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(x^2/x+1)=(x+2/3)

DarkKnight  Jun 26, 2017
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I'm going to assume that this is what the equation is supposed to be. I have two possible interpretations. I'll solve both
 

  1. \(\frac{x^2}{x}+1=x+\frac{2}{3}\)
  2. \(\frac{x^2}{x+1}=\frac{x+2}{3}\)

I'll solve for using the first interpretation:
 

\(\frac{x^2}{x}+1=x+\frac{2}{3}\)I'll utilize an exponent rule to simplify \(\frac{x^2}{x}\) that says that \(\frac{a^b}{a^c}=a^{b-c}\)
\(\frac{x^2}{x}=x^{2-1}=x^1=x\)Reinsert this back into the original equation.
\(x+1=x+\frac{2}{3}\)Subtract on both sides.
\(1=\frac{2}{3}\)\(1\neq\frac{2}{3}\), so there is no value for that satisfies this equation; in other words, there is no solution.
  

 

I'll solve for the second interpretation:

 

\(\frac{x^2}{x+1}=\frac{x+2}{3}\)In order to solve this equation, we must cross multiply. Remember that if \(\text{If}\hspace{1mm}\frac{a}{b}=\frac{c}{d}\hspace{1mm}\text{,then}\hspace{1mm}ad=bc\)
\(3*x^2=(x+2)(x+1)\)Let's simplify the left hand first, which is the easiest.
\(3x^2=(x+2)(x+1)\)Multiply the two binomials together by distributing the x and the 2 to the x+1-term. 
\((x+2)(x+1)=x(x+1)+2(x+1)=x^2+x+2x+2=x^2+3x+2\)Reinsert the simplified version back into the equation.
\(3x^2=x^2+3x+2\)Subtract x^2 from both sides of the equation.
\(2x^2=3x+2\)Bring all the terms to the left side of the equation
\(2x^2-3x-2=0\)Now, use any method to solve for (factoring, completing the square, quadratic equation, etc.) This happens to be factorable, so I will use that method.
  

 

To help visualize what I am doing, I'll attempt to draw an "x."

 

                                                                           \             /

                                                                            \   -4   / 

                                                                             \       / 
                                                                              \     /    

                                                                               \  /

                                                                         A     \/     B

                                                                                /\

                                                                               /  \

                                                                              /     \

                                                                             /       \

                                                                            /   -3   \

                                                                           /            \    

 

The top number is the number of the product of the values of and in a quadratic and the bottom is simply b, the coefficient of the linear term of the equation. Our job is to find 2 numbers for A and B wherein they multiply to get -4 and their sum is -3. Let's create a table:
 

Factors of -4Sum of Factors
\(4*-1\)\(4-1=3\)
\(-1*4\)\(-1+4=3\)
\(-4*1\)\(-4+1=-3\)
\(1*-4\)\(1-4=-3\)
\(2*-2\)\(2-2=0\)
\(-2*2\)\(-2+2=0\)
  

 

Out of all these combinations for factors of 4, which one gives the sum of -3? -4 and 1 or 1 and -4. These are flips of each other. You can use either combination, though. I'll use -4 and 1. 

 

\(2x^2-4x+x-2=0\)

 

What this method does is break up that b-term into 2 parts. Why is this useful? You'll see!

 

\(2x^2-4x+x-2=0\)Solve by grouping. I'll put parentheses aroung the groups I am solving.
\((2x^2-4x)+(x-2)=0\)Factor out the GCF of the first group, which is 2x. The other term, x+2, has a GCF of 1, so it cannot be simplified.
\(2x^2-4x=2x(x-2)\)Reinsert this back into the equation.
\(2x(x-2)+(x-2)=0\)Now, we will utilize the principle that \(ac+bc=c(a+b)\)
\(2x(x-2)+(x-2)=2x(x-2)+1(x-2)=(2x+1)(x-2)\)Ok, reinsert this back into the equation.
\((2x+1)(x-2)=0\)Set both factors equal to 0.
\(2x+1=0\)\(x-2=0\)
\(2x=-1\) 
\(x=-\frac{1}{2}\)\(x=2\)
  

 

Solve for both of the values for x.
  

 

Now, let's look at that original equation again:
\(\frac{x^2}{x+1}=\frac{x+2}{3}\)

 

The solutions work out in the original equation, so your answers are:
 

\(x=-\frac{1}{2}\hspace{1mm}\text{or}\hspace{1mm}x=2\)

TheXSquaredFactor  Jun 26, 2017
edited by TheXSquaredFactor  Jun 26, 2017
 #2
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0

Solve for x:
x^2/(x + 1) = x + 2/3

Multiply both sides by x + 1:
x^2 = (2 (x + 1))/3 + x (x + 1)

Expand out terms of the right hand side:
x^2 = x^2 + (5 x)/3 + 2/3

Subtract x^2 + (5 x)/3 + 2/3 from both sides:
-(5 x)/3 - 2/3 = 0

Bring -(5 x)/3 - 2/3 together using the common denominator 3:
1/3 (-5 x - 2) = 0

Multiply both sides by 3:
-5 x - 2 = 0

Multiply both sides by -1:
5 x + 2 = 0

Subtract 2 from both sides:
5 x = -2

Divide both sides by 5:
Answer: | x = -2/5

Guest Jun 26, 2017

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