To solve this problem, we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. Let's denote [tex]\( x \)[/tex] as the higher grade and [tex]\( y \)[/tex] as the lower grade. The system of equations given is:
[tex]\[ x - y = 16 \tag{1} \][/tex]
[tex]\[ \frac{1}{8}x + \frac{1}{2}y = 52 \tag{2} \][/tex]
We will solve this system step by step.
### Step 1: Solve Equation (1) for [tex]\( x \)[/tex]
From equation (1), we can express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ x - y = 16 \][/tex]
[tex]\[ x = y + 16 \tag{3} \][/tex]
### Step 2: Substitute Equation (3) into Equation (2)
We substitute [tex]\( x \)[/tex] from equation (3) into equation (2):
[tex]\[ \frac{1}{8}(y + 16) + \frac{1}{2}y = 52 \][/tex]
### Step 3: Simplify the Substitution
Distribute [tex]\(\frac{1}{8}\)[/tex] through the parentheses:
[tex]\[ \frac{1}{8}y + 2 + \frac{1}{2}y = 52 \][/tex]
Combine the [tex]\( y \)[/tex] terms:
[tex]\[ \left( \frac{1}{8} + \frac{4}{8} \right) y + 2 = 52 \][/tex]
[tex]\[ \frac{5}{8}y + 2 = 52 \][/tex]
### Step 4: Solve for [tex]\( y \)[/tex]
Subtract 2 from both sides to isolate the [tex]\( y \)[/tex] term:
[tex]\[ \frac{5}{8}y = 50 \][/tex]
Multiply both sides by [tex]\(\frac{8}{5}\)[/tex] to solve for [tex]\( y \)[/tex]:
[tex]\[ y = 50 \times \frac{8}{5} \][/tex]
[tex]\[ y = 10 \times 8 \][/tex]
[tex]\[ y = 80 \][/tex]
### Step 5: Use the Value of [tex]\( y \)[/tex] to Find [tex]\( x \)[/tex]
Substitute [tex]\( y = 80 \)[/tex] back into equation (3):
[tex]\[ x = y + 16 \][/tex]
[tex]\[ x = 80 + 16 \][/tex]
[tex]\[ x = 96 \][/tex]
Thus, the higher grade of Jose's two tests is:
[tex]\[ \boxed{96} \][/tex]
