Exercise 6.2

On the following page we have given the solution of Exercise 6.2 of Mathematics 9 (Science) published by Caravan Book House, Lahore.

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Simplify as rational expression

Question 1:

$\frac{x^2-x-6}{x^2-9}+\frac{x^2+2x-24}{x^2-x-12}$

Solution:

$\begin{align} \frac{x^2-x-6}{x^2-9}+\frac{x^2+2x-24}{x^2-x-12}&=\frac{x^2-3x+2x-6}{(x)^2-(3)^2}+\frac{x^2+6x-4x-24}{x^2-4x+3x-12}\\&= \frac{x(x-3)+2(x-3)}{(x-3)(x+3)}+\frac{x(x+6)-4(x+6)}{x(x-4)+3(x-4)}\\&= \frac{(x-3)(x+2)}{(x-3)(x+3)}+\frac{(x+6)(x-4)}{(x-4)(x+3)}\\&= \frac{(x+2)}{(x+3)}+\frac{(x+6)}{(x+3)}\\&=\frac{x+2+x+6}{x+3}\\&=\frac{2x+8}{x+3}\\&=2\frac{x+4}{x+3}\end{align}$

Question 2:

$\left[\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}\right]+\frac{4x}{x^4-1}$

Solution:

$\begin{align} \left[\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}\right]+\frac{4x}{x^4-1}&=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)-\frac{4x}{x^2+1}+\frac{4x}{x^4-1}\\&= \left(\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}\right)-\frac{4x}{x^2+1}+\frac{4x}{x^4-1}\\&= \frac{x^2+2x+1-x^2+2x-1}{(x^2-1)}-\frac{4x}{x^2+1}+\frac{4x}{x^4-1}\\&= \frac{4x}{x^2-1}-\frac{4x}{x^2+1}+\frac{4x}{x^4-1}\\&=4x\left(\frac{1}{x^2-1}-\frac{1}{x^2+1}\right)+\frac{4x}{x^4-1}\\&=4x\left(\frac{x^2+1-x^2+1}{x^4-1}\right)+\frac{4x}{x^4-1}\\&=4x\frac{2}{x^4-1}+\frac{4x}{x^4-1}\\&=4x\left(\frac{2+1}{x^4-1}\right)\\&=\frac{12x}{x^4-1}\end{align}$

Question 3:

$\frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3}-\frac{2}{x^2-6x+5}$

Solution:

$\begin{align} \frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3}-\frac{2}{x^2-6x+5}&=\frac{1}{x^2-3x-5x+15}+\frac{1}{x^2-x-3x+3}-\frac{2}{x^2-x-5x+5}\\&= \frac{1}{x(x-3)-5(x-3)}+\frac{1}{x(x-1)-3(x-1)}-\frac{2}{x(x-1)-5(x-1)}\\&=\frac{1}{(x-3)(x-5)}+\frac{1}{(x-1)(x-3)}-\frac{2}{(x-1)(x-5)}\\&= \frac{(x-1)+(x-5)-2(x-3)}{(x-1)(x-3)(x-5)}\\&=\frac{x-1+x-5-2x+6}{(x-1)(x-3)(x-5)}\\&=\frac{2x-2x-6+6}{(x-1)(x-3)(x-5)}\\&=0\end{align}$

Question 4:

$\frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3}-\frac{2}{x^2-6x+5}$

Solution:

$\begin{align} \frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3}-\frac{2}{x^2-6x+5}&=\frac{1}{x^2-3x-5x+15}+\frac{1}{x^2-x-3x+3}-\frac{2}{x^2-x-5x+5}\\&= \frac{1}{x(x-3)-5(x-3)}+\frac{1}{x(x-1)-3(x-1)}-\frac{2}{x(x-1)-5(x-1)}\\&=\frac{1}{(x-3)(x-5)}+\frac{1}{(x-1)(x-3)}-\frac{2}{(x-1)(x-5)}\\&= \frac{(x-1)+(x-5)-2(x-3)}{(x-1)(x-3)(x-5)}\\&=\frac{x-1+x-5-2x+6}{(x-1)(x-3)(x-5)}\\&=\frac{2x-2x-6+6}{(x-1)(x-3)(x-5)}\\&=0\end{align}$

Question 5:

$\frac{x+3}{2x^2+9x+9}+\frac{1}{2(2x-3)}-\frac{4x}{4x^2-9}$

Solution:

$\begin{align} \frac{x+3}{2x^2+9x+9}+\frac{1}{2(2x-3)}-\frac{4x}{4x^2-9}&=\frac{x+3}{2x^2+6x+3x+9}+\frac{1}{2(2x-3)}-\frac{4x}{(2x-3)(2x+3)}\\&= \frac{2(2x-3)+(2x+3)-8x}{2(2x+3)(2x-3)}\\&=\frac{4x-6+2x+3-8x}{2(2x+3)(2x-3)}\\&= \frac{-2x-3}{2(2x+3)(2x-3)}\\&=\frac{-(2x+3)}{2(2x+3)(2x-3)}\\&=\frac{-1}{2(2x-3)}\end{align}$

Question 6:

$A-\frac{1}{A}$ if $A=\frac{a+1}{a-1}$

Solution:

$\begin{align} A-\frac{1}{A}&=\frac{a+1}{a-1}-\frac{1}{\frac{a+1}{a-1}}\\&= \frac{a+1}{a-1}-\frac{a-1}{a+1}\\&=\frac{(a+1)^2-(a-1)^2}{(a+1)(a-1)}\\&= \frac{(a^2+2a+1)-(a^2-2a+1)}{(a+1)(a-1)}\\&=\frac{a^2+2a+1-a^2+2a-1}{(a+1)(a-1)}\\&=\frac{4a}{(a^2-1)}\end{align}$

Question 7:

$\left[\frac{x-1}{x-2}+\frac{2}{2-x}\right]-\left[\frac{x+1}{x+2}+\frac{4}{4-x^2}\right]$

Solution:

$\begin{align}\left[\frac{x-1}{x-2}+\frac{2}{2-x}\right]-\left[\frac{x+1}{x+2}+\frac{4}{4-x^2}\right]\\&=\left[\frac{x-1}{x-2}-\frac{2}{x-2}\right]-\left[\frac{x+1}{x+2}-\frac{4}{(x-2)(x+2)}\right]\\&=\frac{x-1-2}{x-2}-\frac{(x+1)(x-2)-4}{(x+2)(x-2)}\\&= \frac{x-3}{x-2}-\frac{x^2-x-2-4}{(x+2)(x-2)}\\&=\frac{x-3}{x-2}-\frac{x^2-x-6}{(x+2)(x-2)}\\&=\frac{x-3}{x-2}-\frac{x^2-3x+2x-6}{(x+2)(x-2)}\\&=\frac{x-3}{x-2}-\frac{x(x-3)+2(x-3)}{(x+2)(x-2)}\\&=\frac{x-3}{x-2}-\frac{(x-3)(x+2)}{(x+2)(x-2)}\\&=\frac{x-3}{x-2}-\frac{(x-3)}{(x-2)}\\&=0 \end{align}$

Question 8:

What rational expression should be subtracted from $\frac{2x^2+2x-7}{x^2+x-6}$ to get $\frac{x-1}{x-2}$
Let $A$ be the required expression.
$\begin{align} \frac{2x^2+2x-7}{x^2+x-6}-A&=\frac{x-1}{x-2}\end{align}$

Solution:

$\begin{align}A\\&=\frac{2x^2+2x-7}{x^2+x-6}-\frac{x-1}{x-2}\\&=\frac{2x^2+2x-7}{x^2+3x-2x-6}-\frac{x-1}{x-2}\\&=\frac{2x^2+2x-7}{x(x+3)-2(x+3)}-\frac{x-1}{x-2}\\&=\frac{2x^2+2x-7}{(x+3)(x-2)}-\frac{x-1}{x-2}\\&=\frac{2x^2+2x-7-(x-1)(x+3)}{(x+3)(x-2)}\\&=\frac{2x^2+2x-7-(x^2+2x-3)}{(x+3)(x-2)}\\&=\frac{2x^2+2x-7-x^2-2x+3}{(x+3)(x-2)}\\&=\frac{x^2-4}{(x+3)(x-2)}\\&=\frac{(x-2)(x+2)}{(x+3)(x-2)}\\&=\frac{(x+2)}{(x+3)}\end{align}$

perfrom the indicated operation and simplify the lowest form

Question 9:

$\frac{x^2+x-6}{x^2-x-6} \times \frac{x^2-4}{x^2-9}$

Solution:

$\begin{align} \frac{x^2+x-6}{x^2-x-6} \times \frac{x^2-4}{x^2-9}&=\frac{x^2+3x-2x-6}{x^2-3x+2x-6} \times \frac{(x-2)(x+2)}{(x-3)(x+3)}\\&=\frac{x(x+3)-2(x+3)}{x(x-3)+2(x-3)} \times \frac{(x-2)(x+2)}{(x-3)(x+3)}\\&=\frac{(x+3)(x-2)}{(x-3)(x+2)} \times \frac{(x-2)(x+2)}{(x-3)(x+3)}\\&=\frac{(x-2)(x-2)}{(x-3)(x-3)}\\&=\frac{(x-2)^2}{(x-3)^2}\end{align}$

Question 10:

$\frac{x^3-8}{x^2-4} \times \frac{x^2+6x+8}{x^2-2x+1}$

Solution:

$\begin{align} \frac{x^3-8}{x^2-4} \times \frac{x^2+6x+8}{x^2-2x+1}&=\frac{x^3-2^3}{x^2-2^2} \times \frac{x^2+4x+2x+8}{x^2-x-x+1}\\&=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} \times \frac{x(x+4)+2(x+4)}{x(x-1)-1(x-1)}\\&=\frac{(x^2+2x+4)}{(x+2)} \times \frac{(x+4)(x+2)}{(x-1)(x-1)}\\&=\frac{(x^2+2x+4)(x+4)}{(x-1)^2}\end{align}$

Question 11:

$\frac{x^4-8x}{2x^2+5x-3} \times \frac{2x-1}{x^2+2x+4} \times \frac{x+3}{x^2-2x}$

Solution:

$\begin{align} \frac{x^4-8x}{2x^2+5x-3} \times \frac{2x-1}{x^2+2x+4} \times \frac{x+3}{x^2-2x}&=\frac{x(x^3-8)}{2x^2+6x-x-3} \times \frac{2x-1}{x^2+2x+4} \times \frac{x+3}{x(x-2)}\\&=\frac{x(x-2)(x^2+2x+4)}{2x(x+3)-1(x+3)} \times \frac{2x-1}{x^2+2x+4} \times \frac{x+3}{x(x-2)}\\&=\frac{x(x-2)(x^2+2x+4)}{(x+3)(2x-1)} \times \frac{2x-1}{x^2+2x+4} \times \frac{x+3}{x(x-2)}\\&=1\end{align}$

Question 12:

$\frac{2y^2+7y-4}{3y^2-13y+4} \div \frac{4y^2-1}{6y^2+y-1}$

Solution:

$\begin{align} \frac{2y^2+7y-4}{3y^2-13y+4} \div \frac{4y^2-1}{6y^2+y-1}&=\frac{2y^2+8y-y-4}{3y^2-12y-y+4} \div \frac{(2y)^2-(1)^2}{6y^2+3y-2y-1}\\&=\frac{2y(y+4)-1(y+4)}{3y(y-4)-1(y-4)} \div \frac{(2y-1)(2y+1)}{3y(2y+1)-1(2y+1)}\\&=\frac{(y+4)(2y-1)}{(y-4)(3y-1)} \div \frac{(2y-1)(2y+1)}{(2y+1)(3y-1)}\\&=\frac{(y+4)(2y-1)}{(y-4)(3y-1)} \times \frac{(3y-1)}{(2y-1)}\\&=\frac{y+4}{y-4}\end{align}$

Question 13:

$\left[\frac{x^2+y^2}{x^2-y^2}- \frac{x^2-y^2}{x^2+y^2}\right] \div \left[\frac{x+y}{x-y}-\frac{x-y}{x+y}\right]$

Solution:

$\begin{align} \left[\frac{x^2+y^2}{x^2-y^2}- \frac{x^2-y^2}{x^2+y^2}\right] \div \left[\frac{x+y}{x-y}-\frac{x-y}{x+y}\right]&=\left[\frac{(x^2+y^2)^2-(x^2-y^2)^2}{(x^2-y^2)(x^2+y^2)}\right] \div \left[\frac{(x+y)^2-(x-y)^2}{(x-y)(x+y)}\right]\\&=\left[\frac{x^4+y^4+2x^2y^2-(x^4+y^4-2x^2y^2)}{(x^2-y^2)(x^2+y^2)}\right] \div \left[\frac{(x^2+y^2+2xy)-(x^2+y^2-2xy)}{(x-y)(x+y)}\right]\\&=\frac{x^4+y^4+2x^2y^2-x^4-y^4+2x^2y^2}{(x^2-y^2)(x^2+y^2)}\div \frac{x^2+y^2+2xy-x^2-y^2+2xy}{(x-y)(x+y)}\\&=\frac{4x^2y^2}{(x^2-y^2)(x^2+y^2)}\div \frac{4xy}{(x-y)(x+y)}\\&=\frac{4x^2y^2}{(x^2-y^2)(x^2+y^2)}\times \frac{(x-y)(x+y)}{4xy}\\&=\frac{xy}{x^2+y^2}\end{align}$

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