+c=9

Example 6: x=1, x=1/2

Further explanation:

<u>Example 2:</u>

\sqrt{4x-4} =6

Taking Square on both sides

(\sqrt{4x-4})^2 =(6)^2\\4x-4=36\\4x=36+4\\4x=40\\\frac{4x}{4}=\frac{40}{4}\\x=10

<u>Example 3:</u>

9+\sqrt{4x+8}=11

Subtracting 9 from both sides

9-9+\sqrt{4x+8}=11-9\\\sqrt{4x+8}=2\\

Taking Square on both sides

(\sqrt{4x+8})^2=(2)^2\\4x+8=4\\4x=4-8\\4x=-4\\\frac{4x}{4}=-\frac{4}{4}\\x=-1

<u>Example 4:</u>

8+\sqrt{x+1}=2

subtracting 8 from both sides

8-8+\sqrt{x+1}=2-8\\\sqrt{x+1}=-6

Taking Square on both sides

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Subtracting 1 from both sides

x+1-1=36-1\\x=35

<u>Example 5:</u>

\sqrt{a-b+c}=3

Taking Square on both sides

(\sqrt{a-b+c})^2=(3)^2\\a-b+c=9\\

<u>Example 6:</u>

2\sqrt{2x-1}=4x-2

Taking square on both sides

(2\sqrt{2x-1})^2=(4x-2)^2\\4(2x-1)=16x^2+4-16x\\8x-4=16x^2-16x+4\\8x=16x^2-16x+4+4\\8x=16x^2-16x+8\\16x^2-16x-8x+8=0\\16x^2-24x+8=0

dividng whole equation by 8

2x^2-3x+1=0\\2x^2-2x-x+1=\\2x(x-1)-1(x-1)=0\\(2x-1)(x-1)=0\\2x-1=0 => x=\frac{1}{2}\\x-1=0 => x=1

Keywords: Radical Expressions, Examples

Learn more about radical expressions at:

  • brainly.com/question/8145673
  • brainly.com/question/10071175

#LearnwithBrainly

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Please help with these?
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Answer:

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