how to solve for x in an inequality of this form: a x + b &lt; c x + d &lt; e x + f (a, b, c, d, e, f are constants; inequality signs may be " ≤") I was taught that we only needed to solve a x + b &lt; c x + d and c x + d &lt; e x + f and not a x + b &lt; e x + f but I do not understand why.

Question

how to solve for x in an inequality of this form:    a  x  +  b  &amp;lt;  c  x  +  d  &amp;lt;  e  x  +  f (a, b, c, d, e, f are constants; inequality signs may be &quot;  ≤&quot;) I was taught that we only needed to solve   a  x  +  b  &amp;lt;  c  x  +  d and   c  x  +  d  &amp;lt;  e  x  +  f and not   a  x  +  b  &amp;lt;  e  x  +  f but I do not understand why.

thoumToofwj · Accepted Answer

Perhaps turning inequality to equality may make things clear.  a  &amp;lt;  b    ⇒          ε    a    =  b  −  a  &amp;gt;  0  b  &amp;lt;  c    ⇒          ε    b    =  c  −  b  &amp;gt;  0  c  −  a  =  (  c  −  b  )  +  (  b  −  a  )  =  (        ε    b    +        ε    a    )  &amp;gt;  0Thus, it is redundant to solve again for   a  &amp;lt;  c, because it is always true.

copafumpv · Accepted Answer

Personally, if I were forced to pick a general rule, I would contend that   a  &amp;lt;  b  &amp;lt;  c means   A  &amp;lt;  B  ∧  B  &amp;lt;  C. For instance, I would read   A  ≠  0  ≠  B as   A  ≠  0  ∧  0  ≠  B, and not as   A  ≠  0  ∧  0  ≠  B  ∧  A  ≠  C, just like I don&#039;t read   A  ∈  B  ∈  C as   A  ∈  B  ∧  B  ∈  C  ∧  A  ∈  C. The fact that   A  &amp;lt;  B  ∧  B  &amp;lt;  C implies   A  &amp;lt;  C would then be incidental. Anyways, since interpreation of   a  ⋄  b  ⋄  c when   ⋄ isn&#039;t a transitive relation is a magnet for nuisances, let&#039;s say that the point of your contention is why you can say that  (  A  &amp;lt;  B  ∧  B  &amp;lt;  C  )  ⇔  (  A  &amp;lt;  B  ∧  B  &amp;lt;  C  ∧  A  &amp;lt;  C  )As you&#039;ve been told, &amp;lt; has the property that for all   A,   B and   C,   A  &amp;lt;  B  ∧  B  &amp;lt;  C implies   A  &amp;lt;  C. Whenever you have a condition   P which is implied by a condition   Q (i.e. that   P is true whenever   Q is), you can safely assume that   (  P  ∧  Q  )  ⇔  P.  [  (  P  ∧  Q  )  ⇒  Q  ] is true because   P  ∧  Q is true if and only if both   P and   Q are, and therefore   P is true whenever   P  ∧  Q is true.  [  Q  ⇒  (  P  ∧  Q  )  ] is true because if   Q is true, then   P is true too by hypothesis. Therefore   P  ∧  Q is true.

how to solve for x in an inequality of this form: <mspace linebreak="newline" /> a x +

Answered question

Answer & Explanation

New Questions in Algebra I