Question

Classify the solutions of $\frac{1}{x+4}$ + $\frac{1}{2}$ = $\frac{1}{x+4}$ as extraneous or non-extraneous.

Answer 1

Answer:

$\huge\star \underline{ \boxed{ \purple{Answer}}}\star$

$\frac{1}{x\:+\:4}\:+\:\frac{1}{2}\:=\:\frac{1}{x\:+\:4}$

$\frac{2\:+\:(x\:+\:4)}{2(x\:+\:4}\:=\:\frac{1}{x\:+\:4}$

$\frac{6\:+\:x}{2x\:+\:8}\:=\:\frac{1}{x\:+\:4}$

$\red{On\:cross\:multiplying\:=}$

$(6\:+\:x)(x\:+\:4)\:=\:2x\:+\:8$

$\blue{Opening\:brakets\:=}$

$6x\:+24\:+\:x^{2}\:+\:4x\:=\:2x\:+\:8$

$x^{2}\:+\:10x\:+\:24\:=\:2x\:+\:8$

$x^{2}\:+\:10x\:-\:2x+\:24\:-\:8\:=\:0$

$x^{2}\:+\:8x\:+\:16\:=\:0$

$x^{2}\:+\:4x\:+\:4x\:+\:16\:=\:0$

$x(x\:+\:4)\:+\:4(x\:+\:4)\:=\:0$

$(x\:+\:4)\:+\:(x\:+\:4)\:=\:0$

$\green{Therefore\:,}$

$x\:=\:-4\:,\:-4$

$-4\:=\:\green{non\:extraneous\:variable}$

Step-by-step explanation:

$\huge\mathcal\pink{Hope \: it \: helps \: you}$

$\huge\mathcal\pink{friend}$

Answer 2

Question :

• Classify the solutions of $\frac{1}{x+4}$ + $\frac{1}{2}$ = $\frac{1}{x+4}$ as extraneous or non-extraneous.

Solution :

$\: \: \: \: \: \large \tt \rightarrow \frac{1}{x + 4} + \frac{1}{2} = \frac{1}{x + 4}$

$\: \: \: \: \: \: \: \large \tt \rightarrow \frac{2 \div (x + 4)}{2(x + 4)} = \frac{1}{x \div 4}$

$\: \: \: \: \: \: \: \large \tt \rightarrow \frac{6 + x}{2x - 8} = \frac{1}{x - 4}$

Cross Multiplying...!

$\boxed{ \tt(6 + x)(x + 4) = 2x - 8}$

Let's Do...!

$\: \: \: \: \: \: \: \tt6x + 24 + {x}^{2} + 4x = 2x + 8$

$\: \: \: \: \: \: \: \: \tt {x}^{2} + 10x + 24 = 2x + 8$

$\: \: \: \: \: \: \: \: \tt {x}^{2} + 10x - 2x + 24 - 8 = 0$

$\: \: \: \: \: \: \: \: \tt{x}^{2} + 8x + 16 = 0$

$\: \: \: \: \: \: \: \: \tt{x}^{2} + 4x + 4x + 16 = 0$

$\: \: \: \: \: \: \: \tt \: x(x + 4) + 4(x + 4) = 0$

$\: \: \: \: \: \: \: \tt(x + 4) + (x + 4) = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: {\boxed {\boxed { \red{ \tt \: x = - 4 \: and \: - 4}}}}$