I’m so confused on how to do it
1 answer:
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Answer:
Part A: The graph of the system is plotted at y-intercepts of 3 and -4. The line of the first equation is dashed and shaded below. The line of the second equation is dashed and shaded above. The solution area is between the second and third quadrants.
Part B: No, the point (-4, 6) is not in the solution are because it only fits into the first equations solution area. [6=2(-4)+3 solved is 6=-5, this is not a true statement.]
Step-by-step explanation:
Part A: Each equation {y>2x+3, y<-3/2x-4} is already in slope intercept form and can be graphed. In the first equation we know that there is a y-intercept of 3 and a slope of 2. This is shown as as a upward moving line with plot points such as (-2,-1), (-1,1), (0,4), (1,6), and so on. In the second equation we know that there is a y-intercept of -4 and a slope of -3/2. This is shown as a downward moving line with plot points such as (0,-4), (-2,-1), (-4, 2), and so on. (See graph below)
Part B: The plot point (-4, 6) is not included in the solution area of the systems. This can be seen on the graph. And, proven mathematically by plugging in the point into the equations:
- y=2x+3 plug in the points (6)=2(-4)+3
- solve (6)=2(-4)+3 ... 2(-4)=-8 ... -8+3=-5 ... 6 = -5 is not a true statement
So, already we know that this plot point is not an answer to the system. Let's continue anyway.
- y=-3/2x-4 plug in the points (6)=-3/2(-4)-4
- solve (6)=-3/2(-4)-4 ... -3/2(-4)= 6 ... 6-4 = 2 ... 6=2 is not a true statement.
