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Sagot :
The proportion that is equivalent to the original equation [tex]\frac{x+1}{2x} =\frac{x^{2} -7x+10}{4x}[/tex]
The true solution to the equation [tex]x_{1} =1, x_{2}=8[/tex]
The extraneous solution to the equation.x=0
You are given the equation
[tex]\frac{1}{2} + \frac{1}{2x}=\frac{x^{2} -7x+10}{4x}[/tex]
we will note that
[tex]\frac{1}{2} + \frac{1}{2x}=\frac{x+1}{2x}[/tex]
this equation can now be rewritten as proportions as
[tex]\frac{x+1}{2x}=\frac{x^{2} -7x+10}{4x}[/tex]
Solving this equation using the main property of proportion:
[tex]4x. (x+1) =2x.(x^{2} -7x+10.\\[/tex]
[tex]2x(2x+2-x^{2} +7x-10) =0,[/tex]
[tex]2x(-x^{2} +9x-8) =0[/tex]
x cannot be equal to 0 (it is placed in the denominator of the initial equation and the denominator cannot be 0), so x=0 is the extraneous solution to the equation.
This will hence translate to
[tex]-x^{2} +9x-8=0[/tex]
[tex]x^{2} +9x+8=0,[/tex]
[tex]x_{1,2} = \frac{\sqrt{(-9)^2-4 . 8} }{2} =[/tex] 9±7/2
=1, 8
Learn more about Equations here: https://brainly.com/question/22688504
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