SOLUTION: I tried using one of your solvers for the following problem and it gave me an error and said solution not found so if you could help that would be great! Find the solution of th

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Question 59767This question is from textbook Alegbra concepts and applications
: I tried using one of your solvers for the following problem and it gave me an error and said solution not found so if you could help that would be great!
Find the solution of the system
y = x^2 + 5
and
y = 2x + 4 This question is from textbook Alegbra concepts and applications

Answer by Cintchr(481) About Me  (Show Source):
You can put this solution on YOUR website!
1. +y+=+x%5E2+%2B+5+
2. +y+=+2x+%2B+4+
Since both are already in "y=" .... set them equal to each other
+x%5E2+%2B+5+=+2x+%2B+4+
Move all parts to the left ... setting the equation equal to zero
+x%5E2+%2B5+-2x+-4+=+0+
combine like terms and put in descending order
+x%5E2+-2x+%2B1+=+0+
Factor
+%28x-1%29%28x-1%29+=+0+
Since both parts are identical ... we only need to work it once
+x-1+=+0+
+x+=+1+

Pulg x = 1 into the equations
+y+=+x%5E2+%2B+5+ and +y+=+2x+%2B+4+
+y+=+1%5E2+%2B+5+ and +y+=+2%281%29+%2B+4+
+y+=+1+%2B+5+ and +y+=+2+%2B+4+
+y+=+6+ and +y+=+6+

Your one solution is
(1,6)
The graphs are a parabola and a line. Since there is only one solution, the point at which they meet is also the "minimum" or vertex of the parabola.