Video: US-SAT04S4-Q26-537149476197

If the systems of inequalities 𝑦 ≀ 3π‘₯ + 2 and 𝑦 > π‘₯ βˆ’ 5 are graphed in the π‘₯𝑦-plane shown, which quadrant would contain no solutions to the system? [A] There are solutions in all four quadrants [B] Quadrant IV [C] Quadrant II [D] Quadrant III

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Video Transcript

If the systems of inequalities 𝑦 is less than or equal to three π‘₯ plus two and 𝑦 is greater than π‘₯ minus five are graphed in the π‘₯𝑦-plane shown, which quadrant would contain no solutions to the system? Is it A) there are solutions in all four quadrants, B) quadrant four, C) quadrant two, or D) quadrant three?

Both of our inequalities are linear and are written in slope-intercept form 𝑦 equals π‘šπ‘₯ plus 𝑏, where π‘š is the gradient of the line and 𝑏 is the 𝑦-intercept. Our first step is to graph the two linear equations 𝑦 equals three π‘₯ plus two and 𝑦 equals π‘₯ minus five. The equation 𝑦 equals three π‘₯ plus two has a slope or gradient of three and a 𝑦-intercept of two. The 𝑦-intercept of two means that it will cross the 𝑦-axis at positive two. A slope or gradient of three means that for every unit we move to the right, we need to move up three units. As this is positive, the slope will be upwards. Joining these two points gives us the graph of the equation 𝑦 equals three π‘₯ plus two.

An alternative method to draw this straight line would be to substitute some π‘₯-coordinates into the equation. In this case, we’ll substitute in π‘₯ equals negative one, π‘₯ equals zero, and π‘₯ equals one. Substituting in π‘₯ equals negative one gives us 𝑦 is equal to three multiplied by negative one plus two. Three multiplied by negative one is negative three. Adding two to this gives us negative one. When π‘₯ is equal to negative one, 𝑦 is equal to negative one. Next, we can substitute in π‘₯ equals zero. Three multiplied by zero is equal to zero. Adding two to this gives us an answer of two. When π‘₯ is equal to zero, 𝑦 is equal to two. Finally, we’ll substitute π‘₯ is equal to one. Three multiplied by one is equal to three. Adding two to this gives us five. When π‘₯ is equal to one, 𝑦 is equal to five.

We can then plot the three coordinates negative one, negative one; zero, two; and one, five. As we can see, these three points all lie on the line 𝑦 equals three π‘₯ plus two. We now need to consider the equation 𝑦 is equal to π‘₯ minus five. This has a slope or gradient equal to one as π‘₯ is the same as one π‘₯. It has a 𝑦-intercept of negative five. This line will intersect the 𝑦-axis at negative five. As the gradient or slope is equal to one, for every one unit we move right, we must move one unit up. This gives us the straight line, as shown on the graph.

At present, it doesn’t look like these two lines intersect. However, they can both be extended. This means that they would intersect further down the third quadrant. The lines are getting closer together in the third quadrant, but moving further apart in the first quadrant. There will be one point of intersection. We now need to consider the inequality signs. The first sign was less than or equal to. This means that the 𝑦-values need to be less than or equal to three π‘₯ plus two. The line will be bold as it is less than or equal to. And we want all the points that are below this line.

Our second inequality was greater than. This means that our line for the inequality 𝑦 is greater than π‘₯ minus five must be a broken one or dashed one. This is because our values cannot lie on the line. As we want the 𝑦-values to be greater than π‘₯ minus five, we want all the points that are above this line. The region that satisfies the system of inequalities is everything between the two lines. It is clear that there are many solutions in quadrant one. Quadrant four and three also have numerous solutions. Quadrant two, whilst having a smaller number of solutions, also has solutions to the system of inequalities.

As there are solutions in quadrant one, two, three, and four, we can conclude that there are solutions in all four quadrants. This means that there is no quadrant that contains no solutions to the system.

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