x2 + 1/ x = 3x + 1/ x
Emily, you might want to rewrite the equation by making sure the exponents and bases are on the right lines...the equal sign should be on the line with the base for the equation to make sense...furthermore, the are certain conditions that need to be met before solving this equation, such as x cannot be "0" because x is a denominator in the equation and cannot take the "0" value.
should your equation look more like: x2 +1/x = 3x + 1/x ?
If yes, then one easy way you go about solving it is this:
- first, we notice that we have 1/x on each side of the equal sign, which allows us to get rid of it by offseting each other. If we move either 1/x on the other side of the equal sign with a negative sign, it will look like this:
x2 + 1/x - 1/x = 3x. the underlined 1/x moved from the right side of the equal sign to the left of the equal sign with a negative sign per the basic rule of sign changes when moving to a side or the other of the equal sign.
Now, on the left sign of the equation we're going to have left just the x2, because 1/x and -1/x will cancel out, so our equation now looks like x2=3x, and can be easily turned into x2 - 3x = 0 by mobving 3x from the right side of equal sign to the left side.
This is a quadratic equation, which is very easy to solve as such, but the easiest way is to factor out "x" which appears in both x2 and 3x.
x(x - 3) = 0
Since this is a multiplication of two terms with a zero product, one of the terms has to be zero in order for the product to be zero.
First term, "x" is zero. But like I said in the beginning, this cannot be because in the original equation "x" is a denominator and cannot be zero.
Second, x - 3 = 0, which means x = 3 (3 -3 = 0)
This is a solution and it is the only solution per the condition "x cannot be zero"
If we are to verify this solution, we can insert x=3 in the original equation:
32 + 1/3 = 3*3 + 1/3 => 9 + 1/3 = 9 +1/3 which is a true expression and verifies the solution x=3
I hope this has helped, assuming my re written equation was your correct question.