Solving Linear Inequalities: Practice

Subtract $6$ from both sides. Subtraction never changes the direction of the symbol.

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

The symbol stays as $<$, since no negative was involved.

Add $3$ to both sides to undo the subtraction. Adding does not change the direction.

$x - 3 + 3 > 5 + 3 \;\Rightarrow\; x > 8$

The symbol stays as $>$.

Divide both sides by the positive number $2$. Dividing by a positive keeps the direction.

$\frac{2x}{2} < \frac{10}{2} \;\Rightarrow\; x < 5$

No flip occurs, because $2$ is positive.

Multiply both sides by the positive number $2$ to undo the division. The direction is unchanged.

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

Multiplying by a positive number keeps the symbol as $>$.

First subtract $1$ from both sides, then divide by the positive number $3$.

$3x < 6 \;\Rightarrow\; x < 2$

Both steps keep the direction, so the answer is $x < 2$.

Here $-x$ is $-1 \cdot x$, so divide both sides by $-1$. The divisor is negative, so reverse $>$ to $<$.

$\frac{-x}{-1} < \frac{4}{-1} \;\Rightarrow\; x < -4$

Keeping the symbol as $>$ would be the forgot-to-flip error.

Divide both sides by the positive number $5$. Dividing by a positive keeps the direction, even though the right side is negative.

$\frac{5x}{5} > \frac{-15}{5} \;\Rightarrow\; x > -3$

The negative on the right does not cause a flip; only a negative divisor would.

Multiply both sides by the positive number $3$. The direction is unchanged.

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

Multiplying by a positive keeps the symbol as $\le$.

Subtract $2$ from both sides to get $5 \ge x$, then read it from the other end.

$5 \ge x \;\Rightarrow\; x \le 5$

Swapping the sides reverses the symbol, giving $x \le 5$.

A solution must be strictly greater than $3$.

$5 > 3 \;\checkmark$

The value $3$ is excluded by the strict symbol, and $0$ and $-4$ are both below $3$, so only $5$ works.

Substitute $x = 2$ into the left side and compare with the right.

$3(2) - 1 = 5, \qquad 5 < 4 \;\text{ is false}$

Since $5 < 4$ does not hold, $x = 2$ is not a solution.

Add $7$ to both sides. Adding does not change the direction.

$x - 7 + 7 \ge -2 + 7 \;\Rightarrow\; x \ge 5$

The symbol stays as $\ge$.