Solving linear inequations

An inequation compares two expressions with $<$, $>$, $\le$ or $\ge$ instead of $=$. A linear inequation in one variable is one where that variable appears only to the first power, such as $2x+3<11$ or $\dfrac{2-3x}{5}\ge -2$. Three properties govern how an inequation behaves under the operations you already use to solve equations. Adding or subtracting the same quantity from both sides never disturbs the direction of the inequality:

where: $a$ and $b$ are the two sides being compared and $c$ is any number added to both. Multiplying or dividing both sides by the same positive number also leaves the direction untouched, but multiplying or dividing by the same negative number reverses it:

where: $c$ is the multiplier — its sign, not its size, decides whether the arrow flips. The working method is then: simplify both sides (clear brackets, clear denominators, combine like terms), transpose so variable terms sit on one side and constants on the other, then divide by the coefficient of the variable — flipping the sign if that coefficient is negative. Finally, state the solution set within whichever number system the question names ($N$, $W$, $Z$ or $R$), and, if asked, mark it on a number line: a filled (solid) circle at a boundary that is included ($\le$, $\ge$), an open (hollow) circle at a boundary that is excluded ($<$, $>$), with an arrow running in the direction of every value that belongs to the set. Unlike an equation, which almost always pins the variable to one number, a linear inequation typically leaves infinitely many numbers satisfying it — every value on one side of the boundary. That is the real shift in thinking this chapter asks for: instead of asking "which single number works," you are asking "which entire stretch of the number line works," and the boundary point itself may or may not be part of the answer depending on whether the original symbol was strict ($<$, $>$) or included equality ($\le$, $\ge$).

Solve $2(x-3)-5\ge 3x-1$ for $x\in Z$, and represent the solution on a number line.

Step 1. Clear the bracket. $2(x-3)-5 = 2x-6-5 = 2x-11$, so the inequation becomes

Step 2. Transpose the variable terms to one side and constants to the other. Subtract $3x$ from both sides and add $11$ to both sides:

Step 3. Divide both sides by $-1$. Because $-1$ is negative, the inequality sign reverses:

Step 4. Restrict to integers and check the boundary. Since $x\in Z$, the solution set is $\{\dots,-13,-12,-11,-10\}$, every integer less than or equal to $-10$. Substitute the boundary value $x=-10$ into the original inequation to verify: left side $2(-10-3)-5=2(-13)-5=-31$; right side $3(-10)-1=-31$. Since $-31\ge -31$ is true, $x=-10$ is correctly included. On the number line this is a filled circle at $-10$ with an arrow extending left. The solution is $x\le -10$, $x\in Z$.

• Forgetting to reverse the inequality sign when multiplying or dividing both sides by a negative number, as happened in Step 3 above — this single slip accounts for most lost marks in the chapter.
• Drawing a filled circle where an open one belongs, or vice versa: $<$ and $>$ need an open (hollow) circle because the boundary is excluded; $\le$ and $\ge$ need a filled (solid) circle because the boundary is included.
• Ignoring the number system stated in the question — if $x$ is restricted to $N$ or $W$, the answer must be listed as specific values like $\{1,2,3,4\}$, not left as a continuous inequality such as $x<5$.
• Losing track of a sign while clearing a bracket, especially a bracket preceded by a minus sign, before the transposition step even begins.

Ask the tutor: If the sign-reversal rule or restricting solutions to $N$, $W$ or $Z$ still feels shaky, ask the AI tutor to work through a few more inequations with you step by step.