Answer:
there is no solution for this equation because the x that make equation true but if you solve it the x would be gone
[tex]x=\frac{10}{3}[/tex]
Step-by-step explanation:
[tex]\frac{4}{x+2} +\frac{1}{x-2} =\frac{5}{x} \\\\(\frac{4}{x+2} +\frac{1}{x-2} =\frac{5}{x})((x+2)(x-2)(x)) \\\\4(x-2)(x)+1(x+2)(x)}= 5(x+2)(x-2)\\4x^{2} -8x+x^{2} +2x=5x^{2} -20\\5x^{2} -6x=5x^{2} -20\\6x=20\\x=\frac{20}{6}=\frac{10}{3}[/tex]
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Answers & Comments
Answer:
there is no solution for this equation because the x that make equation true but if you solve it the x would be gone
Answer:
[tex]x=\frac{10}{3}[/tex]
Step-by-step explanation:
[tex]\frac{4}{x+2} +\frac{1}{x-2} =\frac{5}{x} \\\\(\frac{4}{x+2} +\frac{1}{x-2} =\frac{5}{x})((x+2)(x-2)(x)) \\\\4(x-2)(x)+1(x+2)(x)}= 5(x+2)(x-2)\\4x^{2} -8x+x^{2} +2x=5x^{2} -20\\5x^{2} -6x=5x^{2} -20\\6x=20\\x=\frac{20}{6}=\frac{10}{3}[/tex]