GRE Quantitative Reasoning Question Solved – linear inequalities

First, we need to find all of the values that are in the set S, and then we need to find the values in T. Once we do this, we must find the numbers in the intersection of S and T, which means we must find the values contained in BOTH sets S and T.

S contains all of the values of x such that 2x + 4 < 8. We need to solve this inequality.

2x + 4 < 8

Subtract 4 from both sides.

2x < 4

Divide by 2.

x < 2

Thus, S contains all of the values of x that are less than (but not equal to) 2.

Now, we need to do the same thing to find the values contained in T.

-2x + 3 < 8

Subtract 3 from both sides.

-2x < 5

Divide both sides by -2. Remember, when multiplying or dividing an inequality by a negative number, we must switch the sign.

x > -5/2

Therefore, T contains all of the values of x that are greater than -5/2, or -2.5.

Next, we must find the values that are contained in both S and T. In order to be in both sets, these numbers must be less than 2, but also greater than -2.5. Thus, the intersection of S and T consists of all numbers between -2.5 and 2.

The question asks us to find the sum of the integers in the intersection of S and T. This means we must find all of the integers between -2.5 and 2.

The integers between -2.5 and 2 are the following: -2, -1, 0, and 1. We cannot include 2, because the values in S are LESS than but not equal to 2.

Lastly, we add up the values -2, -1, 0, and 1. The sum of these is -2.

The answer is -2.






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