+2446 I'm going to assume that this is what the equation is supposed to be. I have two possible interpretations. I'll solve both
I'll solve for x 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 a and c 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 -4 | Sum 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. | ||||||||
| 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:
\(x=-\frac{1}{2}\hspace{1mm}\text{or}\hspace{1mm}x=2\)
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