Q.Solve the system of inequalities: $3x - 7 < 5 + x$ ... (1), $11 - 5x \le 1$ ...

Concept: Linear Inequality Solutions — solve each inequality separately, then take the intersection of the solution sets.

Step 1: Solve (1)

$3x - 7 < 5 + x$

$\Rightarrow 3x - x < 5 + 7$

$\Rightarrow 2x < 12$

$\Rightarrow x < 6$

Step 2: Solve (2)

$11 - 5x \le 1$

$\Rightarrow -5x \le 1 - 11$

$\Rightarrow -5x \le -10$

Dividing by $-5$ (reverse inequality):

$\Rightarrow x \ge 2$

Step 3: Intersection …

When you solve a system of inequalities, you're finding all values of $x$ that satisfy every inequality at the same time. Think of it like a set of filters: each inequality is a condition that $x$ must pass. The final solution is the overlap — the region where all conditions hold together.

The key idea is to solve each inequality separately, then combine the results on a number line. The number line isn't just a picture; it's the clearest way to see the intersection of intervals.

1. Solve inequality (1): $3x - 7 < 5 + x$

   Start by bringing the $x$ terms together. Subtract $x$ from both sides:

Now add 7 to both sides:

Finally, divide by 2 (positive, so the inequality direction stays the same):

So the first condition is: all numbers strictly less than 6.

2. Solve inequality (2): $11 - 5x \le 1$

   Subtract 11 from both sides:

Now divide by $-5$. Crucial: when you divide or multiply an inequality by a negative number, the inequality sign reverses direction.

So the second condition is: all numbers greater than or equal to 2.

3. Combine the two conditions

   We need $x$ that satisfies both:

   • $x < 6$
   • $x \ge 2$

   This is the intersection of the two intervals. On the number line, you start at 2 (including 2, because of the $\le$ sign) and go up to, but not including, 6. …