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3 years ago

What are the solutions to the following system of equations? y=x^2+12x+30 and 8x-y=10

1 answer:

6 0

Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:

8x-10 = x^2+12x+30

Group all terms together on the right side. To do this, add -8x+10 to both sides. Then 0 = x^2 +4x +40. You must now solve this quadratic equation for x, if possible. I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.

If you have a graphing calculator, please graph 8x-10 and x^2+12x+30 on the same screen. You will see two separate graphs that do NOT intersect. This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.

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For this case we must follow the steps below:

step 1:

We place each of the given points on a coordinate axis

Step 2:

We join the AC points (represented by the orange line)

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It is observed that the resulting figure after placing the 4 points on a coordinate axis, turns out to be a rhombus.

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Answer:

See attached image

Option D

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