(algebra 1) inequalities. question in the replies

Question

gelphielvr · Answer

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SoundwaveKun · Answer

Do you know how to start this?

SoundwaveKun · Answer

I guess not, I'll help you!

1. Distribute the 5 on the left side
2. Move all terms involving x to one side and constant terms to the other side. Start by adding 5x5x5x to both sides
3. Next, subtract 6 from both sides
4. Divide both sides by 3

Then you should have your answer! ^_^

gelphielvr · Answer

@soundwavekun wrote:

I guess not, I'll help you! 1. Distribute the 5 on the left side 2. Move all terms involving x to one side and constant terms to the other side. Start by adding 5x5x5x to both sides 3. Next, subtract 6 from both sides 4. Divide both sides by 3 Then you should have your answer! ^_^

sorry I was offline. do you know if in the end I have to flip the signs?

ThanosQC · Answer

1) 5(3-x) < -2x + 6
-Distribute the "5"
5*3 = 15, -x*5 = -5x
2) 15-5x < -2x + 6
-Isolate for x
15-6-5x+5x < -2x+5x+6-6
9 < 3x
3)
-Divide and solve for x
9/3 < 3x/3
3 < x, so when properly formatted, x > 3

gelphielvr · Answer

@thanosqc wrote:

1) 5(3-x) < -2x + 6 -Distribute the "5" 5*3 = 15, -x*5 = -5x 2) 15-5x < -2x + 6 -Isolate for x 15-6-5x+5x < -2x+5x+6-6 9 < 3x 3) -Divide and solve for x 9/3 < 3x/3 3 < x, so when properly formatted, x > 3

thank you :)

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