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4 years ago

PLease help!!!!!!!!!!!!!!!!!!!!!!1

Mathematics

2 answers:

Andrew [12]4 years ago

5 0

4(5x-20)=-20
20x-80=-20 ======> Add 80 to both sides
20x-80+80=-20+80
20x=60======> Divide both sides by 20
20/20x=60/20
x=3

schepotkina [342]4 years ago

3 0

4(5x-20)= -20

20x-80= -20
+80 +80
____________

20x=60
_______
20

x=3

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adelina 88 [10]

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}\\\\
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I hope this helps. =)

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

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