Linear Inequalities: questions with answers

Whether the marked number is a solution is shown by the kind of circle. An open circle marks x < a or x > a, so the number is left out. A dark circle marks x ≤ a or x ≥ a, so it is included. The thick line only shows which side the other solutions lie on.

The symbol is strict in every binding, so the boundary itself fails and the extreme integer is the one just inside it. Answering the boundary treats the strict symbol as though it allowed equality.

This is the chapter's opening inequality with different numbers. An equation would claim the money is spent exactly, which the whole-packet rule makes impossible. The chapter writes 30x < 200 for this situation.

The boundary is the same number whichever way the symbol ends up, so this variant isolates the arithmetic from the sign rule. Expanding the bracket on its first term only changes 5 back to 1 and moves the boundary to a different number entirely.

The total after the last test is 150+x and there are 4 tests. Therefore the average is $(150+x)/4$. “At least 65” means greater than or equal to 65. Choose (150 + x)/4 ≥ 65.

The chapter's own reading of 3 < 5 < 7 is '5 is greater than 3 and less than 7'. The conjunction is the source's own word and it is the whole content of the notation: a value has to satisfy both halves to satisfy the statement.

Every letter looks alike on the page, which is why counting them all is the natural mistake. The chapter names its forms (5) to (8) as inequalities in ONE variable x. It names (9) to (12) as inequalities in TWO variables x and y, whatever letters their coefficients happen to use.

₹130 ÷ ₹40 is deliberately not a whole number, so the answer is the whole number just below it. The untested next value is the natural wrong answer and costs 160, which is more than the shopper has.

Answering five means the equals sign has been counted in; answering two means only the strict pair was remembered. The chapter uses all four throughout, and sorts them into strict and slack in the very next paragraph.

The option leaving out 0 is the near miss: the negative integers really are added, but so is 0, which is an integer and not a natural number. Which set x is drawn from changes the answer from a finite list to an infinite one.

Subtract $12x$ and 1 from both sides. The inequality becomes $(4-12)x>10-1$, or $-8x>9$.

The coefficient of $x$ is -8, which is negative. Dividing by -8 reverses the inequality: $x<-9/8$.

The earlier algebra is valid. Line 4 is the first wrong line because it keeps $>$ instead of changing it to $<$.

Answer: Line 4. The correct solution is $x<-9/8$.

Half the bindings carry a SQUARE bracket at the finite end, which is exactly what makes matching them look right. The chapter writes [8, ∞) and (-∞, 2), never [8, ∞].

After a reversal the working reads backwards, so writing the interval takes a deliberate extra step. The chapter does exactly that in Example 10: having reached 5 ≥ x ≥ -11/3 it rewrites the line as -11/3 ≤ x ≤ 5 before reading the interval off.

The term removed is NEGATIVE here, which is what makes a reversal look justified. Rule 1 says nothing about the sign of the number added; it is Rule 2, and only its negative half, that ever turns the symbol.

The figure draws the two rays and stops there, so the overlap has to be worked out, not read off. The two circles are set independently, so each bracket is decided on its own. The chapter's Fig 5.3 marks the overlap with a bold segment; this figure does not.