ONE MORE PROBLEM TO HELP ME WITH!!!!
How do you graph the inequality on a plane?
3x+4y is less than or equal to 12
Plot the straight line
y = -(3/4)x -3
which is the same formula, in standard format.
For the inequality, solutions are below that line. You might black out the area above the line where solutions do not exist, or use cross-hatching. Your textbook probably suggests a way to highlight the solution region.
To graph the inequality 3x + 4y ≤ 12 on a plane, follow these steps:
Step 1: Solve the inequality for y to get it in the form y ≤ mx + b.
In this case, subtract 3x from both sides to isolate the y term:
4y ≤ -3x + 12
Step 2: Divide both sides of the inequality by 4 to solve for y:
y ≤ (-3/4)x + 3
Step 3: Now we have the inequality in slope-intercept form (y ≤ mx + b). This form tells us that y is less than or equal to a line with a slope of -3/4 and a y-intercept of 3.
Step 4: Graph the line: Start by plotting the y-intercept, which is the point (0,3). Then use the slope to find other points and draw a straight line through them. Since the inequality is inclusive (y ≤), the line should be solid.
Step 5: Determine the region to shade: Since we are looking for points that are less than or equal to the line, shade the region below the line.
Now you have successfully graphed the inequality 3x + 4y ≤ 12 on a plane!