If \frac{2}{x} + \frac{3}{x - 2} = 2 and x > 1, what is the value of x - 1? — SAT

If $\frac{2}{x} + \frac{3}{x - 2} = 2$ and $x > 1$ , what is the value of $x - 1$ ?

Answer: 3

Answer

The correct answer is 3.

By multiplying both sides of the equation by the common denominator

x(x - 2)

, we obtain the quadratic equation

2x^2 - 9x + 4 = 0

. Factoring this equation yields the solutions

x = 0.5

and

x = 4

. Since the problem specifies that

x > 1

, we select

x = 4

. Substituting this value into the expression

x - 1

gives the final answer of 3.

Step-by-Step Solution

Multiply the equation by the common denominator

x(x - 2)

to eliminate the denominators.

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

This clears the fractions and allows us to work with a polynomial equation.

Expand and simplify both sides of the equation.

5x - 4 = 2x^2 - 4x

Combining like terms simplifies the expression before solving.

Rearrange the terms to set the quadratic equation equal to zero.

2x^2 - 9x + 4 = 0

Putting the equation in standard form

ax^2 + bx + c = 0

is necessary for factoring.

Factor the quadratic expression.

(2x - 1)(x - 4) = 0

Factoring allows us to find the roots of the quadratic equation.

Solve for

x

and apply the condition

x > 1

x = 4

The equation yields two solutions:

x = 0.5

and

x = 4

. The condition

x > 1

excludes

x = 0.5

, leaving

x = 4

as the only valid solution.

Calculate the value of the requested expression

x - 1

3

Substituting

x = 4

into

x - 1

gives

4 - 1 = 3

Key Concept

Solving rational equations by clearing denominators and solving the resulting quadratic equation while considering domain constraints.

Alternative Method

Instead of factoring, the quadratic formula can be used to solve

2x^2 - 9x + 4 = 0

, where

x = \frac{9 \pm \sqrt{(-9)^2 - 4(2)(4)}}{2(2)} = \frac{9 \pm 5}{4}

, giving

x = 0.5

and

x = 4

. Applying

x > 1

leaves

x = 4

, so

x - 1 = 3

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Topics

More questions from this topic