Question

Part 1

Answer 1

Answer:

Part 1. $\dfrac{x+1}{2x}=\dfrac{x^2-7x+10}{4x}.$

Part 2. Solutions: $x_1=1,\ x_2=8.$

Extraneous solution: $x=0.$

Step-by-step explanation:

Part 1. You are given the equation

$\dfrac{1}{2}+\dfrac{1}{2x}=\dfrac{x^2-7x+10}{4x}.$

Note that

$\dfrac{1}{2}+\dfrac{1}{2x}=\dfrac{x+1}{2x},$

then the equation  rewritten as proportion is

$\dfrac{x+1}{2x}=\dfrac{x^2-7x+10}{4x}.$

Part 2. Solve this equation using the main property of proportion:

$4x\cdot (x+1)=2x\cdot (x^2-7x+10),\\ \\2x(2x+2-x^2+7x-10)=0,\\ \\2x(-x^2+9x-8)=0.$

x cannot be equal 0 (it is placed in the denominator of the initial equation and denominator cannot be 0), so $x=0$ is  extraneous solution to the equation.

Thus,

$-x^2+9x-8=0,\\ \\x^2-9x+8=0,\\ \\x_{1,2}=\dfrac{9\pm\sqrt{(-9)^2-4\cdot 8}}{2}=\dfrac{9\pm7}{2}=1,\ 8.$

Answer 2

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