Question

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WITH SOLUTION.​

Answer 1

`1) \:\:\: (14x)/(y)`

`2) \:\:\:(16x^2)/(9y^2)`

`3)\:\:\:10`

`4)\:\:\:(x-2)/(x+4)`

Explanation:

I will provide detailed explanation for the first problem and quicker steps to solve the others. You should be able to figure them out

Dividing a number(integer or fraction) by a fraction is equivalent to multiplying that number by the reciprocal of the divisor fraction

So if you have an expression

`(a)/(b) / (c)/(d)\\\\\implies (a)/(b) * (d)/(c)`

Question 1

`(14x^5)/(5y^2)\:/ (2x^4)/(10y)\\\\= (14x^5)/(5y^2) * (10y)/(2x^4)\\\\`

`=(14x^5\cdot \:10y)/(10y^2x^4)`

10 in the numerator and 10 in the denominator cancel:

`=(14x^5y)/(y^2x^4)`

`(x^5)/(x^4) = x^(5-4) = x \;\;\;\;\;\;\text{Since }(x^a)/(x^b)=x^(a-b)}`

Likewise

`(y)/(y^2) = y^(1-2) = y ^(-1) = (1)/(y)`

Therefore

`=(14x^5\cdot \:10y)/(10y^2x^4) = 14 (x)/(y) = (14x)/(y)`
Answer: Question 1

Question 2

`\left((x)/(y)\right)\left((4y)/(3x)\right)\:/ (6y^2)/(8x^2)`

`(x)/(y) \cdot (4y)/(3x) = (4)/(3)`

`(6y^2)/(8x^2) = (4y^2)/(3x^2)`'

`\left((x)/(y)\right)\left((4y)/(3x)\right)\:/ (6y^2)/(8x^2) \\\\= (4)/(3) * (3x^2)/(4y^2)\\\\= (16x^2)/(9y^2)`

Question 3

`(\left(5\left(y+6\right)\right))/(2\left(xy-3\right))\:/ (\left(y\:+\:6\right))/(4\left(xy\:-\:3\right))\\\\\\(5(y + 6))/(2(xy-3)) * (4(xy-3))/(y+ 6)\\\\`

The y + 6 terms cancel out from numerator and denominator as do the xy - 3terms

`= (5)/(2) * (4)/(1)\\\\= (20)/(2) \\\\= 10`

Question 4

`(x^2-4)/(x^2+5x+4)\:/ (x+2)/(x+1)\\\\=(x^2-4)/(x^2+5x+4)\:* (x+1)/(x+2)`

`x^2-4 = (x + 2)(x-2) \;\;\;\; \text{Since $a^2-b^2 = (a +b)(a-b)$}`

`x^2 + 5x + 4\;\;\; \text{can be factored as: }\\\\(x + 1)(x + 4)`

Therefore

`(x^2-4)/(x^2+5x+4)\:/ (x+2)/(x+1)\\\\\\=(x^2-4)/(x^2+5x+4)\:* (x+1)/(x+2)\\\\\\= ((x+2)(x-2))/((x+1)(x+4)) * (x+1)/(x+2)\\\\\\\\`

The x+2 terms cancel out as do the x + 1 terms giving the result

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

Answer 2

`\textsf{1. \quad $(14x)/(y)$}`

`\textsf{2. \quad $(16x^3)/(9y^2)$}`

`\textsf{3. \quad $10$}`

`\textsf{4. \quad $(x-2)/(x+4)$}`

Explanation:

When dividing two fractions, multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of a fraction is its inverse (swap the numerator and denominator).

Question 1

`\boxed{\begin{minipage}{3 cm}\underline{Exponent Rules}\\\\$(a^b)/(a^c)=a^(b-c)$\\\\$a^(-n)=(1)/(a^n)$\\\end{minipage}}`

`\begin{aligned}(14x^5)/(5y^2) / (2x^4)/(10y) & = (14x^5)/(5y^2) * (10y)/(2x^4)\\\\& = (14x^5 * 10y)/(5y^2* 2x^4)\\\\& = (140x^5y)/(10x^4y^2)\\\\& = (14x^5y)/(x^4y^2)\\\\& = 14x^(5-4)y^(1-2)\\\\& = 14x^1y^(-1)\\\\& = (14x)/(y)\end{aligned}`

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Question 2

`\begin{aligned}\left((x)/(y) \right)\left((4y)/(3x) \right) / (6y^2)/(8x^3)&=(4xy)/(3xy) / (6y^2)/(8x^3)\\\\&=(4)/(3) / (6y^2)/(8x^3)\\\\&=(4)/(3) * (8x^3)/(6y^2)\\\\&=(4 * 8x^3)/(3 * 6y^2)\\\\&=(32x^3)/(18y^2)\\\\&=(16x^3)/(9y^2)\end{aligned}`

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Question 3

`\begin{aligned}\left[(5(y+6))/(2(xy-3)) \right] / \left((y+6)/(4(xy-3))\right) & = \left[(5(y+6))/(2(xy-3)) \right] * \left((4(xy-3))/(y+6)\right)\\\\& = (5(y+6) * 4(xy-3))/(2(xy-3) * (y+6))\\\\& = (20(y+6)(xy-3))/(2(xy-3)(y+6))\\\\& = (20)/(2)\\\\&=10\end{aligned}`

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Question 4

Factor the numerator and denominator of the first fraction.

Apply the Difference of Two Squares to the numerator:

`\implies x^2-4=x^2-2^2=(x-2)(x+2)`

Factor the denominator:

`\begin{aligned}\implies x^2+5x+4 & =x^2+x+4x+4\\ & =x(x+1)+4(x+1)\\ & =(x+1)(x+4)\end{aligned}`

Therefore:

`\begin{aligned}\left((x^2-4)/(x^2+5x+4)\right) / \left( (x+2)/(x+1) \right) & = \left((x^2-4)/(x^2+5x+4)\right) * \left( (x+1)/(x+2) \right)\\\\& = \left( ((x-2)(x+2))/((x+1)(x+4)) \right) * \left( (x+1)/(x+2) \right)\\\\& = ((x-2)(x+2)(x+1))/((x+1)(x+4)(x+2))\\\\& = (x-2)/(x+4)\\\\\end{aligned}`