1. ## questions

help on this one Solve: 4 – 6(x + 2)> or = to 5x + 36

2. Originally Posted by RashanHoliday
help on this one Solve: 4 – 6(x + 2)> or = to 5x + 36
Hello,

expand the bracket and collext all x on the RHS of the inequality, the constants on the LHS:

$4-6(x+2) \geq 5x+36$

$4-6x-12 \geq 5x+36$

$-44 \geq 11x$. Divide by 11.

$-4 \geq x$

EB

3. ## ?

hey,

how did you get x is more than or equal to 4? Shouldn't it be x is less than or equal to 4? Here's my reasoning.
1. Distribute the x
2.Add 12 to both sides
3. subtract 4 from both sides
4.subtract 5x from both sides
5. divide by negative 11 (you flip the sign when you divide by a negative)
6. You wind up getting x is less than or equal to negative 4.

4. Originally Posted by mscalculus
hey,

how did you get x is more than or equal to 4? Shouldn't it be x is less than or equal to 4? Here's my reasoning.

$4-6(x+2) \ge 5x+36$

1. Distribute the x
I presume you mean distribute the 6

$4-6x-12 \ge 5x+36$

2.Add 12 to both sides
$4-6x \ge 5x+48$

3. subtract 4 from both sides
$-6x \ge 5x+44$

4.subtract 5x from both sides
$-11x \ge 44$

5. divide by negative 11 (you flip the sign when you divide by a negative)
$x \le -4$

6. You wind up getting x is less than or equal to negative 4.
Yes, but that is what earboth has:

$x \le -4$ and $-4 \ge x$ are the same thing.

RonL

5. Originally Posted by mscalculus
hey,

how did you get x is more than or equal to 4? Shouldn't it be x is less than or equal to 4? Here's my reasoning.
1. Distribute the x
2.Add 12 to both sides
3. subtract 4 from both sides
4.subtract 5x from both sides
5. divide by negative 11 (you flip the sign when you divide by a negative)
6. You wind up getting x is less than or equal to negative 4.
Hello,

sorry for this late reply:

1. if you read my answer from right to left you'll get exactly your answer

2. I collected the x at the RHS to avoid the division by a negative number. As you have pointed out you have to change the $\geq$-sign into a $\leq$-sign, which I don't have to do.

EB