Video: US-SAT05S4-Q25-527186295759

Consider the system of inequalities. π¦ > 3π₯ + 2, π¦ > (1/2) π₯ + 3. If the system is graphed on the π₯π¦-plane below, which quadrant would not contain any solutions to the system?

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Video Transcript

Consider the system of inequalities. π¦ is greater than three π₯ plus two. And π¦ is greater than a half π₯ plus three. If the system is graphed on the π₯π¦-plane below, which quadrant would not contain any solutions to the system?

Both of the inequalities are written in slope intercept form. π¦ equals ππ₯ plus π. Where π is the slope or gradient and π is the π¦-intercept. Letβs firstly consider the equation π¦ equals three π₯ plus two. And what it would look like on the graph. This equation would have a π¦-intercept of two. This means that it crosses the π¦-axis at the point two. It has a slope or gradient of three. This means that for every one unit we move to the right, we need to move up three units. The line π¦ equals three π₯ plus two cuts through the first second and third quadrant. We could also have drawn this straight line by substituting values of π₯ into the equation. For example, π₯ equals negative one. π₯ equals zero. And π₯ equals one.

Substituting in these values into the equation three π₯ plus two gives us corresponding π¦-values of negative one, two, and five. Plotting these three points and drawing a straight line through them would give the same line as we found earlier. We now need to repeat the process for the equation π¦ equals a half π₯ plus three. This equation has a π¦-intercept of three. It has a slope or gradient of a half. This means that for every two units we move to the right, we need to move up one unit. Once again, we can draw the straight line through these two points.

Both of our inequalities had greater than signs. The π¦-value needed to be greater than three π₯ plus two and greater than a half π₯ plus three. This means that the overall solution would need to be above both of the lines. It is clear from our graph that some of this region would be contained in quadrant one and in quadrant two. There is also a small area of our graph in quadrant three. If we continued our π₯-axis to the left. We would see that there is a larger area that is greater than the line π¦ equals a half π₯ plus three. We can therefore say that there are solutions in quadrants one, two, and three. There are no solutions in quadrant four as neither of the lines passes through quadrant four. And both of our inequalities are greater than the two equations.