Respuesta :
Answer:
Part 1. [tex]\dfrac{x+1}{2x}=\dfrac{x^2-7x+10}{4x}.[/tex]
Part 2. Solutions: [tex]x_1=1,\ x_2=8.[/tex]
Extraneous solution: [tex]x=0.[/tex]
Step-by-step explanation:
Part 1. You are given the equation
[tex]\dfrac{1}{2}+\dfrac{1}{2x}=\dfrac{x^2-7x+10}{4x}.[/tex]
Note that
[tex]\dfrac{1}{2}+\dfrac{1}{2x}=\dfrac{x+1}{2x},[/tex]
then the equation rewritten as proportion is
[tex]\dfrac{x+1}{2x}=\dfrac{x^2-7x+10}{4x}.[/tex]
Part 2. Solve this equation using the main property of proportion:
[tex]4x\cdot (x+1)=2x\cdot (x^2-7x+10),\\ \\2x(2x+2-x^2+7x-10)=0,\\ \\2x(-x^2+9x-8)=0.[/tex]
x cannot be equal 0 (it is placed in the denominator of the initial equation and denominator cannot be 0), so [tex]x=0[/tex] is extraneous solution to the equation.
Thus,
[tex]-x^2+9x-8=0,\\ \\x^2-9x+8=0,\\ \\x_{1,2}=\dfrac{9\pm\sqrt{(-9)^2-4\cdot 8}}{2}=\dfrac{9\pm7}{2}=1,\ 8.[/tex]
Part A: i forgot the question but i know its the third option (c)
Part B:
Solve the original equation by solving the proportion.
The solutions are: (1,6)
Part c:
Name the extraneous solution(s) to the equation
answer: Neither
These are for edu2020