x + y = 92
x/5 + y/2 = 34 /* both sides *10 *10.x/5 + 10.y/2 = 340 2x + 5y = 340
* x + y = 92 /both sides -2 2x + 5y = 340
-2x -2y = -184 + 2x +5y = 340 3y = 156 ** y = 32 **
x + y = 92 x + 32 = 92 x = 60
x + y = 92 5 1 x + 2 1 y = 34 x = 92 − y 2 x + 5 y = 340 2 ( 92 − y ) + 5 y = 340 184 − 2 y + 5 y = 340 3 y = 156 y = 52 x + 52 = 92 x = 40
The two numbers are 40 and 52. The first number is 40, and the second number is 52, satisfying both conditions given in the problem. The solution was found by setting up and solving a system of equations.
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[tex] \frac{1}{2} + 1 \frac{1}{4} + \frac{1}{4} \\ \\ = \frac{2}{4} + \frac{5}{4} + \frac{1}{4} \\ \\ = \frac{2 + 5 + 1}{4} = \frac{8}{4} = 2[/tex]Jawabannya C. 2