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Zorluk: Çok zorSystems of Linear Equations
For each real number kk except 22, the system of equations below has a unique solution (x,y)(x, y).
kx+(k+2)y=3k+1(k1)x+ky=2k1\begin{aligned} kx + (k + 2)y &= 3k + 1 \\ (k - 1)x + ky &= 2k - 1 \end{aligned}
If the solution (x,y)(x, y) to the system also satisfies the equation x+2y=6x + 2y = 6, what is the value of kk?

Cevap: 4

Cevap

The value of kk is 4.
Subtracting the second equation from the first equation yields x+2y=k+2x + 2y = k + 2. Since the solution must also satisfy x+2y=6x + 2y = 6, we set k+2=6k + 2 = 6, which gives k=4k = 4.

Adım Adım Çözüm

1
Subtract the second equation from the first equation.
x+2y=k+2x + 2y = k + 2
To find a direct linear combination of xx and yy that can be compared directly to the target equation.
2
Equate the resulting expression to the target equation x+2y=6x + 2y = 6.
k+2=6k + 2 = 6
Since the solution (x,y)(x, y) must satisfy x+2y=6x + 2y = 6, the value of the linear combination x+2yx + 2y from the system must equal 66.
3
Solve for kk by subtracting 2 from both sides.
k=4k = 4
Isolating the variable kk yields the final solution.

Anahtar Kavram

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

Alternatif Yöntem

Solve the system for xx and yy in terms of kk using elimination. Multiplying the first equation by (k1)(k-1) and the second by kk, and then subtracting them yields y=k2k1k2y = \frac{k^2 - k - 1}{k - 2}. Substituting this back gives x=k2+2k2k2x = \frac{-k^2 + 2k - 2}{k - 2}. Substituting these expressions into the equation x+2y=6x + 2y = 6 results in k24k2=6\frac{k^2 - 4}{k - 2} = 6. For k2k \neq 2, factoring k24k^2 - 4 as (k2)(k+2)(k-2)(k+2) allows simplification to k+2=6k + 2 = 6, which yields k=4k = 4.
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