Question

Hi can someone help me here​

Answer 1

see explanation

Explanation:

`(x+2)/(3)` -
`(2x-4)/(2)` = 0

multiply through by 6 ( the LCM of 3 and 2 ) to clear the fractions

2(x + 2) - 3(2x - 4) = 0 ← distribute parenthesis on left side and simplify

2x + 4 - 6x + 12 = 0

- 4x + 16 = 0 ( subtract 16 from both sides )

- 4x = - 16 ( divide both sides by - 4 )

x = 4

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`(5x)/(2)` +
`(x-2)/(8)` = 2 ( multiply through by 8 to clear the fractions )

4(5x) + x - 2 = 16

20x + x - 2 = 16

21x - 2 = 16 ( add 2 to both sides )

21x = 18 ( divide both sides by 21 )

x =
`(18)/(21)` =
`(6)/(7)`

-----------------------------------------------------------

`(x-5)/(4)` -
`(3x-1)/(5)` = -
`(3)/(2)`

multiply through by 20 ( the LCM of 4, 5, 2 ) to clear the fractions

5(x - 5) - 4(3x - 1) = - 30 ← distribute parenthesis on left side and simplify

5x - 25 - 12x + 4 = - 30

- 7x - 21 = - 30 ( add 21 to both sides )

- 7x = - 9 ( divide both sides by - 7 )

x =
`(9)/(7)`

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`(1)/(x-1)` +
`(1)/(x+1)` =
`(4x+2)/(x^2-1)` ← x² - 1 is a difference of squares , then

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

multiply through by (x - 1)(x + 1) to clear the fractions

x + 1 + x - 1 = 4x + 2

2x = 4x + 2 ( subtract 4x from both sides )

- 2x = 2 ( divide both sides by - 2 )

x = - 1

If we substitute this value back into the original equation, we find

`(1)/(-1-1)` +
`(1)/(-1+1)` =
`(4(-1)+2)/((-1)^2-1)`

`(1)/(-2)` +
`(1)/(0)` =
`(-2)/(1-1)`

`(1)/(-2)` +
`(1)/(0)` =
`(-2)/(0)`

division by zero is undefined

thus x = - 1 is an extraneous solution

the equation has no solution

Answer 2

`\textsf{1.} \quad x=4`

`\textsf{2.} \quad x=(6)/(7)`

`\textsf{3.} \quad x=(9)/(7)`

`\textsf{4.} \quad \textsf{No solution}`

Explanation:

Question 1

Given equation:

`(x+2)/(3)-(2x-4)/(2)=0`

`\textsf{Add \; $(2x-4)/(2)$ \; to both sides of the equation}:`

`\implies (x+2)/(3)-(2x-4)/(2)+(2x-4)/(2)=0+(2x-4)/(2)`

`\implies (x+2)/(3)=(2x-4)/(2)`

Cross multiply:

`\implies 2(x+2)=3(2x-4)`

`\implies 2x+4=6x-12`

Subtract 2x from both sides:

`\implies 2x+4-2x=6x-12-2x`

`\implies 4=4x-12`

Add 12 to both sides:

`\implies 4+12=4x-12+12`

`\implies 16=4x`

`\implies 4x=16`

Divide both sides by 4:

`\implies (4x)/(4)=(16)/(4)`

`\implies x=4`

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Question 2

Given equation:

`(5x)/(2)+(x-2)/(8)=2`

Multiply the numerator and denominator of the first fraction by 4:

`\implies (5x \cdot 4)/(2\cdot 4)+(x-2)/(8)=2`

`\implies (20x)/(8)+(x-2)/(8)=2`

`\textsf{Apply the fraction rule} \quad (a)/(c)+(b)/(c)=(a+b)/(c):`

`\implies (20x+x-2)/(8)=2`

`\implies (21x-2)/(8)=2`

Multiply both sides by 8:

`\implies ((21x-2) \cdot 8)/(8)=2 \cdot 8`

`\implies 21x-2=16`

Add 2 to both sides:

`\implies 21x-2+2=16+2`

`\implies 21x=18`

Divide both sides by 21:

`\implies (21x)/(21)=(18)/(21)`

`\implies x=(18)/(21)`

Reduce the fraction by dividing the numerator and denominator by 3:

`\implies x=(18 / 3)/(21 / 3)`

`\implies x=(6)/(7)`

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Question 3

Given equation:

`(x-5)/(4)-(3x-1)/(5)=-(3)/(2)`

Multiply the numerator and denominator of the first fraction by 5, and the numerator and denominator of the second fraction by 4:

`\implies ((x-5) \cdot 5)/(4\cdot 5)-((3x-1)\cdot 4)/(5\cdot 4)=-(3)/(2)`

`\implies (5x-25)/(20)-(12x-4)/(20)=-(3)/(2)`

`\textsf{Apply the fraction rule} \quad (a)/(c)-(b)/(c)=(a-b)/(c):`

`\implies (5x-25-(12x-4))/(20)=-(3)/(2)`

`\implies (5x-25-12x+4)/(20)=-(3)/(2)`

`\implies (-7x-21)/(20)=-(3)/(2)`

Cross multiply:

`\implies 2(-7x-21)=-3(20)`

`\implies -14x-42=-60`

Add 42 to both sides:

`\implies -14x-42+42=-60+42`

`\implies -14x=-18`

Divide both sides by -14:

`\implies (-14x)/(-14)=(-18)/(-14)`

`\implies x=(18)/(14)`

Reduce the fraction by dividing the numerator and denominator by 2:

`\implies x=(18 / 2)/(14 / 2)`

`\implies x=(9)/(7)`

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Question 4

Given equation:

`(1)/(x-1)+(1)/(x+1)=(4x+2)/(x^2-1)`

Factor the denominator of the fraction on the right side of the equation:

`\implies (1)/(x-1)+(1)/(x+1)=(4x+2)/((x+1)(x-1))`

Multiply the numerator and denominator of the first fraction by (x+1), and the numerator and denominator of the second fraction by (x-1):

`\implies (x+1)/((x+1)(x-1))+(x-1)/((x+1)(x-1))=(4x+2)/((x+1)(x-1))`

`\textsf{Apply the fraction rule} \quad (a)/(c)+(b)/(c)=(a+b)/(c):`

`\implies (x+1+x-1)/((x+1)(x-1))=(4x+2)/((x+1)(x-1))`

`\implies (2x)/((x+1)(x-1))=(4x+2)/((x+1)(x-1))`

Multiply both sides by (x+1)(x-1):

`\implies (2x(x+1)(x-1))/((x+1)(x-1))=((4x+2)((x+1)(x-1))/((x+1)(x-1))`

`\implies 2x=4x+2`

Subtract 4x from both sides:

`\implies 2x-4x=4x+2-4x`

`\implies -2x=2`

Divide both sides by -2:

`\implies (-2x)/(-2)=(2)/(-2)`

`\implies x=-1`

However, if we substitute x = -1 into the original equation we get:

`\implies (1)/((-1)-1)+(1)/((-1)+1)=(4(-1)+2)/((-1)^2-1)`

`\implies -(1)/(2)+(1)/(0)=(-2)/(0)`

A number divided by zero is undefined. Therefore, there is no solution for this equation.