Solved on Feb 27, 2024

Multiply the given equation by the LCD to eliminate denominators. The resulting equation is $(x+2)(x-3) = (x-2)(x-3)$

STEP 1

Assumptions
1. The equation given is: $\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}$
2. The least common denominator (LCD) is $3x(x-2)$ .
3. We will multiply each term by the LCD to eliminate the denominators.

STEP 2

Multiply the first term by the LCD: $\left(\frac{x+2}{3x}\right)(3x)(x-2)$

STEP 3

Cancel the common factors in the first term: $\frac{x+2}{\cancel{3x}}(\cancel{3x})(x-2)$

STEP 4

After canceling, the first term becomes: $(x+2)(x-2)$

STEP 5

Multiply the second term by the LCD: $\left(-\frac{1}{x-2}\right)(3x)(x-2)$

STEP 6

Cancel the common factors in the second term: $-\frac{1}{\cancel{x-2}}(3x)(\cancel{x-2})$

STEP 7

After canceling, the second term becomes: $-3x$

STEP 8

Multiply the third term by the LCD: $\left(\frac{x-3}{3x}\right)(3x)(x-2)$

STEP 9

Cancel the common factors in the third term: $\frac{x-3}{\cancel{3x}}(\cancel{3x})(x-2)$

STEP 10

After canceling, the third term becomes: $(x-3)(x-2)$

STEP 11

Now, rewrite the original equation with the denominators eliminated: $(x+2)(x-2) - 3x = (x-3)(x-2)$
The resulting equation is: $x^2 - 4 - 3x = x^2 - 5x + 6$

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Multiply the given equation by the LCD to eliminate denominators. The resulting equation is $(x+2)(x-3) = (x-2)(x-3)$

STEP 1

Assumptions
1. The equation given is: $\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}$
2. The least common denominator (LCD) is $3x(x-2)$ .
3. We will multiply each term by the LCD to eliminate the denominators.

STEP 2

Multiply the first term by the LCD: $\left(\frac{x+2}{3x}\right)(3x)(x-2)$

STEP 3

Cancel the common factors in the first term: $\frac{x+2}{\cancel{3x}}(\cancel{3x})(x-2)$

STEP 4

After canceling, the first term becomes: $(x+2)(x-2)$

STEP 5

Multiply the second term by the LCD: $\left(-\frac{1}{x-2}\right)(3x)(x-2)$

STEP 6

Cancel the common factors in the second term: $-\frac{1}{\cancel{x-2}}(3x)(\cancel{x-2})$

STEP 7

After canceling, the second term becomes: $-3x$

STEP 8

Multiply the third term by the LCD: $\left(\frac{x-3}{3x}\right)(3x)(x-2)$

STEP 9

Cancel the common factors in the third term: $\frac{x-3}{\cancel{3x}}(\cancel{3x})(x-2)$

STEP 10

After canceling, the third term becomes: $(x-3)(x-2)$

STEP 11

Now, rewrite the original equation with the denominators eliminated: $(x+2)(x-2) - 3x = (x-3)(x-2)$
The resulting equation is: $x^2 - 4 - 3x = x^2 - 5x + 6$

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Multiply the given equation by the LCD to eliminate denominators. The resulting equation is (x+2)(x−3)=(x−2)(x−3)(x+2)(x-3) = (x-2)(x-3)(x+2)(x−3)=(x−2)(x−3)

STEP 1

Assumptions1. The equation given is: x+23x−1x−2=x−33x\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}3xx+2​−x−21​=3xx−3​2. The least common denominator (LCD) is 3x(x−2)3x(x-2)3x(x−2).3. We will multiply each term by the LCD to eliminate the denominators.

STEP 2

Multiply the first term by the LCD: (x+23x)(3x)(x−2)\left(\frac{x+2}{3x}\right)(3x)(x-2)(3xx+2​)(3x)(x−2)

STEP 3

Cancel the common factors in the first term: x+23x(3x)(x−2)\frac{x+2}{\cancel{3x}}(\cancel{3x})(x-2)3xx+2​(3x)(x−2)

STEP 4

After canceling, the first term becomes: (x+2)(x−2)(x+2)(x-2)(x+2)(x−2)

STEP 5

Multiply the second term by the LCD: (−1x−2)(3x)(x−2)\left(-\frac{1}{x-2}\right)(3x)(x-2)(−x−21​)(3x)(x−2)

STEP 6

Cancel the common factors in the second term: −1x−2(3x)(x−2)-\frac{1}{\cancel{x-2}}(3x)(\cancel{x-2})−x−2​1​(3x)(x−2​)

STEP 7

After canceling, the second term becomes: −3x-3x−3x

STEP 8

Multiply the third term by the LCD: (x−33x)(3x)(x−2)\left(\frac{x-3}{3x}\right)(3x)(x-2)(3xx−3​)(3x)(x−2)

STEP 9

Cancel the common factors in the third term: x−33x(3x)(x−2)\frac{x-3}{\cancel{3x}}(\cancel{3x})(x-2)3xx−3​(3x)(x−2)

STEP 10

After canceling, the third term becomes: (x−3)(x−2)(x-3)(x-2)(x−3)(x−2)

STEP 11

Now, rewrite the original equation with the denominators eliminated: (x+2)(x−2)−3x=(x−3)(x−2)(x+2)(x-2) - 3x = (x-3)(x-2)(x+2)(x−2)−3x=(x−3)(x−2)The resulting equation is: x2−4−3x=x2−5x+6x^2 - 4 - 3x = x^2 - 5x + 6x2−4−3x=x2−5x+6

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Multiply the given equation by the LCD to eliminate denominators. The resulting equation is $(x+2)(x-3) = (x-2)(x-3)$

Assumptions
1. The equation given is: $\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}$
2. The least common denominator (LCD) is $3x(x-2)$ .
3. We will multiply each term by the LCD to eliminate the denominators.

Multiply the first term by the LCD: $\left(\frac{x+2}{3x}\right)(3x)(x-2)$

Cancel the common factors in the first term: $\frac{x+2}{\cancel{3x}}(\cancel{3x})(x-2)$

After canceling, the first term becomes: $(x+2)(x-2)$

Multiply the second term by the LCD: $\left(-\frac{1}{x-2}\right)(3x)(x-2)$

Cancel the common factors in the second term: $-\frac{1}{\cancel{x-2}}(3x)(\cancel{x-2})$

After canceling, the second term becomes: $-3x$

Multiply the third term by the LCD: $\left(\frac{x-3}{3x}\right)(3x)(x-2)$

Cancel the common factors in the third term: $\frac{x-3}{\cancel{3x}}(\cancel{3x})(x-2)$

After canceling, the third term becomes: $(x-3)(x-2)$

Now, rewrite the original equation with the denominators eliminated: $(x+2)(x-2) - 3x = (x-3)(x-2)$
The resulting equation is: $x^2 - 4 - 3x = x^2 - 5x + 6$