Inequalities: Understanding inequality

The task is to interpret the inequality shown by a number line diagram.

First, observe the number line diagram provided. The numbers -4 and 3 are highlighted with vertical dashed lines. A critical point is at 3, where the circle is filled, indicating the inclusion of this point in the set. The line then extends from 3 to the right, suggesting that any point greater than or equal to 3 is included.

This indicates the inequality for $x$ is $x \geq 3$. The filled circle means 3 itself is part of the solution.

Thus, the solution to the inequality represented by the diagram is:

$3 \leq x$

This matches with choice number 3 in the provided options: $3 ≤ x$.

To solve the problem and determine the inequality represented by the number line, follow these steps:

• Examine the endpoints of the interval on the number line. At $-7$, there is an open circle, indicating that $-7$ is not included in the interval.
• At $0$, there is a closed circle, indicating that $0$ is included in the interval.
• This gives us the inequality for the interval: open at $-7$ ($<$) and closed at $0$ ($\leq$).

Therefore, the inequality represented by the number line is $-7 < x \leq 0$.

This is consistent with option 1 in the provided choices. The inequality is represented by $-7 < x \leq 0$.

First, we will move the elements:

$5-8x>7x+3$

$5-3>7x+8x$
$2>15x$

We divide the answer by 13, and we get:

$x > \frac{2}{15}$

In the exercise, we have an inequality equation.

We treat the inequality as an equation with the sign -=,

And we only refer to it if we need to multiply or divide by 0.

 $10x-4 ≤ -3x-8$

We start by organizing the sections:

$10x+3x-4 ≤ -8$

$13x-4 ≤ -8$

$13x ≤ -4$

Divide by 13 to isolate the X

$x≤-\frac{4}{13}$

Let's look again at the options we were asked about:

Answer A is with different data and therefore was rejected.

Answer C shows a case where X is greater than$-\frac{4}{13}$, although we know it is small, so it is rejected.

Answer D shows a case (according to the white circle) where X is not equal to$-\frac{4}{13}$, and only smaller than it. We know it must be large and equal, so this answer is rejected.

Therefore, answer B is the correct one!

To solve the compound inequality $40x + 57 \leq 5x - 13 \leq 25x + 7$, we first break it down into two inequalities and solve them separately.

Step 1: Solve the inequality $40x + 57 \leq 5x - 13$.

Subtract $5x$ from both sides: $40x - 5x + 57 \leq -13$.

This simplifies to $35x + 57 \leq -13$.

Subtract 57 from both sides to isolate the term with $x$: $35x \leq -70$.

Divide by 35 to solve for $x$: $x \leq -2$.

Step 2: Solve the inequality $5x - 13 \leq 25x + 7$.

Subtract $5x$ from both sides: $-13 \leq 20x + 7$.

Subtract 7 from both sides to isolate the term with $x$: $-20 \leq 20x$.

Divide by 20 to solve for $x$: $-1 \leq x$.

Step 3: Determine the solution by finding the intersection of the two solutions.

The first solution is $x \leq -2$ and the second solution is $-1 \leq x$.

There is no overlap between $x \leq -2$ and $-1 \leq x$. Therefore, there is no value for $x$ that satisfies both conditions simultaneously.

Therefore, the solution is "No solution."

The correct diagram corresponds to choice 1, which indicates a solution of "No solution."