Solve the compound inequality 3818 ≤ 499x − 20y < 4018 − 2734 ≤ 33x − 78y < −2534.

Step-by-step solution

1. 1

   Simplify the constant buried in the middle of the chain. The term $4018-2734$ is not a variable expression at all, so evaluate it before anything else:

   $4018-2734=1284$

   The chain now reads

   $3818 \le 499x-20y < 1284 \le 33x-78y < -2534$

2. 2

   Remember what a chain of inequalities actually asserts. A written chain $p \le q < r \le s < t$ is shorthand for every neighbouring comparison holding at once. So this one asserts four separate conditions simultaneously:

   $3818 \le 499x-20y,\qquad 499x-20y < 1284,$
   $1284 \le 33x-78y,\qquad 33x-78y < -2534$

   That is why you must check the constants against each other, not just solve each piece for $x$ and $y$.

3. 3

   Test the first two links against each other. They both constrain the same quantity $499x-20y$:

   $3818 \le 499x-20y < 1284$

   This needs one number to be at least $3818$ and strictly less than $1284$ at the same time. Since $3818 > 1284$, no real value of $499x-20y$ can do it. The interval $[3818,\,1284)$ is empty.

4. 4

   Test the last two links the same way. They both constrain $33x-78y$:

   $1284 \le 33x-78y < -2534$

   Again the lower bound exceeds the upper bound, because $1284 > -2534$. The interval $[1284,\,-2534)$ is also empty.

5. 5

   Conclude with the empty solution set. Two independent parts of the chain are self-contradictory, so no choice of $x$ and $y$ can satisfy the whole statement. There is nothing left to solve for:

   $\{(x,y)\}=\varnothing$

   The useful habit here: whenever a compound inequality is written as one long chain, read off the constant bounds first. If any lower bound exceeds the upper bound sitting to its right, you are done immediately — no algebra on $x$ and $y$ is needed.