Answer :
The value of [tex]x[/tex] for which [tex]f(x, 2x) = 27[/tex] is A) 3.
To solve the problem, we start with the functional equation given:
[tex]f(3x + 2y, 2x - 5y) = 19x[/tex]
This tells us how the function [tex]f(x, y)[/tex] behaves for the given transformations of [tex]x[/tex] and [tex]y[/tex]. We need to find the function [tex]f(x, y)[/tex] itself. Let's assume a simpler form for the function:
[tex]f(x, y) = Ax + By + C[/tex]
Here, [tex]A, B,[/tex] and [tex]C[/tex] are constants. By substituting the transformed variables into this equation, we have:
[tex]f(3x + 2y, 2x - 5y) = A(3x + 2y) + B(2x - 5y) + C[/tex]
Simplifying this, we get:
[tex]= 3Ax + 2Ay + 2Bx - 5By + C[/tex]
Combining like terms:
[tex]= (3A + 2B)x + (2A - 5B)y + C[/tex]
We need this to equal [tex]19x[/tex], which has no terms in [tex]y[/tex] and a constant term of zero. Therefore, we set up the following equations by matching coefficients:
- [tex]3A + 2B = 19[/tex]
- [tex]2A - 5B = 0[/tex]
- [tex]C = 0[/tex]
From the second equation [tex]2A - 5B = 0[/tex], we can express [tex]A[/tex] in terms of [tex]B[/tex]:
[tex]A = \frac{5B}{2}[/tex]
Next, we substitute [tex]A[/tex] back into the first equation:
[tex]3\left(\frac{5B}{2}\right) + 2B = 19[/tex]
This simplifies to:
[tex]\frac{15B}{2} + 2B = 19[/tex]
Converting [tex]2B[/tex] into halves:
[tex]\frac{15B}{2} + \frac{4B}{2} = 19[/tex]
[tex]\frac{19B}{2} = 19[/tex]
Multiplying both sides by 2, we get:
[tex]19B = 38[/tex]
[tex]B = 2[/tex]
Now substituting back to find [tex]A[/tex]:
[tex]A = \frac{5(2)}{2} = 5[/tex]
Now we have determined that:
[tex]A = 5, B = 2, C = 0[/tex]
Thus, our function is:
[tex]f(x, y) = 5x + 2y[/tex]
Next, we need to solve for [tex]x[/tex] such that [tex]f(x, 2x) = 27[/tex]. Substituting into our function, we have:
[tex]f(x, 2x) = 5x + 2(2x) = 5x + 4x = 9x[/tex]
Setting this equal to 27:
[tex]9x = 27[/tex]
[tex]x = 3[/tex]
