How do you solve #3/x+x=4#?
1 Answer
May 22, 2018
Explanation:
#"multiply all terms by "x#
#cancel(x)xx3/cancel(x)+x^2=4x#
#rArr3+x^2=4x#
#"subtract "4x" from both sides"#
#rArrx^2-4x+3=0larrcolor(blue)"in standard form"#
#"the factors of + 3 which sum to - 4 are - 1 and - 3"#
#rArr(x-1)(x-3)=0#
#"equate each factor to zero and solve for x"#
#x-1=0rArrx=1#
#x-3=0rArrx=3#
