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Luba_88 [7]
3 years ago
15

Simplify (x + 2/ x^2 + 2x -3) / (x + 2/x^2 - x)

Mathematics
2 answers:
rjkz [21]3 years ago
5 0

Answer:

The simplest form is x/(x + 3)

Step-by-step explanation:

* To simplify the rational Expression lets revise the factorization

  of the quadratic expression

*  To factor a quadratic in the form x² ± bx ± c:

- First look at the c term  

# If the c term is a positive number, and its factors are r and s they

  will have the same sign and their sum is b.

#  If the c term is a negative number, then either r or s will be negative

   but not both and their difference is b.

- Second look at the b term.  

# If the c term is positive and the b term is positive, then both r and

  s are positive.  

Ex: x² + 5x + 6 = (x + 3)(x + 2)  

# If the c term is positive and the b term is negative, then both r and s

  are negative.  

Ex:  x² - 5x + 6 = (x -3)(x - 2)

# If the c term is negative and the b term is positive, then the factor

  that is positive will have the greater absolute value. That is, if

  |r| > |s|, then r is positive and s is negative.  

Ex: x² + 5x - 6 = (x + 6)(x - 1)

# If the c term is negative and the b term is negative, then the factor

  that is negative will have the greater absolute value. That is, if

  |r| > |s|, then r is negative and s is positive.

Ex: x² - 5x - 6 = (x - 6)(x + 1)

* Now lets solve the problem

- We have two fractions over each other

- Lets simplify the numerator

∵ The numerator is \frac{x+2}{x^{2}+2x-3}

- Factorize its denominator

∵  The denominator = x² + 2x - 3

- The last term is negative then the two brackets have different signs

∵ 3 = 3 × 1

∵ 3 - 1 = 2

∵ The middle term is +ve

∴ -3 = 3 × -1 ⇒ the greatest is +ve

∴ x² + 2x - 3 = (x + 3)(x - 1)

∴ The numerator = \frac{(x+2)}{(x+3)(x-2)}

- Lets simplify the denominator

∵ The denominator is \frac{x+2}{x^{2}-x}

- Factorize its denominator

∵  The denominator = x² - 2x

- Take x as a common factor and divide each term by x

∵ x² ÷ x = x

∵ -x ÷ x = -1

∴ x² - 2x = x(x - 1)

∴ The denominator = \frac{(x+2)}{x(x-1)}

* Now lets write the fraction as a division

∴ The fraction = \frac{x+2}{(x+3)(x-1)} ÷ \frac{x+2}{x(x-1)}

- Change the sign of division and reverse the fraction after it

∴ The fraction = \frac{(x+2)}{(x+3)(x-1)}*\frac{x(x-1)}{(x+2)}

* Now we can cancel the bracket (x + 2) up with same bracket down

 and cancel bracket (x - 1) up with same bracket down

∴ The simplest form = \frac{x}{x+3}

Amanda [17]3 years ago
5 0

ANSWER

\frac{x}{x + 3}

EXPLANATION

We want to simplify:

\frac{x +2 }{ {x}^{2}  + 2x - 3}  \div  \frac{x + 2}{ {x}^{2}- x}

Multiply by the reciprocal of the second fraction:

\frac{x +2 }{ {x}^{2}  + 2x - 3}   \times   \frac{{x}^{2}- x}{ x + 2}

Factor;

\frac{x +2 }{ (x + 3)(x - 1)}   \times   \frac{x(x - 1)}{ x + 2}

We cancel out the common factors to get:

\frac{x}{x + 3}

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Answer:

<em>(a) x=2, y=-1</em>

<em>(b)  x=2, y=2</em>

<em>(c)</em> \displaystyle x=\frac{5}{2}, y=\frac{5}{4}

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.

We call the determinant of the system

\Delta=\begin{vmatrix}a &b \\c  &d \end{vmatrix}

We also define:

\Delta_x=\begin{vmatrix}p &b \\q  &d \end{vmatrix}

And

\Delta_y=\begin{vmatrix}a &p \\c  &q \end{vmatrix}

The solution for x and y is

\displaystyle x=\frac{\Delta_x}{\Delta}

\displaystyle y=\frac{\Delta_y}{\Delta}

(a) The system to solve is

\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.

Calculating:

\Delta=\begin{vmatrix}1 &1 \\1  &-2 \end{vmatrix}=-2-1=-3

\Delta_x=\begin{vmatrix}1 &1 \\4  &-2 \end{vmatrix}=-2-4=-6

\Delta_y=\begin{vmatrix}1 &1 \\1  &4 \end{vmatrix}=4-3=3

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1

The solution is x=2, y=-1

(b) The system to solve is

\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.

Calculating:

\Delta=\begin{vmatrix}4 &-1 \\1  &-1 \end{vmatrix}=-4+1=-3

\Delta_x=\begin{vmatrix}6 &-1 \\0  &-1 \end{vmatrix}=-6-0=-6

\Delta_y=\begin{vmatrix}4 &6 \\1  &0 \end{vmatrix}=0-6=-6

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2

The solution is x=2, y=2

(c) The system to solve is

\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.

Calculating:

\Delta=\begin{vmatrix}-1 &2 \\1  &2 \end{vmatrix}=-2-2=-4

\Delta_x=\begin{vmatrix}0 &2 \\5  &2 \end{vmatrix}=0-10=-10

\Delta_y=\begin{vmatrix}-1 &0 \\1  &5 \end{vmatrix}=-5-0=-5

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}

The solution is

\displaystyle x=\frac{5}{2}, y=\frac{5}{4}

(d) The system to solve is

\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.

Calculating:

\Delta=\begin{vmatrix}6 &-1 \\4  &-2 \end{vmatrix}=-12+4=-8

\Delta_x=\begin{vmatrix}-5 &-1 \\6  &-2 \end{vmatrix}=10+6=16

\Delta_y=\begin{vmatrix}6 &-5 \\4  &6 \end{vmatrix}=36+20=56

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7

The solution is x=-2, y=-7

4 0
3 years ago
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