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givi [52]
3 years ago
6

Find the difference between8/15 and 2/3 Show all calculations in your final answer.

Mathematics
1 answer:
kherson [118]3 years ago
3 0

Answer:

  • \frac{8}{15}-\left(\frac{-2}{3}\right)=\frac{6}{5}          

Step-by-step explanation:

  • Let the value of a number 'a' be = 8/15
  • Let the value of a number 'b' be = 2/3

The difference between the two numbers can be calculated by subtracting the numbers

a-b=\frac{8}{15}-\left(\frac{-2}{3}\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

         =\frac{8}{15}-\frac{-2}{3}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

          =\frac{8}{15}-\left(-\frac{2}{3}\right)

\mathrm{Apply\:rule}\:-\left(-a\right)=a

          =\frac{8}{15}+\frac{2}{3}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

          =\frac{8+10}{15}

          =\frac{18}{15}

\mathrm{Cancel\:the\:common\:factor:}\:3

           =\frac{6}{5}

Thus,

  • \frac{8}{15}-\left(\frac{-2}{3}\right)=\frac{6}{5}                    
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8/x-5 - 9/x-4 = 5/x^2-9x+20
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2752942

_______________


Solve the equation:

\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}


Reduce the fractions at the left side so that they have the same denominator:

\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}


Numerators must be equal:

\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
\mathsf{-x=-8}\\\\
\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}


I hope this helps. =)


Tags:  <em>rational equation fraction solution algebra</em>

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