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

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Answer 1 · 2018-05-27T19:28:25

See explanation

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

Condition 1: $9 - 2 x \leq 3$ $9-2x<=3$
Condition 2: $2 x + 10 \leq 6 - x$ $2x+10<=6-x$

Consider condition 1

Add $2 x$ $2x$ 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

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

$3<=x$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider condition 2

Add $x$ to both sides

$3x+10<=6$

Subtract 10 from both sides

$3x<=-4$

Divide both sides by 3

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

$-4/3 >= x>=3$

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

Tony B

How do you solve 9-2x \le 3 or 3x+10 \le 6-x9−2x≤3or3x+10≤6−x9-2x \le 3 or 3x+10 \le 6-x?