Question

2y-5x=≥3x 2x+y≤6 y+3≥0

Answer 1

Ответ:

у=-

$\frac{1}{3}$

Answer 2

It looks like you've provided a system of linear inequalities with three equations involving the variables x and y. To solve this system, we'll need to find the values of x and y that satisfy all three inequalities simultaneously. Let's work through each inequality step by step:

1. $2y - 5x \geq 3x$
2. $2x + y \leq 6$
3. $y + 3 \geq 0$

Let's start by solving the third inequality:

$y + 3 \geq 0$
Subtract 3 from both sides:
$y \geq -3$

This means that any value of y that is greater than or equal to -3 will satisfy this inequality.

Next, let's solve the second inequality:

$2x + y \leq 6$
Subtract $2x$ from both sides:
$y \leq 6 - 2x$

This gives us a boundary for y in terms of x. Now, let's combine this with the first inequality:

$2y - 5x \geq 3x$
Add $5x$ to both sides:
$2y \geq 8x$
Divide by 2:
$y \geq 4x$

So, for the first inequality, any value of y that is greater than or equal to 4x will satisfy it.

Now we have boundaries for y in terms of x from the second and third inequalities, and we want to find the values of x and y that satisfy all three inequalities. Combining the inequalities:

1. $y \geq 4x$
2. $y \leq 6 - 2x$
3. $y \geq -3$

From these inequalities, we can deduce that the common region where all three inequalities are satisfied lies where the conditions overlap:

1. $y \geq 4x$ (Region above the line)
2. $y \leq 6 - 2x$ (Region below the line)
3. $y \geq -3$ (Region above the line)

Graphically, if you were to plot these inequalities on a graph, the overlapping region of the shaded areas would be the solution to the system of inequalities.

Keep in mind that the solution to this system of inequalities is not a single point, but rather a region in the coordinate plane where all three conditions are met.