Given 2x - y = 480 and 2y - x + m/3 = 0, find x and y in terms of m, and the values of m for which x + y <= 8m/3.

Step-by-step solution

1. 1

   Separate the two equations from the inequality. The two equations determine $x$ and $y$ uniquely for each $m$; the inequality then filters which values of $m$ are acceptable. Solving first and testing second is the only order that works, since the inequality alone cannot pin down $x$ and $y$.

2. 2

   Put the second equation in standard form. From $2y-x+\dfrac{m}{3}=0$,

   $-x+2y=-\frac{m}{3}$

3. 3

   Substitute and solve for $x$, watching the sign. From the first equation $y=2x-480$; substituting,

   $-x+2(2x-480)=-\frac{m}{3}\ \Longrightarrow\ 3x-960=-\frac{m}{3}\ \Longrightarrow\ 9x-2880=-m$

   Now isolate carefully: $9x=2880-m$, so

   $x=\frac{2880-m}{9}$

   The right-hand side of $9x-2880=-m$ is $-m$, so adding $2880$ gives $2880-m$, not $m-2880$. This one sign is the whole difficulty of the problem.

4. 4

   Back-substitute for $y$.

   $y=2x-480=\frac{5760-2m}{9}-\frac{4320}{9}=\frac{1440-2m}{9}$

   Check with $m=0$: $x=320$, $y=160$, and indeed $2(320)-160=480$ ✓ and $2(160)-320+0=0$ ✓.

5. 5

   Form $x+y$ and apply the inequality.

   $x+y=\frac{2880-m+1440-2m}{9}=\frac{4320-3m}{9}=\frac{1440-m}{3}$

   so the condition $x+y\le\dfrac{8m}{3}$ becomes

   $\frac{1440-m}{3}\le\frac{8m}{3}\ \Longrightarrow\ 1440-m\le 8m\ \Longrightarrow\ 1440\le 9m\ \Longrightarrow\ m\ge 160$

6. 6

   Verify at the boundary and on both sides. At $m=160$: $x=\tfrac{2720}{9}$, $y=\tfrac{1120}{9}$, $x+y=\tfrac{1280}{3}$ and $\tfrac{8m}{3}=\tfrac{1280}{3}$ — equality, so $160$ is included ✓. At $m=900$: $x+y=180\le 2400$ ✓. At $m=0$: $x+y=480$ but $\tfrac{8m}{3}=0$, so the inequality fails ✓, confirming that small $m$ is excluded.