A system of linear equations is given by \begin{aligned} kx - 4y &= 12 \\ 3x + ky &= 5 \en… — SAT

A system of linear equations is given by

\begin{aligned} kx - 4y &= 12 \\ 3x + ky &= 5 \end{aligned}

where

k

is a constant. If the system has a unique solution

(x, y)

such that

x > 0

and

y < 0

, how many possible integer values of

k

exist?

A
7
B
8
9Cevap
D
10

Cevap

To find the number of integer values of

k

for which the system has a solution with

x > 0

and

y < 0

, we first express

x

and

y

in terms of

k

. Eliminating

y

by multiplying the first equation by

k

and the second by

4

and adding them yields

(k^2 + 12)x = 12k + 20

, or

x = \frac{12k+20}{k^2+12}

. Similarly, eliminating

x

yields

y = \frac{5k-36}{k^2+12}

. Since the denominator

k^2 + 12

is strictly positive for all real

k

, the sign of

x

and

y

depends solely on their numerators. For

x > 0

, we require

12k + 20 > 0

, which gives

k > -\frac{5}{3} \approx -1.67

. For

y < 0

, we require

5k - 36 < 0

, which gives

k < \frac{36}{5} = 7.2

. Combining these constraints gives the interval

-\frac{5}{3} < k < \frac{36}{5}

. The integers in this interval are

-1, 0, 1, 2, 3, 4, 5, 6,

and

7

, which is a total of 9 integers.

Adım Adım Çözüm

Solve the system of equations for

x

in terms of

k

by eliminating

y

x = \frac{12k + 20}{k^2 + 12}

Multiply the first equation by

k

and the second by 4, then add them:

k(kx - 4y) + 4(3x + ky) = 12k + 20 \implies (k^2 + 12)x = 12k + 20

Solve the system of equations for

y

in terms of

k

by eliminating

x

y = \frac{5k - 36}{k^2 + 12}

Multiply the first equation by 3 and the second by

k

, then subtract the first from the second:

k(3x + ky) - 3(kx - 4y) = 5k - 36 \implies (k^2 + 12)y = 5k - 36

Apply the condition

x > 0

to find a constraint on

k

k > -\frac{5}{3}

Since the denominator

k^2 + 12

is positive for all real values of

k

, the expression for

x

is positive if and only if its numerator is positive:

12k + 20 > 0 \implies k > -\frac{20}{12} = -\frac{5}{3}

Apply the condition

y < 0

to find another constraint on

k

k < \frac{36}{5}

Similarly, since

k^2 + 12 > 0

, the expression for

y

is negative if and only if its numerator is negative:

5k - 36 < 0 \implies k < \frac{36}{5}

Combine the constraints and count the number of integer values of

k

in the resulting interval.

9 integer values

The combined inequality is

-\frac{5}{3} < k < \frac{36}{5}

, which simplifies to approximately

-1.67 < k < 7.2

. The integers in this range are

-1, 0, 1, 2, 3, 4, 5, 6, 7

, giving a total of 9 integers.

Anahtar Kavram

Solving systems of linear equations with parameters under inequality constraints

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