DOUBLE CHECK ANSWERS

[MARIE]

Just looking for a second opinion to see if I did these problems correctly.

1. 2x-(7+x) < or equal to x
2x-7-x < or equal to x
x-x-7 < or equal to x-x
-7 < or equal to x
(x, -7)

2. -2x(x+6) > x+4
-2x-12+12 > x + 4 + 12
-2x > x + 16
-2x - x > x - x + 16
-2x - x > 16
-3x > 16
-3/-3x > -16/3
(x, -16/3)

3. 2(4-3/5q) < 5
8 -30q < 5
8 - 8 - 30q < 5 - 8
-30q < -3
-30/-30q < -3/-30
q < 3/30
q < 1/10
(q, 1/10)

4. t/3 -t/2 > 3/4
t > 3/4
(t, 3/4)

5. l2x-1l < 3
l2x-1+1l < 3 + 1
l2xl < 4
2x/2 < 4/2
x < 2
(x, 2)

I think I have the hang of this, but a second opinion would be greatly appreciated. Thanks!!

 

[tkhunny]

Two things jump out:

1) I have no idea what your final notation means. (x,-7)?? Normally, there would be no 'x' in there, just numbers or infinity. It should be the Real Numbers that satisfy the original statement.

2) You appear to have ignored the absolute value entirely. That's no good. It isn't just funny looking parentheses.

|5| = 5 -- That's great, as long as you KNOW what is inside is POSITIVE.
|-5| = 5 -- Well, that's quite a bit different, isn't it?

 

[soroban]

Hello, Marie!

Here's #1 . . .

\(\displaystyle 1)\;\;2x\,-\,(7\,+\,x)\:\leq\:x\)

We have: \(\displaystyle \:2x\,-\,7\,-\,x\:\leq\:x\)

. . . . . . . . . . . .\(\displaystyle x\,-\,7\:\leq\:x\)

Subtract \(\displaystyle x\) from both sides:

. . . . . . . . . . . . .\(\displaystyle \:-7 \:\leq \:0\)

This is true for all values of \(\displaystyle x.\)

Solution set: \(\displaystyle \-\infty,\,\infty)\)