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morpeh [17]
2 years ago
6

Solve (2/3)-(1/x)=5/6

Mathematics
1 answer:
Keith_Richards [23]2 years ago
5 0
\frac{2}{3} - \frac{1}{x} = \frac{5}{6}   Multiply both sides by 3
2 - \frac{1(3)}{x} = \frac{5(3)}{6}   Simplify the numerators
2 - \frac{3}{x} = \frac{15}{6}   Multiply both sides by 6
12 - \frac{3(6)}{x} = 15   Simplify the numerator
12 - \frac{18}{x} = 15   Multiply both sides by x
12x - 18 = 15x   Subtract 12x from both sides
       -18 = 3x    Divide both sides by 3
        -6 = x      Switch the sides to make it easier to read
         x = -6

Check your answer by plugging -6 in for the x in the original problem:

\frac{2}{3} - \frac{1}{x} = \frac{5}{6}   Plug in -6 for x
\frac{2}{3} - \frac{1}{-6} = \frac{5}{6}   Change \frac{2}{3} to \frac{4}{6} so all denominators are the same
\frac{4}{6} - \frac{1}{-6} = \frac{5}{6}   Change \frac{1}{-6} to  \frac{-1}{6} for easier work
\frac{4}{6} - \frac{-1}{6} = \frac{5}{6}   Subtract the numerators (4 - (-1) )
\frac{5}{6} = \frac{5}{6}

So, x = -6 .
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Answer:

\large\boxed{1.\ (4)^{-3x^2}=\left(\dfrac{1}{4}\right)^{3x^2}}

\large\boxed{2.\ ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x}

Step-by-step explanation:

Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\\------------\\\\(4)^{-3x^2}=\left[(4)^{-1}\right]^{3x^2}=\left(\dfrac{1}{4}\right)^{3x^2}

Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\ and\ (a^n)^m=a^{nm}\\--------------------\\\\ab^{-3x}=a\cdot b^{-3x}=a\left[(b)^{-1}\right]^{3x}=a\left(\dfrac{1}{b}\right)^{3x}\\\\ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x

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2 years ago
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If a+b+c=0 for a,b,c then prove a²/bc+b²/ca+c²/ab=3.
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According to the identity if a+b+c=0
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<span>a²/bc+b²/ca+c²/ab=3. 
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3 years ago
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