ANSWER TO QUESTION 1
[tex]\frac{\frac{y^2-4}{x^2-9}} {\frac{y-2}{x+3}}[/tex]
Let us change middle bar to division sign.
[tex]\frac{y^2-4}{x^2-9}\div \frac{y-2}{x+3}[/tex]
We now multiply with the reciprocal of the second fraction
[tex]\frac{y^2-4}{x^2-9}\times \frac{x+3}{y-2}[/tex]
We factor the first fraction using difference of two squares.
[tex]\frac{(y-2)(y+2)}{(x-3)(x+3)}\times \frac{x+3}{y-2}[/tex]
We cancel common factors.
[tex]\frac{(y+2)}{(x-3)}\times \frac{1}{1}[/tex]
This simplifies to
[tex]\frac{(y+2)}{(x-3)}[/tex]
ANSWER TO QUESTION 2
[tex]\frac{1+\frac{1}{x}} {\frac{2}{x+3}-\frac{1}{x+2}}[/tex]
We change the middle bar to the division sign
[tex](1+\frac{1}{x}) \div (\frac{2}{x+3}-\frac{1}{x+2})[/tex]
We collect LCM to obtain
[tex](\frac{x+1}{x})\div \frac{2(x+2)-1(x+3)}{(x+3)(x+2)}[/tex]
We expand and simplify to obtain,
[tex](\frac{x+1}{x})\div \frac{2x+4-x-3}{(x+3)(x+2)}[/tex]
[tex](\frac{x+1}{x})\div \frac{x+1}{(x+3)(x+2)}[/tex]
We now multiply with the reciprocal,
[tex](\frac{(x+1)}{x})\times \frac{(x+2)(x+3)}{(x+1)}[/tex]
We cancel out common factors to obtain;
[tex](\frac{1}{x})\times \frac{(x+2)(x+3)}{1}[/tex]
This simplifies to;
[tex]\frac{(x+2)(x+3)}{x}[/tex]
ANSWER TO QUESTION 3
[tex]\frac{\frac{a-b}{a+b}} {\frac{a+b}{a-b}}[/tex]
We rewrite the above expression to obtain;
[tex]\frac{a-b}{a+b}\div {\frac{a+b}{a-b}}[/tex]
We now multiply by the reciprocal,
[tex]\frac{a-b}{a+b}\times {\frac{a-b}{a+b}}[/tex]
We multiply out to get,
[tex]\frac{(a-b)^2}{(a+b)^2}[/tex]
ANSWER T0 QUESTION 4
To solve the equation,
[tex]\frac{m}{m+1} +\frac{5}{m-1} =1[/tex]
We multiply through by the LCM of [tex](m+1)(m-1)[/tex]
[tex](m+1)(m-1) \times \frac{m}{m+1} + (m+1)(m-1) \times \frac{5}{m-1} =(m+1)(m-1) \times 1[/tex]
This gives us,
[tex](m-1) \times m + (m+1) \times 5}=(m+1)(m-1)[/tex]
[tex]m^2-m+ 5m+5=m^2-1[/tex]
This simplifies to;
[tex]4m-5=-1[/tex]
[tex]4m=-1-5[/tex]
[tex]4m=-6[/tex]
[tex]\Rightarrow m=-\frac{6}{4}[/tex]
[tex]\Rightarrow m=-\frac{3}{2}[/tex]
ANSWER TO QUESTION 5
[tex]\frac{3}{5x}+ \frac{7}{2x}=1[/tex]
We multiply through with the LCM of [tex]10x[/tex]
[tex]10x \times \frac{3}{5x}+10x \times \frac{7}{2x}=10x \times1[/tex]
We simplify to get,
[tex]2 \times 3+5 \times 7=10x[/tex]
[tex]6+35=10x[/tex]
[tex]41=10x[/tex]
[tex]x=\frac{41}{10}[/tex]
[tex]x=4\frac{1}{10}[/tex]
Method 1: Simplifying the expression as it is.
[tex]\frac{\frac{3}{4}+\frac{1}{5}}{\frac{5}{8}+\frac{3}{10}}[/tex]
We find the LCM of the fractions in the numerator and those in the denominator separately.
[tex]\frac{\frac{5\times 3+ 4\times 1}{20}}{\frac{(5\times 5+3\times 4)}{40}}[/tex]
We simplify further to get,
[tex]\frac{\frac{15+ 4}{20}}{\frac{25+12}{40}}[/tex]
[tex]\frac{\frac{19}{20}}{\frac{37}{40}}[/tex]
With this method numerator divides(cancels) numerator and denominator divides (cancels) denominator
[tex]\frac{\frac{19}{1}}{\frac{37}{2}}[/tex]
Also, a denominator in the denominator multiplies a numerator in the numerator of the original fraction while a numerator in the denominator multiplies a denominator in the numerator of the original fraction.
That is;
[tex]\frac{19\times 2}{1\times 37}[/tex]
This simplifies to
[tex]\frac{38}{37}[/tex]
Method 2: Changing the middle bar to a normal division sign.
[tex](\frac{3}{4}+\frac{1}{5})\div (\frac{5}{8}+\frac{3}{10})[/tex]
We find the LCM of the fractions in the numerator and those in the denominator separately.
[tex](\frac{5\times 3+ 4\times 1}{20})\div (\frac{(5\times 5+3\times 4)}{40})[/tex]
We simplify further to get,
[tex](\frac{15+ 4}{20})\div (\frac{(25+12)}{40})[/tex]
[tex]\frac{19}{20}\div \frac{(37)}{40}[/tex]
We now multiply by the reciprocal,
[tex]\frac{19}{20}\times \frac{40}{37}[/tex]
[tex]\frac{19}{1}\times \frac{2}{37}[/tex]
[tex]\frac{38}{37}[/tex]
