+128474
3/x + 6 / [ x + 3] = 8 / [ x + 1]
I prefer to simplify the left side first, and then cross-multiply...so we have
(3 [x + 3] + 6x ) / [ x(x + 3) ] = 8 / [x+ 1]
[9x + 9] / [ x ( x + 3) ] = 8 / [x + 1]
[ 9 ( x + 1) ] [x + 1] = 8[ x (x + 3)]
9 [x^2 + 2x + 1] = 8x^2 + 24x
9x^2 + 18x + 9 = 8x^2 + 24x
x^2 - 6x + 9 = 0 this factors as
(x - 3)^2 = 0 and its obvious that x = 3 must make this true
P.S. - the error was made when we forgot to mulltiply the right side by the LCD, too !!!
+2441 One of the greatest ways to detect an error in solving is to do it yourself. I, too, have trouble spotting errors that others make--even if it is very obvious. However, if you do the problem and compare your answer with the given result, it is much easier to spot the error. I will also solve \(\frac{3}{x}+\frac{6}{x+3}=\frac{8}{x+1}\) because it is good practice.
| \(\frac{3}{x}+\frac{6}{x+3}=\frac{8}{x+1}\) | First, the solver in this problem states that the LCD is \(x(x+3)(x+1)\), which is correct. Now, let's multiply both sides by the LCD | ||
| \(\frac{3x(x+3)(x+1)}{x}+\frac{6x(x+3)(x+1)}{x+3}=\frac{8x(x+3)(x+1)}{x+1}\) | Simplify every fraction. Notice that, in every fraction, you can simplify it such that the fraction disappears. Let's do that. | ||
| \(3(x+3)(x+1)+6x(x+1)=8x(x+3)\) | Right away, I already see a disagreement between my answer and the solver's answer. They are both different. This is the error. Since I like going the extra mile of solving, I will do that! First, I'll distribute 3(x+3) first. | ||
| \((3x+9)(x+1)+6x(x+1)=8x(x+3)\) | You'll notice that it is not worth looking at the method that the solver uses, as I have already spotted the error. I can now use my own method to solve. Remember that \(ac+bc=(a+b)c\) by the inverse of the distributive property. I will use this in the next step. | ||
| \((6x+3x+9)(x+1)=8x(x+3)\) | Combine like terms in the first set of parentheses and subtract \(8x(x+3)\) from both sides of the equation. | ||
| \((9x+9)(x+1)=8x(x+3)\) | Unfortunately, I do not see a way to use the above reverse-distributive-property trick to simplify any further. I can factor out a 9 from both sides. | ||
| \(9(x+1)(x+1)=8x(x+3)\) | Now, do \((a+b)^2=a^2+2ab+b^2\). | ||
| \(9(x^2+2x+1)=8x(x+3)\) | Now, expand both sides. | ||
| \(9x^2+18x+9=8x^2+24x\) | Now, subtract 8x^2+24 from both sides. | ||
| \(x^2-6x+9=0\) | This trinomial happens to be a perfect-square trinomial. Let's use that to our advantage. | ||
| \((x-3)^2=0\) | Take the square root of both sides. Remember that taking the square root means taking the absolute value. | ||
| \(|x-3|=0\) | The absolute value splits your answer into the negative and positive answers. Luckily, the absolute value of 0 is always 0--whether positive or negative. | ||
| For the left equation, add 3. For the right equation, divide by -1. | ||
| For the second equation, we can stop because we already calculated the value for x when x-3=0. Of course, it's 3. | ||
This value for x is outside of the undefined singularity point, so this answer is correct.
For your information, you were solving this equation using the quadratic formula incorrectly. Your first line using the quadratic equation is \(x = {-6 \pm \sqrt{(-6)^2-4(1)(4)} \over 2(1)}\), but it should be \(x = {\textcolor{red}{-}(-6) \pm \sqrt{b^2-4ac} \over 2a}\). Also, \(-3\) happens to be an undefined singularity point because it is a solution to \(x+3=0\).
Another thing to mention is that I agree with Cphill that it is easier to simplify the left-hand side first, but it might be harder to detect the error if you do not do the problem as the solver did it.
+2441 I have decided to split #11 and #12 into 2 problems. I think it will be more pleasing to view that way. This time, I need not solve it to spot the error; this time, I spotted it immediately. The error is line is in line 3. The transition from division and multiplication is incorrect.
When you divide by a fraction, the following rule applies. I am sure you are aware of this, but I am doing this to recap!
\(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}*\frac{d}{c}\)
The third line is \(\frac{r+4}{(r-3)(r-4)}*\frac{(r-4)(r-3)}{(r+2)(r+4)}\), but it should be \(\frac{r+4}{(r-3)(r-4)}*\color{red}{\frac{(r+2)(r+4)}{(r-4)(r-3)}}\). And of course, \(\frac{r+4}{(r-3)(r-4)}*\frac{(r-4)(r-3)}{(r+2)(r+4)}\neq\frac{r+4}{(r-3)(r-4)}*\color{red}{\frac{(r+2)(r+4)}{(r-4)(r-3)}}\).
The solving for this one is much simpler, thankfully. This is because the same canceling will occur. It is just that one of the denominators that used to be is now a numerator. This means that the answer is the inverse of \(\frac{1}{r+2}\), which means \(\frac{1}{\frac{1}{r+2}}\). The inverse of the inverse of a number is itself, so the expression simplifies to \(r+2\).
