For each real number k except 2, the system of equations below has a unique solution (x, y… — SAT

For each real number

k

except

2

, the system of equations below has a unique solution

(x, y)

\begin{aligned} kx + (k + 2)y &= 3k + 1 \\ (k - 1)x + ky &= 2k - 1 \end{aligned}

If the solution

(x, y)

to the system also satisfies the equation

x + 2y = 6

, what is the value of

k

Answer: 4

Answer

The value of

k

is 4.

Subtracting the second equation from the first equation yields

x + 2y = k + 2

. Since the solution must also satisfy

x + 2y = 6

, we set

k + 2 = 6

, which gives

k = 4

Step-by-Step Solution

Subtract the second equation from the first equation.

x + 2y = k + 2

To find a direct linear combination of

x

and

y

that can be compared directly to the target equation.

Equate the resulting expression to the target equation

x + 2y = 6

k + 2 = 6

Since the solution

(x, y)

must satisfy

x + 2y = 6

, the value of the linear combination

x + 2y

from the system must equal

6

Solve for

k

by subtracting 2 from both sides.

k = 4

Isolating the variable

k

yields the final solution.

Key Concept

Solving systems of linear equations with parameter coefficients by identifying algebraic structure and linear combinations.

Alternative Method

Solve the system for

x

and

y

in terms of

k

using elimination. Multiplying the first equation by

(k-1)

and the second by

k

, and then subtracting them yields

y = \frac{k^2 - k - 1}{k - 2}

. Substituting this back gives

x = \frac{-k^2 + 2k - 2}{k - 2}

. Substituting these expressions into the equation

x + 2y = 6

results in

\frac{k^2 - 4}{k - 2} = 6

. For

k \neq 2

, factoring

k^2 - 4

(k-2)(k+2)

allows simplification to

k + 2 = 6

, which yields

k = 4

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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