Prove that (2n+1)^2 - (2n+1)^2 is a multiple of 8 for all positive integer values of n
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Answer:
Step-by-step explanation:
y > 3x + 10
y < -3x - 1
<h3> Part A:</h3>You'd graph the given systems of linear inequalities the same way as you graph the linear equations.
To graph y > 3x + 10, plot the y-intercept, (0, 10), then use the slope, m = 3 (rise 3, run 1), to plot other points on the graph. Use a dashed line (because of the ">" symbol).
Follow the same steps for the other linear inequality. Plot the y-intercept, (0, -1), then use the slope, m = -3 (down 3, run 1) to plot other points. Use a dashed line (because of the "<" symbol).
Pick a test point on either side of the boundary line and plug it into the original problem. This will help determine which side of the boundary line is the solution. Plug in a test point that is not on the boundary line.
Use the point of origin, (0, 0) as the test point. Plug in these values into the given systems of linear inequalities to see whether it will provide a true statement.
y > 3x + 10
0 > 3(0) + 10
0 > 0 + 10
0 > 10 (False statement).
y < -3x - 1
0 < -3(0) - 1
0 < 0 - 1
0 < -1 (false statement).
Since the point of origin provided a false statement to the given systems of linear inequalities, you must shade the half-plane region where it doesn't contain the test point.
<h3>Part B: </h3>
You'll do the same process as what I've done for the test point. Plug in the values of (8, 10) into the given systems of linear inequalities. If it provides a false statement, then it means that it is not a solution to the system.
y > 3x + 10
10 > 3(8) + 10
10 > 24 + 10
10 > 34 (False statement).
y < -3x - 1
10 < -3(8) - 1
10 < -24 - 1
10 < -25 (false statement).
Therefore, (8, 10) is not a solution to the system.
