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+1 vote
solve 3x - 4 _<_ 2x + 11 _>_ -1

solve -2/3x > 8 or -2/3x < 4
asked Feb 19, 2013 in ALGEBRA 2 by futai Scholar

3 Answers

+2 votes
3x - 4 ≤ 2x + 11 ≥ -1

3x - 4 ≤ 2x + 11  or 2x + 11 ≥ -1

Take 3x - 4 ≤ 2x + 11

Add 4 to each side.

3x ≤ 2x + 15

Subtract 2x from each side.

x ≤ 15

And
answered Feb 20, 2013 by britally Apprentice
0 votes

1). 3x - 4 ≤ 2x + 11 ≥ -1

3x - 4 ≤ 2x + 11  or 2x + 11 ≥ -1

Take 3x - 4 ≤ 2x + 11

Add 4 to each side.

3x ≤ 2x + 15

Subtract 2x from each side.

x ≤ 15

And 2x + 11 ≥ -1

Subtract 11 from each side.

2x ≥ -12

Divide each side by 2.

x ≥ -6

Therefore x ≤ 15 or x ≥ -6

Graph the solution set on a number line.

answered Feb 20, 2013 by britally Apprentice
0 votes

2). -(2/3)x > 8 or -(2/3)x < 4

Take -2/3 > 8

Multiply each side by 3.

-2x > 24

Divide each side by 2.

- x > 12.

Multiply each side by negative one and flip the symbol.

x < - 12.

And -(2/3)x < 4

Multiply each side by 3.

-2x < 12

Divide each side by 2.

- x <  6.

Multiply each side by negative one and flip the symbol.

x > - 6.

Therefore x < - 12 or x > - 6.

Graph the solution set on a number line.

answered Feb 20, 2013 by britally Apprentice

2) -(2/3)x > 8 or -(2/3)x < 4

Your solution is correct.

But the solution on number line is


Observe the graph the open circle means -6 and -12 are not included in the solution set.

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