Question #f0505

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Question #f0505

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Answer 1 · 2018-02-02T00:19:55

$x = 1$ $x=1$

$x = - \frac{5}{2}$ $x=-5/2$

Explanation:

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The Process
"_________"

**"We need to turn this rational equation into a non-rational one to be able to solve it. The easiest way to do that, is to perform operations that give us a quotient (fraction) on each side of the equation. The steps are:

$1)$ $1)$ add $4$ $4$ and $\frac{3}{x - 2}$ $3/(x-2)$ together by taking a common denominator and turning them into one fraction. The common denominator is $x - 2$ $x-2$ and $4$ $4$ is actually $\frac{4}{1}$ $4/1$ .

$\frac{4}{1} + \frac{3}{x - 2} = \frac{4 (x - 2) + 3}{x - 2}$ $4/1+3/(x-2)=(4(x-2)+3)/(x-2)$

What we did was divide the common denominator ( $x - 2$ $x-2$ ) by the denominator of the first term, i.e. $4$ $4$ , and got $(x - 2)$ $(x-2)$ . Then we multiplied it by the numerator, i.e. $4$ $4$ and put the result on top.
Then we divided the common denominator by the denominator of the second term and got $1$ $1$ , i.e. $\frac{x - 2}{x - 2}$ $(x-2)/(x-2)$ . Then we multiplied it by the numerator of the second term, i.e. $3$ $3$ and put the result on top. This is how we ended up with the fraction on the right hand side.

$2)$ $2)$ After simplifying the new fraction, i.e. combining like terms and removing parentheses, we set it equal to the fraction we had on the left hand side.

$3)$ $3)$ Then we cross-multiplied the fractions to get a regular equation which, after simplifying, gave us a quadratic equation. This means an equation with the $x$ $x$ having the power of $2$ $2$ .

$4)$ $4)$ Then we used the quadratic formula to solve for $x$ $x$ . If you have an equation in the form of:

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

The formula for $x$ $x$ is:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

In this equation:

$a = 2$ $a=2$ , $b = 3$ $b=3$ , and $c = - 5$ $c=-5$

We plugged them into the formula and got two answers for $x$ $x$ .
I hope this helps."**

I am going to assume you meant the following:

$\frac{5}{x + 4} = 4 + \frac{3}{x - 2}$ $5/(x+4)=4+3/(x-2)$

$\frac{5}{x + 4} = \frac{4 (x - 2) + 3}{x - 2}$ $5/(x+4)=(4(x-2)+3)/(x-2)$

$\frac{5}{x + 4} = \frac{4 x - 8 + 3}{x - 2}$ $5/(x+4)=(4x-8+3)/(x-2)$

$\frac{5}{x + 4} = \frac{4 x - 5}{x - 2}$ $5/(x+4)=(4x-5)/(x-2)$

Cross-multiplying, we get:

$(x + 4) (4 x - 5) = 5 (x - 2)$ $(x+4)(4x-5)=5(x-2)$

$4 x^{2} - 5 x + 16 x - 20 = 5 x - 10$ $4x^2-5x+16x-20=5x-10$

$4 x^{2} + 11 x - 20 - 5 x + 10 = 0$ $4x^2+11x-20-5x+10=0$

$4 x^{2} + 6 x - 10 = 0$ $4x^2+6x-10=0$

$2 (2 x^{2} + 3 x - 5) = 0$ $2(2x^2+3x-5)=0$

$2 x^{2} + 3 x - 5 = 0$ $2x^2+3x-5=0$

Using the quadratic formula:

$x = \frac{- 3 \pm \sqrt{9 - 4 (2) (- 5)}}{4} = \frac{- 3 \pm \sqrt{49}}{4} = \frac{- 3 \pm 7}{4}$ $x=(-3+-sqrt(9-4(2)(-5)))/4=(-3+-sqrt49)/4=(-3+-7)/4$

$x = 1$ $x=1$

$x = - \frac{5}{2}$ $x=-5/2$

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Question #f0505

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