SAT inequalities: solve ranges and test feasible points

01Reverse the sign for a negative multiplier

For $3x+4<19$, subtract 4 and divide by 3 to obtain $x<5$. Neither operation reverses the inequality because 3 is positive.

For $-2x+7\ge 15$, subtract 7 to get $-2x\ge 8$. Dividing by $-2$ reverses the direction, giving $x\le -4$. Test $x=-5$: the original left side is 17, which is at least 15. Test $x=0$: the left side is 7, so it fails.

The reason for reversing is order. Although $2<5$, multiplying both numbers by $-1$ gives $-2>-5$. Adding or subtracting a negative number does not reverse an inequality; multiplication or division by a negative does.

If a multiplier contains an unknown variable, you cannot assume its sign. Split into cases or use a method that avoids multiplying by an expression whose sign you do not know.

02Keep strict and inclusive boundaries distinct

The symbols $<$ and $>$ exclude the boundary value. The symbols $\le$ and $\ge$ include it. On a number line, an open point excludes the endpoint and a filled point includes it.

If $x<5$, the real-number solutions approach 5 but do not include it, so there is no greatest real solution. If the prompt restricts $x$ to integers, the greatest allowed integer is 4. This restriction changes the answer; do not assume it merely because the choices are whole numbers.

For $x\le 5$, the boundary value is included. Check the exact language in word problems: "at most" means inclusive, while "less than" is strict. "At least" gives a lower inclusive boundary.

03Solve a compound condition as written

For $2<3x+5\le 14$, subtract 5 from all three parts: $-3<3x\le 9$. Divide all parts by 3 to get $-1<x\le 3$. The lower endpoint is excluded and the upper endpoint included.

An "and" condition requires both inequalities. If $x>2$ and $x\le 6$, the solution is $2<x\le 6$. An "or" condition allows either one. If $x<-2$ or $x>3$, the solution consists of two separated ranges.

Test the boundaries and one interior value. For $-1<x\le 3$, $x=-1$ fails, $x=0$ works, and $x=3$ works. Those checks catch a flipped endpoint or an accidental change from strict to inclusive.

04Translate a budget or capacity into a constraint

Suppose a club has $50 for supplies, pays a fixed $8 shipping charge, and buys notebooks at $6 each. With $n$ notebooks, the budget condition is $6n+8\le 50$. Solving gives $n\le 7$.

Because notebooks are counted, $n$ must be a nonnegative integer. Seven notebooks cost $50; eight cost $56 and exceed the budget. If the budget were $49, the algebra would give $n\le \frac{41}{6}$, so the greatest whole-number count would be 6, not a rounded 7.

Write the fixed cost separately from the per-item cost. "At most $50" permits spending exactly $50. "Under $50" would exclude it. The interpretation of the inequality is part of the solution.

05For two variables, test both inequalities

A system describes the overlap of its constraints. Consider $y\ge 2x+1$ and $y<-x+7$. The point $(1,4)$ satisfies both: $4\ge 3$ and $4<6$. The point $(2,5)$ satisfies the first at equality, but fails the second because $5<5$ is false.

For a graph, a strict inequality has an excluded boundary, shown as a dashed line; an inclusive inequality has an included boundary, shown as a solid line. Desmos can graph the inequalities and show the overlapping region. If a point appears close to a boundary, substitute its coordinates to settle the result exactly.

One feasible point does not establish that an entire region is feasible. A boundary intersection also needs both inclusion checks. If one condition is strict at that intersection, the intersection itself is excluded.

06Avoid using a linear rule on every expression

For a quadratic inequality, first find the roots and determine where the expression is positive or negative. In $(x-1)(x-4)<0$, the product is negative between the roots, so $1<x<4$. It is positive outside that interval. The strict symbol excludes both roots.

For an absolute-value inequality, think in terms of distance. $|x-2|\le 3$ means $x$ lies within 3 units of 2, so $-1\le x\le 5$. An inequality asking for a distance greater than 3 gives outside ranges instead. Do not remove absolute-value bars without considering both directions.

07Finish with a range and a check

State the full range, identify any endpoint restrictions, and test a value in the original inequality. For a multiple-choice feasible-point question, direct substitution may be shorter than graphing every boundary.

Practice inequalities in the SAT question bank, then compare linear equations and systems of equations. Keep your review specific: negative division, endpoint inclusion, "and" versus "or," or a context restriction. Those labels tell you what to check on the next problem.