How do you solve the system #x^2+y^2+8x+7=0# and #x^2+y^2+4x+4y-5=0# and #x^2+y^2=1#?

Answer 1

Cora Alexander

There are no points where all three equations intersect.

We have the following equations:

#E1: x^2+y^2+8x+7=0# #E2: x^2+y^2+4x+4y-5=0# #E3: x^2+y^2=1#

Let's substitute in E3 into E1 and E2, making the #x^2 and y^2# terms equal to 1:

#E1: 1+8x+7=0# #E2: 1+4x+4y-5=0#

Now let's solve E1:

#E1: 8x+8=0#

#E1: 8x=-8#

#E1: x=-1#

Now let's solve E2:

#E2: 1+4x+4y-5=0#

#E2: 1+4(-1)+4y-5=0#

#E2: 1-4+4y-5=0#

#E2: -8+4y=0#

#E2: 4y=8#

#E2: y=2#

And now let's check our work by substituting into E3:

#E3: x^2+y^2=1#

#E3: (-1)^2+(2)^2=1#

#E3: 1+4!=1#

#E3: 5!=1#

So there is no solution in this system that satisfies all three equations.

We can see this in the following graphs:

This is the graph of #E1: x^2+y^2+8x+7=0#:

graph{x^2+y^2+8x+7=0 [-20, 20, -10, 10]}

This is the graph of #E2: x^2+y^2+4x+4y-5=0#:

graph{x^2+y^2+4x+4y-5=0 [-20, 20, -10, 10]}

This is the graph of #E3: x^2+y^2=1#:

graph{x^2+y^2=1 [-20, 20, -10, 10]}

As you can see, there are no points where all three graphs intersect.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.