1.1 Linear Functions Solve each equation for the unknown letters [1.1.73] The letters 𝑎, 𝑏, 𝑐 are constants.

1.1 Linear Functions

Solve each equation for the unknown letters [1.1.73]

The letters $a, b, c$ are constants.

$\displaystyle \frac{1}{x-a}+\frac{1}{x+a}=\frac{2}{x-1}$

$\displaystyle \Rightarrow (x-a)(x+a)(x-1) \left[ \frac{1}{x-a}+\frac{1}{x+a}=\frac{2}{x-1} \right]$

$\displaystyle \Rightarrow \frac{(x-a)(x+a)(x-1)}{x-a} + \frac{(x-a)(x+a)(x-1)}{x+a} = \frac{2(x-a)(x+a)(x-1)}{x-1}$

$\displaystyle \Rightarrow \frac{(x-a)(x+a)(x-1)}{(x-a)} + \frac{(x-a)(x+a)(x-1)}{(x+a)} = \frac{2(x-a)(x+a)(x-1)}{(x-1)}$

$\displaystyle \Rightarrow (x+a)(x-1)+(x-a)(x-1)=2(x-a)(x+a)$

DIFFERENCE OF TWO SQUARES

$\displaystyle (x-a)(x+a)=x^2-a^2$

$\displaystyle \Rightarrow x(x+a)-(x+a)+[x^2-x-ax+1]=2[x^2-a^2]$

$\displaystyle \Rightarrow x^2+ax-x-a+x^2-x-ax+1=2x^2-2a^2$

$\displaystyle \Rightarrow x^2-2x-a+x^2+1-2x^2+2a^2=0$

$\displaystyle \Rightarrow -2x-a+1+2a^2=0 \Rightarrow 2a^2-2x-a+1=0$

$\displaystyle \Rightarrow 2a^2-2x-a+1=0 \Rightarrow -2x=-2a^2+a-1$

$\displaystyle \Rightarrow \left[-\frac{1}{2}\right](-2)x = -\frac{1}{2}[-2a^2+a-1]$

$\displaystyle \Rightarrow x=a^2-\frac{1}{2}a+\frac{1}{2}.$

$\displaystyle \therefore x=a^2-\frac{1}{2}a+\frac{1}{2}.$

TYPO ↑ [We checked Wolfram and ChatGPT for the answer and then found the typo.]

$\displaystyle (x+a)(x-1)+(x-a)(x-1)=2(x-a)(x+a)$

$\displaystyle \Rightarrow (x^2-x+ax-a)+(x^2-x-ax+a) = 2x^2-2a^2 \Rightarrow -2x=-2a^2$

$\displaystyle \therefore x=a^2.$

[RP] Show the answer is correct.

Continue with PLEM Academy

Join the PLEM Academy Research Position to review mathematics, physics, engineering, and related STEM lessons, identify typos and corrections, and participate in the research and editing process.

Learn More About the Research Position

For fiction, member releases, mailers, books, and other bonus material, join the Fiction Fan Club at AuthorJonD.com .

1.1 Linear Functions

Continue with PLEM Academy

Share this: