Subtracting Inequalities

Method 1 : Test out some values.

Consider the case when $a=8,b=6, c=5,d=4$ ,
then $a>b$ is indeed true and $c>d$ in indeed true as well, thus the constraints $a>b$ and $c>d$ are satisfied. And also $a-c = 2$ and $b-d= 1$ , so $a-c> b-d$ is also true as well.

So the inequality is true for this instance when $a=8,b=6, c=5,d=4$ .

Consider another case when $a=8,b=7,c=6,d=4$ , then $a>b$ is indeed true and $c>d$ in indeed true as well, thus the constraints $a>b$ and $c>d$ are satisfied. However $a-c = 1$ and $b-d= 2$ , so $a-c> b-d$ can't be true.

So the inequality is not true for this instance when $a=8,b=6, c=5,d=4$ .

In other words, the inequality is not always true and not always false either. And thus the answer is "No, it's not always true".

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Method 2 : Rearranging the terms in the inequality.

Looking at answer choices tells us that we want to determine whether the inequality , $a-c > b-d$ is always true, or always false, or not necessarily true.

Determining the truth of the inequality $a-c> b-d$ is similar of determining the truth of the inequality $a-b> c-d$ .

Since we know that the constraints $a>b$ and $c>d$ must be satisfied, then $a-b>0$ and $c-d> 0$ must be satisfied also. So we know that $a-b$ and $c-d$ are strictly positive but there isn't sufficient information to know whether which of these numbers $(a-b,c-d)$ , this is because we could have

$a-b> c-d>0 \qquad \text{ or } \qquad c-d > a-b > 0 \qquad \text{ or } \qquad a-b = c-d > 0.$

Since at least of the inequation/equation above must be true, then our claim of $a-b>c-d$ must not be strictly true or false only, hence we will arrive at the same conclusion as Method 1.