Q.Solve for $x$: $4x + 3 \ge 2x + 17$, $3x - 5 < -2$.

The key idea is to find the intersection of the solution sets for each linear inequality.

First, solve the inequality $4x + 3 \ge 2x + 17$:

$4x - 2x \ge 17 - 3$

$2x \ge 14$

$x \ge 7$

Next, solve the inequality $3x - 5 < -2$:

$3x < -2 + 5$

$3x < 3$

$x < 1$ …

When we are asked to "solve for $x$" in a system of inequalities, it means we need to find all values of $x$ that satisfy all the given inequalities simultaneously. This is different from solving a single equation, where we typically find one or a few discrete values. For inequalities, the solution is usually a range or an interval of numbers.

The core idea is to treat each inequality independently first, isolating $x$ on one side. Once we have the solution set for each individual inequality, we then look for the values of $x$ that are common to all those solution sets. This common region is the intersection of the individual solution sets.

Let's break down the problem:

1. Solve the first inequality: $4x + 3 \ge 2x + 17$

   Our goal is to isolate $x$. We can do this by moving all terms involving $x$ to one side and all constant terms to the other side.

   Subtract $2x$ from both sides:

   $4x - 2x + 3 \ge 2x - 2x + 17$

   $2x + 3 \ge 17$

   Now, subtract $3$ from both sides:

   $2x + 3 - 3 \ge 17 - 3$

   $2x \ge 14$

   Finally, divide both sides by $2$. Since $2$ is a positive number, the direction of the inequality sign remains unchanged.

   $\frac{2x}{2} \ge \frac{14}{2}$

   $x \ge 7$

   This means the first inequality is satisfied by all real numbers $x$ that are greater than or equal to $7$. In interval notation, this is $[7, \infty)$.

   Watch out

   A common mistake when solving inequalities is forgetting to reverse the inequality sign when multiplying or dividing both sides by a negative number. For example, if you had $-2x \ge 14$, dividing by $-2$ would yield $x \le -7$. This rule is crucial for maintaining the correctness of the solution. In this step, we divided by a positive number ($2$), so the sign did not flip.

2. Solve the second inequality: $3x - 5 < -2$

   Again, we want to isolate $x$.

   Add $5$ to both sides:

   $3x - 5 + 5 < -2 + 5$

   $3x < 3$

   Now, divide both sides by $3$. Since $3$ is a positive number, the inequality sign remains unchanged.

   $\frac{3x}{3} < \frac{3}{3}$

   $x < 1$ …