Question

Half of the sum of two numbers is 12, while one-fourth of their products is 35. Find the numbers

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Answer 1

Answer: 10 and 14.

Explanation:

Let the numbers be x and y.

Make and solve a system of equations:

`\displaystyle\\\left \{ {{(x+y)/(2)=12\ \ \ \ (1) } \atop {(1)/(4)xy=35\ \ \ \ (2) }} \right. \\\\\\`

Multiply equation (1) by 2 and equation (2) by 4:

`\displaystyle\\\left \{ {{x+y=24} \atop {xy=140}} \right. \\\\\\\left \{ {{x=24-y} \atop {(24-y)*y=140\right.\\\\\\\left \{ {{x=24-y} \atop {24y-y^2=140}} \right. \\\\\\\left \{ {{x=24-y} \atop {-y^2+24y-140=0\ \ \ \ (3)}} \right.`

Multiply equation (1) by (-1):

`y^2-24x+140=0\\\\y^2-10x-14x+140=0\\\\y(y-10)-14(y-10)=0\\\\(y-10)(y-14)=0\\\\y-10=0\\\\y=10.\ \ \ \ \Rightarrow\\\\x=24-10\\\\x=14\\\\y-14=0\\\\y=14.\ \ \ \ \Rightarrow\\\\x=24-14\\\\x=10.`

Answer 2

The sum of two numbers = 12

x+y=12

product of two numbers = 35

xy=35

The sum of reciprocals = x

1

​

+

y

1

​

=

xy

x+y

​

=

35

12

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Explanation: