Question

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Answer 1

ANSWER TO QUESTION 1


`\frac{(y^2-4)/(x^2-9)} {(y-2)/(x+3)}`

Let us change middle bar to division sign.


`(y^2-4)/(x^2-9)/ (y-2)/(x+3)`

We now multiply with the reciprocal of the second fraction


`(y^2-4)/(x^2-9)* (x+3)/(y-2)`

We factor the first fraction using difference of two squares.


`((y-2)(y+2))/((x-3)(x+3))* (x+3)/(y-2)`

We cancel common factors.


`((y+2))/((x-3))* (1)/(1)`

This simplifies to


`((y+2))/((x-3))`

ANSWER TO QUESTION 2


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

We change the middle bar to the division sign


`(1+(1)/(x)) / ((2)/(x+3)-(1)/(x+2))`

We collect LCM to obtain


`((x+1)/(x))/ (2(x+2)-1(x+3))/((x+3)(x+2))`

We expand and simplify to obtain,


`((x+1)/(x))/ (2x+4-x-3)/((x+3)(x+2))`


`((x+1)/(x))/ (x+1)/((x+3)(x+2))`

We now multiply with the reciprocal,


`(((x+1))/(x))* ((x+2)(x+3))/((x+1))`

We cancel out common factors to obtain;


`((1)/(x))* ((x+2)(x+3))/(1)`

This simplifies to;


`((x+2)(x+3))/(x)`

ANSWER TO QUESTION 3


`\frac{(a-b)/(a+b)} {(a+b)/(a-b)}`

We rewrite the above expression to obtain;


`(a-b)/(a+b)/ {(a+b)/(a-b)}`

We now multiply by the reciprocal,


`(a-b)/(a+b)* {(a-b)/(a+b)}`

We multiply out to get,


`((a-b)^2)/((a+b)^2)`

ANSWER T0 QUESTION 4

To solve the equation,


`(m)/(m+1) +(5)/(m-1) =1`

We multiply through by the LCM of
`(m+1)(m-1)`


`(m+1)(m-1) * (m)/(m+1) + (m+1)(m-1) * (5)/(m-1) =(m+1)(m-1) * 1`

This gives us,


`(m-1) * m + (m+1) * 5}=(m+1)(m-1)`


`m^2-m+ 5m+5=m^2-1`

This simplifies to;


`4m-5=-1`


`4m=-1-5`


`4m=-6`


`\Rightarrow m=-(6)/(4)`


`\Rightarrow m=-(3)/(2)`

ANSWER TO QUESTION 5


`(3)/(5x)+ (7)/(2x)=1`

We multiply through with the LCM of
`10x`


`10x * (3)/(5x)+10x * (7)/(2x)=10x *1`

We simplify to get,


`2 * 3+5 * 7=10x`


`6+35=10x`


`41=10x`


`x=(41)/(10)`


`x=4(1)/(10)`

Method 1: Simplifying the expression as it is.


`((3)/(4)+(1)/(5))/((5)/(8)+(3)/(10))`

We find the LCM of the fractions in the numerator and those in the denominator separately.


`((5* 3+ 4* 1)/(20))/(((5* 5+3* 4))/(40))`

We simplify further to get,


`((15+ 4)/(20))/((25+12)/(40))`


`((19)/(20))/((37)/(40))`

With this method numerator divides(cancels) numerator and denominator divides (cancels) denominator


`((19)/(1))/((37)/(2))`

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;


`(19* 2)/(1* 37)`

This simplifies to


`(38)/(37)`

Method 2: Changing the middle bar to a normal division sign.


`((3)/(4)+(1)/(5))/ ((5)/(8)+(3)/(10))`

We find the LCM of the fractions in the numerator and those in the denominator separately.


`((5* 3+ 4* 1)/(20))/ (((5* 5+3* 4))/(40))`

We simplify further to get,


`((15+ 4)/(20))/ (((25+12))/(40))`


`(19)/(20)/ ((37))/(40)`

We now multiply by the reciprocal,


`(19)/(20)* (40)/(37)`


`(19)/(1)* (2)/(37)`


`(38)/(37)`