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## solve 3y<6x+4

3y≤6x+4

Q: solve 3y<6x+4
3y≤6x+4

A: The question asks for all the values of x and y that fulfill the equations.

First solve for 3y = 6x + 4 -> y = 2x + (4/3).

On a cartisean coordinate system this equation splits the graph into three regions. The line graphed by the above equation, all the points below it and all the point above it. y < 2x + (4/3) represents all the points below it. y <= 2x + (4/3) represents all the points below the line and including the line.

You can graph the line by finding the x and y intercepts. The y intercept is found by setting x to zero giving: (0, 4/3). The x intercept is found by setting y to zero giving: (-2/3,0).

A. 3y<6x+4
B. 3y≤6x+4

A.

3y<6x+4

Divide both sides by 3:

y < 2x + 4/3

The line y = 2x + 4/3 defines 3 sets of points on the coordinate plane:
1.) all the points above the line, y > 2x + 4/3;
2.) all the points on the line, y = 2x + 4/3; and
3.) all the points below the line, y < 2x + 4/3;

In this case you have case 3; so you would graph the line y = 2x + 4/3 with a dashed line and shade everything below it. All points in the shaded region are solutions to the inequality y < 2x + 4/3.

B.

3y ≤ 6x + 4

y ≤ 2x + 4/3

Here you have to include both the "less than" and the "equal", both cases 2 & 3 apply from above.

So you would graph the line y = 2x + 4/3 with a solid line and shade everything below it. All points on the line and in the shaded region are solutions to the inequality y ≤ 2x + 4/3.
3y ≤ 6x + 4

In this equation, we'll treat the inequality sign as an equals sign....

First, we'll solve for y

3y ≤ 6x + 4

divide both sides by 3

(3y/3) ≤ (6x + 4)/3

y ≤ (6x + 4)/3

Now substitute y in the original equation

3(6x + 4)/3 ≤ 6x + 4

(18x + 12)/3 ≤ 6x + 4

6x + 4 ≤ 6x + 4

6x - 6x ≤ 4 - 4

0 ≤ 0

From the resulting answer, x can be any integer between -∞ (negative infinity) and zero (0)