For a real constant k \neq -1, consider the linear equation in one variable x: \frac{2(x -… — GRE General Test

For a real constant

k \neq -1

, consider the linear equation in one variable

x

\frac{2(x - 3)}{k + 1} - \frac{x + 1}{3} = 1

Which of the following values of

k

result in a solution

x

that is a positive integer? Select all such values.

-2Cevap
B
0
C
1
2Cevap
3Cevap

Cevap

The values of

k

that yield a positive integer solution for

x

are

-2

2

, and

3

Solving the linear equation for

x

in terms of

k

gives

x = \frac{4k + 22}{5 - k}

. Substituting

k = -2

yields

x = 2

, substituting

k = 2

yields

x = 10

, and substituting

k = 3

yields

x = 17

. All three resulting values of

x

are positive integers.

Adım Adım Çözüm

Clear denominators by multiplying the entire equation by

3(k + 1)

6(x - 3) - (x + 1)(k + 1) = 3(k + 1)

Eliminating fractional terms simplifies isolation of the variable

x

Expand all terms on both sides of the equation.

6x - 18 - (kx + x + k + 1) = 3k + 3 \implies 5x - kx - 19 - k = 3k + 3

Carefully distribute negative signs and combine like terms.

Group terms containing

x

on the left side and constant/k terms on the right side.

(5 - k)x = 4k + 22 \implies x = \frac{4k + 22}{5 - k}

Express

x

explicitly as a rational function of the parameter

k

Evaluate

x

for each given option to determine which produce positive integers.

For

k = -2

x = 14/7 = 2

(integer);
For

k = 0

x = 22/5 = 4.4

(not integer);
For

k = 1

x = 26/4 = 6.5

(not integer);
For

k = 2

x = 30/3 = 10

(integer);
For

k = 3

x = 34/2 = 17

(integer).

Test each candidate value of

k

against the condition that

x

must be a positive integer.

Anahtar Kavram

Solving parametric linear equations in one variable and evaluating integer solutions

Tahmini Süre:2m 30s

Soru

Cevap

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