Is the simultaneous equation x2 plus y2 equals 25 y equals x plus 1 even possible?

Answer 1

Yes, it's possible

(1) x2 + y2 = 25

(2) y = x + 1

Substituting (2) into (1) gives:

(3) x2 + (x + 1)2 = 25

(4) x2 + x2 + 2x + 1 = 25

(5) 2x2 + 2x - 24 = 0

(6) x2 + x - 12 = 0

Plugging 1, 1, and -12 into the quadratic equation gives us two answers:

x = 3, y - 4 and x = -4, y = -3

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Another contributor was working this at exactly the same time as the first

contributor, and posts his work here, just to demonstrate how great minds

run in the same gutter:

Shouldn't be a big problem. When you think about this for a while, it might seem like it might not be very difficult; but then, when you dig into it a little further, it turns out to be downright easy.

The first glance:

x2+y2=25 is a circle, centered at the origin, with radius=25.

y=x+1 is a 45-degree line, shifted one unit above the origin.

They have two common points ... the two points where the 45-degree line cuts

the circle. Those two points are the solutions to the simultaneous equations.

Since they're on a circle, we can't tell right away where the points will be, and

with fear and trembling, we hesitantly write them down and try to solve them

algebraically.

The drudge work:

Substitute [y=x+1] into [x2 + y2 = 25]:

x 2 + (x+1)2 = 25

x2 + x2 + 2x + 1 = 25

2x2 + 2x -24 = 0

x2 + x - 12 = 0

We don't even need to resort to the quadratic formula to solve this! We can

just look at it and factor it:

(x-4)(x+3) = 0

x = 4, (y = -3)

x = -3, (y = 4)

Well, you and I can factor it just by looking at it, but the middle-schooler whose homework we're doing probably needs to show his (our) work.