Solving Linear Inequalities

Undo the operations from the outside in. First subtract $4$ from both sides, which does not change the direction:

Now divide both sides by $3$. The divisor is positive, so the symbol stays put:

Read the answer three matching ways. As an inequality it is $x < 5$. On a number line it is an open circle on $5$ (strict, so $5$ is excluded) with the shading running left toward the smaller numbers. In interval notation it is $(-\infty, 5)$.

Confirm the direction by testing one value inside the solution set and one outside it, always against the original inequality. The value $x = 0$ should be inside, since $0 < 5$: the original gives $3(0) + 4 = 4$, and $4 < 19$ is true. The value $x = 6$ should be outside, since $6 < 5$ is false: the original gives $3(6) + 4 = 22$, and $22 < 19$ is false, exactly as the solution predicts.

Start by moving the constant. Subtract $5$ from both sides, which does not touch the direction:

Now isolate $x$ by dividing both sides by $-3$. The divisor is negative, so reverse $\ge$ to $\le$:

The flip happened at the divide-by-negative step, not at the earlier subtraction. As an inequality the answer is $x \le -2$. On a number line it is a filled circle on $-2$ (inclusive, so $-2$ is included) with the shading running left; in interval notation it is $(-\infty, -2]$.

Test a value on each side. Inside, $x = -5$ satisfies $x \le -2$: the original gives $5 - 3(-5) = 5 + 15 = 20$, and $20 \ge 11$ is true. Outside, $x = 0$ fails $x \le -2$: the original gives $5 - 3(0) = 5$, and $5 \ge 11$ is false. Both checks agree with $x \le -2$.

The variable terms are $4x$ and $7x$. Collecting on the side with the larger coefficient, the right, keeps that coefficient positive and avoids any flip. Subtract $4x$ from both sides:

Subtract $9$ from both sides, then divide by the positive number $3$:

which reads $x < -4$ from the other end. No flip was ever needed.

Collecting on the left instead reaches the same place through a flip. Subtract $7x$ from both sides to get $-3x - 3 > 9$, add $3$ to get $-3x > 12$, then divide by $-3$ and reverse the symbol:

Both routes give $x < -4$, the interval $(-\infty, -4)$. A quick check confirms it: $x = -5$ is inside, and the original reads $4(-5) - 3 = -23$ against $7(-5) + 9 = -26$, where $-23 > -26$ is true; $x = 0$ is outside, and $-3 > 9$ is false.

Distribute the $-2$ across the parentheses, being careful with signs: $-2 \cdot x = -2x$ and $-2 \cdot (-4) = +8$, so

Subtract $8$ from both sides to get $-2x < 2$. Now divide both sides by $-2$; the divisor is negative, so reverse $<$ to $>$:

The answer is $x > -1$, an open circle on $-1$ with the shading running right, the interval $(-1, \infty)$.

As a check, $x = 0$ is inside and gives $-2(0 - 4) = 8$, with $8 < 10$ true, while $x = -2$ is outside and gives $-2(-2 - 4) = 12$, with $12 < 10$ false.

Work on all three parts together. First subtract $1$ from each part, which does not change either symbol:

Now divide all three parts by $-2$. The divisor is negative, so reverse both symbols. Dividing $-8$ by $-2$ gives $4$, dividing $-2x$ by $-2$ gives $x$, and dividing $4$ by $-2$ gives $-2$:

Rewrite it with the smaller number on the left, which is the standard reading:

On a number line this is a filled circle on $-2$, an open circle on $4$, and a shaded band between them; in interval notation it is $[-2, 4)$.

Test the endpoints. At $x = -2$ the expression is $-2(-2) + 1 = 5$, and $-7 < 5 \le 5$ holds, so $-2$ is in, matching the filled circle. At $x = 4$ the expression is $-2(4) + 1 = -7$, and $-7 < -7$ is false, so $4$ is out, matching the open circle.

Solve each inequality by itself. The first gives

and the second gives

The solution set is every number that is below $-1$ or at least $2$. The two pieces do not overlap, so the graph is two separate rays: an open circle on $-1$ shaded left, and a filled circle on $2$ shaded right. In interval notation, joined by the union symbol, that is $(-\infty, -1) \cup [2, \infty)$.

A spot check confirms the reading: $x = -2$ satisfies the first piece, $x = 5$ satisfies the second, and $x = 0$ satisfies neither, so it is correctly left out of the shading.

Solve $2x + 5 < 2x + 1$. Subtract $2x$ from both sides, and the variable disappears:

That leftover is false, and there is no $x$ left to change it, so the inequality has no solution. This is the shape of $x < x - 1$: no number can be less than one below itself.

Now solve $3x - 4 \le 3x + 2$. Subtract $3x$ from both sides:

That leftover is true and free of $x$, so the original was true for every value of $x$. The solution is all real numbers, the interval $(-\infty, \infty)$. This is the shape of $x \le x + 1$: a number is always at most one more than itself.