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# Forgot how to solve

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

DarkKnight  Jun 26, 2017
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#1
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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 x on both sides. $$1=\frac{2}{3}$$ $$1\neq\frac{2}{3}$$, so there is no value for x 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 x (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
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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