Question

topic 2 → solving rational equations
directions: write the excluded value for each equation (the x-value(s) that make the denominator = 0.
solve for the value(s) of x. write your final answer in the space provided. show work.

1. $\frac{x+3}{x-5} = \frac{x}{x-2}$

excluded values x ≈ _______
answer: x = _______

1. $\frac{12}{x} = \frac{x-2}{3}$

excluded values x ≈ _______
answer: x = _______

1. $\frac{1}{x-3} + \frac{x}{x-1} = \frac{2}{x^2-4x+3}$

excluded values x ≈ _______
answer: x = _______

Explanation:

Problem 1

Step1: Find excluded values

Set denominators to 0:
$x-5=0 \implies x=5$; $x-2=0 \implies x=2$

Step2: Cross-multiply to eliminate fractions

$(x+3)(x-2) = x(x-5)$

Step3: Expand both sides

$x^2 -2x +3x -6 = x^2 -5x$

Step4: Simplify equation

$x^2 +x -6 = x^2 -5x$
Subtract $x^2$ from both sides:
$x -6 = -5x$

Step5: Solve for x

$x +5x = 6$
$6x = 6$
$x = \frac{6}{6} = 1$

Step1: Find excluded values

Set denominator to 0:
$x=0$

Step2: Cross-multiply to eliminate fractions

$12 \times 3 = x(x-2)$

Step3: Expand and simplify

$36 = x^2 -2x$
Rearrange to standard quadratic:
$x^2 -2x -36 = 0$

Step4: Solve quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here $a=1, b=-2, c=-36$:
$x=\frac{2\pm\sqrt{(-2)^2-4(1)(-36)}}{2(1)} = \frac{2\pm\sqrt{4+144}}{2} = \frac{2\pm\sqrt{148}}{2} = \frac{2\pm2\sqrt{37}}{2} = 1\pm\sqrt{37}$

Step1: Factor denominator

$x^2-4x+3=(x-3)(x-1)$

Step2: Find excluded values

Set denominators to 0:
$x-3=0 \implies x=3$; $x-1=0 \implies x=1$

Step3: Multiply by common denominator

$(x-1) + x(x-3) = 2$

Step4: Expand and simplify

$x-1 + x^2 -3x = 2$
$x^2 -2x -1 = 2$
$x^2 -2x -3 = 0$

Step5: Factor quadratic

$(x-3)(x+1)=0$

Step6: Solve and exclude invalid x

$x=3$ (excluded, makes denominator 0), $x=-1$

Answer:

Excluded Values $x
eq 2, 5$