Consider the system of linear equations in two variables x and y shown below, where k is a… — GRE General Test

Consider the system of linear equations in two variables

x

and

y

shown below, where

k

is a constant:

\begin{aligned} 2x + ky &= 10 \\ 3x - 6y &= 15 \end{aligned}

Which of the following statements must be true? Select all such statements.

If $k = -4$ , the system has infinitely many solutions.Answer
If $k = 0$ , the system has a unique solution $(x, y) = (5, 0)$ .Answer
C
If $k = 4$ , the system has no solution.
D
If $k = 2$ , the solution to the system satisfies $x + y = 7$ .
E
There is no real value of $k$ for which the system has a solution with $y > 0$ .

The correct statements are the ones stating that if

k = -4

, the system has infinitely many solutions, and if

k = 0

, the system has a unique solution

(x, y) = (5, 0)

Analyzing the simplified relation

(k + 4)y = 0

demonstrates that setting

k = -4

makes the equation identity

0y = 0

, yielding infinitely many solutions. For any other value of

k

, including

k = 0

y

must equal

0

, which gives

x = 5

, establishing a unique solution at

(5, 0)

Simplify the second equation to express

x

in terms of

y

3x - 6y = 15 \implies x - 2y = 5 \implies x = 2y + 5

Expressing

x

explicitly allows direct substitution into the first linear equation.

Substitute

x = 2y + 5

into the first equation

2x + ky = 10

2(2y + 5) + ky = 10 \implies 4y + 10 + ky = 10 \implies (k + 4)y = 0

This reduces the 2x2 system to a single linear equation in

y

parameterized by

k

Analyze the conditions for

y

based on the parameter

k

k = -4

, the equation becomes

0y = 0

, which is true for all real

y

(infinitely many solutions). If

k \neq -4

, then

y = 0

and

x = 5

(a unique solution).

Determines system consistency and solution multiplicity across all values of

k

Parametric Analysis of 2x2 Linear Systems

Question