Inequalities: Practice

The symbol $\le$ means 'less than or equal to', which in words is 'at most'.

$n \le 10 \;\Leftrightarrow\; n \text{ is at most } 10$

So $n$ can be $10$ or anything smaller, but nothing larger.

Substitute $6$ for $x$ and check whether the statement is true.

$6 > 5 \;\checkmark$

Since $6$ really is greater than $5$, the value $x = 6$ is a solution.

A solution must be a number strictly less than $3$. Test each option.

$1 < 3 \;\checkmark$

The value $3$ is not less than itself, and $5$ and $4$ are larger than $3$, so only $1$ works.

Subtract $4$ from both sides. Adding or subtracting does not change the symbol.

$x + 4 - 4 < 10 - 4 \;\Rightarrow\; x < 6$

Check inside the range: $x = 0$ gives $0 + 4 = 4 < 10$, which is true.

Divide both sides by the positive coefficient $3$. Dividing by a positive number keeps the symbol.

$\frac{3x}{3} \le \frac{12}{3} \;\Rightarrow\; x \le 4$

Check the boundary: $x = 4$ gives $3(4) = 12 \le 12$, which is true.

Divide both sides by the positive coefficient $5$. The symbol is unchanged.

$\frac{5x}{5} > \frac{30}{5} \;\Rightarrow\; x > 6$

Check inside the range: $x = 7$ gives $5(7) = 35 > 30$, which is true.

Subtract $7$ from both sides. The symbol is unchanged by subtraction.

$x + 7 - 7 \le 7 - 7 \;\Rightarrow\; x \le 0$

Check the boundary: $x = 0$ gives $0 + 7 = 7 \le 7$, which is true.

'At least $48$' means $48$ is allowed and so is anything taller, but nothing shorter.

$h \ge 48$

The phrase 'at least' always points to a $\ge$ symbol, which includes the boundary value $48$.

The variable is divided by the positive number $3$, so multiply both sides by $3$. The symbol stays.

$3 \cdot \frac{x}{3} \ge 3 \cdot 2 \;\Rightarrow\; x \ge 6$

Check the boundary: $x = 6$ gives $\frac{6}{3} = 2 \ge 2$, which is true.

On the number line, the number farther to the left is the smaller one. Compare each pair.

$-5 < -1$

Since $-5$ sits to the left of $-1$, it is the smaller number, so $-5 < -1$ is true. The others reverse the correct order.

The symbol $\ge$ allows the boundary, so $4$ itself counts, along with anything larger.

$4 \ge 4 \;\checkmark$

The values $2$, $3$, and $0$ are all smaller than $4$, so only $4$ is a solution among the choices.

Add $5$ to both sides. The symbol stays the same for addition.

$x - 5 + 5 > -2 + 5 \;\Rightarrow\; x > 3$

Check inside the range: $x = 4$ gives $4 - 5 = -1 > -2$, which is true.