Question

How do you solve `9-2x \le 3 or 3x+10 \le 6-x`?

Answer 1

See explanation

Explanation:

We have two conditions that are combined to define the limit of values that may be assigned to `x`

Condition 1: `9-2x<=3`
Condition 2: `2x+10<=6-x`

Consider condition 1

Add `2x` to both sides

`9color(white)("dd")ubrace(-2x+2x)<=3+2x`

`9 color(white)("ddd")+ 0color(white)("dddd")<=2x+3`

Subtract 3 from both sides

`9-3<=2x + 0`

Divide both sides by 2

`(9-3)/2<=2/2color(white)(.)x`

`3<=x`
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Consider condition 2

Add `x` to both sides

`3x+10<=6`

Subtract 10 from both sides

`3x<=-4`

Divide both sides by 3

`x<=-4/3`
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Combining these we have:

`-4/3 >= x>=3`

In other words `x` does not take on any values between and excluding `-4/3 and 3`