Question

Solve for the value of x in the given rational inequalities.​

Answer 1

see explanation

Explanation:

1

1 +
`(x-5)/(2)` -
`(x+3)/(5)` ≤ 0

multiply through by 10 ( the LCM of 2 and 5 ) to clear the fractions

10 + 5(x - 5) - 2(x + 3) ≤ 0 ← distribute parenthesis on left side and simplify

10 + 5x - 25 - 2x - 6 ≤ 0

3x - 21 ≤ 0 ( add 21 to both sides )

3x ≤ 21 ( divide both sides by 3 )

x ≤ 7

2

`(6x)/(5)` -
`(2)/(3)` ≥ 4

multiply through by 15 ( the LCM of 5 and 3 ) to clear the fractions

18x - 10 ≥ 60 ( add 10 to both sides )

18x ≥ 70 ( divide both sides by 18 )

x ≥
`(70)/(18)` , that is

x ≥
`(35)/(9)`

3

`(x)/(10)` -
`(2)/(8)` ≥ 1

multiply through by 40 ( the LCM of 10 and 8 ) to clear the fractions

4x - 10 ≥ 40 ( add 10 to both sides )

4x ≥ 50 ( divide both sides by 4 )

x ≥
`(50)/(4)` , that is

x ≥
`(25)/(2)`

Answer 2

`\textsf{1.} \quad x \leq 7`

`\textsf{2.} \quad x \geq (35)/(9)`

`\textsf{3.} \quad x \geq (25)/(2)`

Explanation:

Question 1

Given inequality:

`1+(x-5)/(2)-(x+3)/(5) \leq 0`

`\textsf{Add \; $(x+3)/(5)$ \; to both sides}:`

`\implies 1+(x-5)/(2)-(x+3)/(5) +(x+3)/(5)\leq 0+(x+3)/(5)`

`\implies 1+(x-5)/(2)\leq (x+3)/(5)`

`\textsf{Rewrite $1$ as $(2)/(2)$}:`

`\implies (2)/(2)+(x-5)/(2)\leq (x+3)/(5)`

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

`\implies (2+x-5)/(2)\leq (x+3)/(5)`

`\implies (x-3)/(2)\leq (x+3)/(5)`

Cross multiply:

`\implies 5(x-3)\leq 2(x+3)`

`\implies 5x-15\leq 2x+6`

Subtract 2x from both sides:

`\implies 5x-15-2x\leq 2x+6-2x`

`\implies 3x-15\leq 6`

Add 15 to both sides:

`\implies 3x-15+15\leq 6+15`

`\implies 3x\leq 21`

Divide both sides by 3:

`\implies (3x)/(3)\leq (21)/(3)`

`\implies x \leq 7`

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Question 2

Given inequality:

`(6x)/(5)-(2)/(3) \geq 4`

`\textsf{Add \; $(2)/(3)$ \; to both sides}:`

`\implies (6x)/(5)-(2)/(3) +(2)/(3)\geq 4+(2)/(3)`

`\implies (6x)/(5)\geq (12)/(3)+(2)/(3)`

`\implies (6x)/(5)\geq (12+2)/(3)`

`\implies (6x)/(5)\geq (14)/(3)`

Cross multiply:

`\implies 3(6x) \geq 14(5)`

`\implies 18x \geq 70`

Divide both sides by 18:

`\implies (18x)/(18) \geq (70)/(18)`

`\implies x \geq (70)/(18)`

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

`\implies x \geq (70 / 2)/(18 / 2)`

`\implies x \geq (35)/(9)`

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Question 3

Given inequality:

`(x)/(10)-(2)/(8) \geq 1`

`\textsf{Add \; $(2)/(8)$ \; to both sides}:`

`\implies (x)/(10)-(2)/(8) +(2)/(8)\geq 1+(2)/(8)`

`\implies (x)/(10)\geq (8)/(8)+(2)/(8)`

`\implies (x)/(10)\geq (8+2)/(8)`

`\implies (x)/(10)\geq (10)/(8)`

Cross multiply:

`\implies 8(x) \geq 10(10)`

`\implies 8x \geq 100`

Divide both sides by 8:

`\implies (8x)/(8) \geq (100)/(8)`

`\implies x \geq (100)/(8)`

Reduce the fraction by dividing the numerator and the denominator by 4:

`\implies x \geq (100 / 4)/(8 / 4)`

`\implies x \geq (25)/(2)`