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The sum of two numbers exceeds a third number by four. If the sum of the three numbers is at least 20 and 28,find any three intgral values satisfying the inequality .

User Kaufman
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Answer:

Let the three numbers be x, y, and z.

We are given that x + y = z + 4.

We are also given that x + y + z >= 20 and x + y + z <= 28.

Combining these two inequalities, we get:

z + 4 + z >= 20

2z >= 16

z >= 8

Since z is an integer, z can be 8, 9, 10, 11, 12, 13, 14, 15, 16, or 17.

For each value of z, we can find the corresponding values of x and y using the equation x + y = z + 4.

For example, if z = 8, then x + y = 12.

So, the three integral values satisfying the inequality are 8, 4, and 0.

Another example is z = 15.

In this case, x + y = 19.

So, the three integral values satisfying the inequality are 15, 2, and 2.

There are many other possible solutions.

User Jonathan Hill
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