SOLUTION: Solve by substitution {{{x^2+y^2=26}}} {{{y=-x+8}}}

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Question 140321: Solve by substitution

x%5E2%2By%5E2=26
y=-x%2B8




Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the given equations

x%5E2%2By%5E2=26
y=-x%2B8




x%5E2%2By%5E2=26 Start with the first equation

x%5E2%2B%28-x%2B8%29%5E2=26 Plug in y=-x%2B8


x%5E2%2Bx%5E2-16x%2B64=26 Foil


x%5E2%2Bx%5E2-16x%2B64-26=0 Subtract 26 from both sides


2x%5E2-16x%2B38=0 Combine like terms


Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve 2%2Ax%5E2-16%2Ax%2B38=0 ( notice a=2, b=-16, and c=38)




x+=+%28--16+%2B-+sqrt%28+%28-16%29%5E2-4%2A2%2A38+%29%29%2F%282%2A2%29 Plug in a=2, b=-16, and c=38



x+=+%2816+%2B-+sqrt%28+%28-16%29%5E2-4%2A2%2A38+%29%29%2F%282%2A2%29 Negate -16 to get 16



x+=+%2816+%2B-+sqrt%28+256-4%2A2%2A38+%29%29%2F%282%2A2%29 Square -16 to get 256 (note: remember when you square -16, you must square the negative as well. This is because %28-16%29%5E2=-16%2A-16=256.)



x+=+%2816+%2B-+sqrt%28+256%2B-304+%29%29%2F%282%2A2%29 Multiply -4%2A38%2A2 to get -304



x+=+%2816+%2B-+sqrt%28+-48+%29%29%2F%282%2A2%29 Combine like terms in the radicand (everything under the square root)



x+=+%2816+%2B-+4%2Ai%2Asqrt%283%29%29%2F%282%2A2%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%2816+%2B-+4%2Ai%2Asqrt%283%29%29%2F%284%29 Multiply 2 and 2 to get 4



After simplifying, the quadratic has roots of

x=4+%2B+sqrt%283%29i or x=4+-+sqrt%283%29i



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Answer:

Since we get a complex solution for the equation 2x%5E2-16x%2B38=0, this means that the equations x%5E2%2By%5E2=26 and y=-x%2B8 do not intersect. So there are no solutions