SOLUTION: Solve. If equation has no solution, write no solution.
x over x+1 (+) x+1 over x = 5 over 2
{{{2x^2+2(x+1)(x+1) = 5[x(x+1)}}}
2x^2+2(x^2+x+x+2) = 5(x^2+x)
2x^2 + 2x^
Question 69184This question is from textbook Algebra Structure and Method
: Solve. If equation has no solution, write no solution.
x over x+1 (+) x+1 over x = 5 over 2
2x^2+2(x^2+x+x+2) = 5(x^2+x)
2x^2 + 2x^2 + 2x + 2x+4 = 5x^2+5x
0 = x^2 + x-4
that is as far as I got, please help! This question is from textbook Algebra Structure and Method
Cartoon (animation) form: For tutors: simplify_cartoon( 0=x^2+x-4 )
If you have a website, here's a link to this solution.
DETAILED EXPLANATION
Look at . Moved these terms to the left ,, It becomes . Look at . Added fractions or integers together It becomes . Look at . Moved to the right of expression It becomes . Look at . Equation is a quadratic equation: -x^2-x+4 =0, and has solutions -2.56155281280883,1.56155281280883 It becomes . Result: This is an equation! Solutions: x=-2.56155281280883,x=1.56155281280883.