SAT Linear Inequalities: The Complete Guide

On a number line, $x>4$ has an open dot at four and extends to the right. The condition $x\le4$ has a closed dot at four and extends left. In interval notation, these would use an open parenthesis or closed bracket at four, but you can understand the entire solution without memorizing interval symbols.

For a two-variable condition, $y\le2x+5$ includes the boundary and the points below it. For $x\ge3$, the boundary is vertical and the allowed points lie on or to its right. Do not force a vertical boundary into a slope-intercept equation; its geometry is simpler in the original form.

If a boundary contains the origin, $(0,0)$ cannot tell you which side to shade because it is on the boundary. Choose another easy point, such as $(1,0)$ or $(0,1)$, provided that point is off the line. A strict inequality is false at every boundary point; an inclusive inequality is true there but still needs an off-line test for the side.

The word and requires both conditions, so use their intersection. The word or permits values satisfying at least one condition, so use their union. The pair $x>2$ and $x<8$ gives the interval between two and eight. The pair $x<2$ or $x>8$ gives two separated outer regions.

A compound inequality written on one line is usually an and statement. For $-2<3x+1\le10$, subtract one from all three parts to obtain $-3<3x\le9$, then divide by positive three to get $-1<x\le3$. The left endpoint stays excluded and the right endpoint stays included. Check a value inside, such as zero, and each endpoint separately.

If a compound inequality is multiplied by a negative number, reverse both comparisons. For $-4<-2x\le6$, dividing every part by negative two gives $2>x\ge-3$, usually rewritten $-3\le x<2$. Reordering the final statement from least to greatest makes the number-line graph easier to read. It does not create another sign change; it restates the same comparisons from the other direction.

When translating a range from words, retain both limits. A temperature described as above negative five and no higher than twelve gives $-5<T\le12$, with different endpoint treatment. The word between by itself can be ambiguous in casual speech; the question's stated inclusion or comparison language determines which endpoints are allowed.

Translate a budget by building total cost first. If two item counts are $a$ and $b$, with prices six and nine dollars and a budget of 120 dollars, write $6a+9b\le120$. If the buyer needs at least 15 items, add $a+b\ge15$. Nonnegative count restrictions are part of the model too.

Desmos can graph inequalities directly, including their shaded regions. Enter each condition separately to inspect the overlap. A budget boundary and a minimum-count boundary can make it easy to see which combinations are feasible. Include $x\ge0$ and $y\ge0$ when the variables represent nonnegative quantities.

Graphing is especially useful for comparing candidate points or locating an intersection that might set an upper limit. But a graph is continuous: it does not automatically turn a fractional coordinate into a valid count. After identifying a likely integer maximum, test the actual integer pair in all constraints.

For a strict upper bound such as $n<8$, the greatest integer is seven. For $n\le8$, it is eight. For $n\ge8.2$, the least integer is nine. Interpret the boundary before rounding; ordinary rounding to the nearest whole number does not answer a minimum-capacity or maximum-budget question.

Do not confuse at least with exactly. A condition $x+y\ge18$ permits 18, 19, or more items. You may eventually show that buying exactly 18 is best for a particular optimization, but that must follow from the costs and objective rather than replacing the condition at the start.

A shading shortcut such as above for greater than works only after the inequality is solved for $y$. The inequality $-2y>4x+6$ becomes $y<-2x-3$, which shades below. The original greater-than symbol by itself does not identify above or below.

The allowed set can be empty. The conditions $x>9$ and $x\le4$ have no common values. On a graph, two individually large half-planes can also fail to overlap after additional restrictions. Do not force an answer merely because each condition has many solutions by itself.

A feasible point is not automatically an optimum. To prove a maximum, establish an upper bound and exhibit a point that reaches it. The hard example does both: it shows $y\le6$ for every allowed purchase and then provides a purchase with $y=6$.

An inequality with a parameter needs sign cases before division. For $ax<12$, if $a>0$, then $x<12/a$. If $a<0$, then $x>12/a$. If $a=0$, the statement becomes $0<12$, true for every real $x$. One formula cannot silently cover all three cases.

When a question asks for a greatest possible whole-number value, first find the exact bound. If $3n<25$, then $n<25/3$, so the greatest integer is eight. If $3n\le24$, eight is still allowed; if $3n<24$, the maximum becomes seven. Exact boundary arithmetic prevents an incorrect rounding rule.

Two-variable optimization often depends on where boundary lines meet, but a contextual answer may require neighboring integer points. A fractional intersection can suggest a bound without providing a feasible count pair. Test candidate integer combinations in all original conditions, including nonnegativity, instead of rounding both coordinates independently.

You can combine inequalities to create a bound, as long as every operation preserves direction. Adding two true inequalities pointing the same way gives another true inequality. Multiplying one by a positive constant is safe. Subtracting inequalities without checking direction can produce an invalid conclusion, so rewrite it as addition of a correctly negated inequality if needed.

If a word problem includes a minimum earning requirement and maximum time limit, the slopes have different meanings. The feasible region contains combinations meeting both. Use the units to write each condition independently before comparing their boundary intersection.

Use a 24-minute session. Spend five minutes on one-variable inequalities, including negative coefficients and strict boundaries. For every result, test one allowed value and the boundary in the original expression. This builds a quick check that does not depend on repeating your algebra.

Spend seven minutes in Solar on graph and candidate-point questions. Identify boundary orientation, inclusion, and shaded side before selecting a choice. Include a vertical boundary and a boundary through the origin so you practice choosing an alternative test point.

Use six minutes for word-problem constraints. Label variables and units, translate all limits, and list domain restrictions. Spend the remaining six minutes reviewing a maximum or minimum question, proving both its bound and the feasibility of the proposed extreme.

On your next mixed set, compare a Desmos region with your algebraic description for one missed item. If they disagree, check the translation first, then the sign operation, then the boundary symbol. Fix the earliest wrong step rather than adjusting the graph to resemble an expected answer.