Let's address each part of this question step-by-step.
### Question 36: Finding the Values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
We are given:
[tex]\[ 2^{x+y} = 2^{2x-y} = \sqrt{8} \][/tex]
First, we note that:
[tex]\[ \sqrt{8} = 2^{3/2} \][/tex]
Thus, the equations can be set up as follows:
[tex]\[ 2^{x+y} = 2^{3/2} \][/tex]
[tex]\[ 2^{2x-y} = 2^{3/2} \][/tex]
From these expressions, we can equate the exponents since the bases are the same (both are powers of 2):
[tex]\[ x + y = \frac{3}{2} \quad \text{(Equation 1)} \][/tex]
[tex]\[ 2x - y = \frac{3}{2} \quad \text{(Equation 2)} \][/tex]
Now, let's solve the system of linear equations:
1. [tex]\( x + y = \frac{3}{2} \)[/tex]
2. [tex]\( 2x - y = \frac{3}{2} \)[/tex]
First, we add both equations to eliminate [tex]\( y \)[/tex]:
[tex]\[ (x + y) + (2x - y) = \frac{3}{2} + \frac{3}{2} \][/tex]
[tex]\[ 3x = 3 \][/tex]
[tex]\[ x = 1 \][/tex]
Next, substitute [tex]\( x = 1 \)[/tex] back into Equation 1 to find [tex]\( y \)[/tex]:
[tex]\[ 1 + y = \frac{3}{2} \][/tex]
[tex]\[ y = \frac{3}{2} - 1 \][/tex]
[tex]\[ y = \frac{1}{2} \][/tex]
Thus, the respective values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are:
[tex]\[ x = 1, \quad y = \frac{1}{2} \][/tex]
Therefore, the correct answer is:
(a) [tex]\( 1, \frac{1}{2} \)[/tex]
