Show that kt + xy > ky + tx is equivalent to (k - x)(t - y) > 0.

Step-by-step solution

1. 1

   Move everything to one side. Subtract $ky$ and $tx$ from both sides so the comparison is against zero:

   $kt-tx+xy-ky>0$

2. 2

   Group the terms in pairs sharing a factor. The first two share $t$, the last two share $y$:

   $t(k-x)+y(x-k)>0$

3. 3

   Make the two brackets identical. The second bracket is the negative of the first, since $x-k=-(k-x)$:

   $t(k-x)-y(k-x)>0$

   Spotting this sign flip is the whole trick; without it the grouping looks like a dead end.

4. 4

   Factor out the common binomial.

   $(k-x)(t-y)>0$

5. 5

   State the condition and check it. A product of two reals is positive exactly when both are positive or both are negative:

   $\left(k>x\ \text{and}\ t>y\right)\quad\text{or}\quad\left(k<x\ \text{and}\ t<y\right)$

   Test $k=4,t=3,x=1,y=2$ (both differences positive): left $=12+2=14$, right $=8+3=11$, and $14>11$ $\checkmark$, with $(3)(1)=3>0$ $\checkmark$. Test $k=1,t=3,x=4,y=2$ (mixed signs): left $=3+8=11$, right $=2+12=14$, so the inequality fails $\checkmark$, matching $(-3)(1)=-3<0$.