• May 21st 2010, 01:27 PM
OneidaFL
The post reads: If you replace the equal sign of an equation with the inequality sign, there is never a time when the same value will be a solution to both, namely due to the fact that a solution less than or greater than never be equal to the same solution.

Solve: 6x + 7 > 15 x = 1 Can this be solved with the factor given for X. I do not think so, How am I suppose to solve this can someone help me to understand the steps that are needed to solve this inequality
• May 21st 2010, 01:34 PM
skeeter
Quote:

Originally Posted by OneidaFL
The post reads: If you replace the equal sign of an equation with the inequality sign, there is never a time when the same value will be a solution to both, namely due to the fact that a solution less than or greater than never be equal to the same solution.

Solve: 6x + 7 > 15 x = 1 Can this be solved with the factor given for X. I do not think so, How am I suppose to solve this can someone help me to understand the steps that are needed to solve this inequality

I'm assuming the "equal sign" is a typo ... do you mean the inequality

$\displaystyle 6x+7 > 15x+1$ ?

if so ...

subtract $\displaystyle 6x$ and $\displaystyle 1$ from both sides ...

$\displaystyle 6 > 9x$

divide both sides by $\displaystyle 9$

$\displaystyle \frac{6}{9} > x$

reduce and rewrite ...

$\displaystyle x < \frac{2}{3}$
• May 21st 2010, 01:35 PM
rtblue
Sorry. I didn't not recognize that the = sign would be a +. I agree with skeeter.
• May 21st 2010, 01:40 PM
OneidaFL
unfortunately I think that the student was trying to give the value of x=1 for the following
• May 21st 2010, 01:42 PM
OneidaFL
Correction for inequality question
This is the actual problem: 6x + 7 > 15

I need to know how to figure this out. The student had given the value of x to be 1. That just doesn't make sense to me.

Thank You!
• May 21st 2010, 02:09 PM
pickslides
$\displaystyle 6x + 7 > 15$

$\displaystyle 6x + 7 {\color{red}-7}> 15{\color{red}-7}$

$\displaystyle 6x > 8$

$\displaystyle \frac{6x}{{\color{red}6}} > \frac{8}{{\color{red}6}}$

$\displaystyle x > \frac{8}{6} \approx 1.33$
• May 21st 2010, 02:40 PM
Keep
Quote:

Originally Posted by OneidaFL
unfortunately I think that the student was trying to give the value of x=1 for the following

The solution is the same as you would do for equation. If you had $\displaystyle 6x +7 = 15$ how would you solve?

I guess you would do this.

$\displaystyle 6x+7 = 15$
$\displaystyle 6x = 15-7$

$\displaystyle x = 8/6 = 4/3$

Now it is the same for inequality.

$\displaystyle 6x+7 > 15$ meanining $\displaystyle 6x > 15-7$, meaning $\displaystyle 6x > 8$, meaning $\displaystyle x > 4/3$
• May 21st 2010, 03:14 PM
skeeter
most probably, the longest thread I've seen over a linear inequality. (Wondering)
• May 21st 2010, 03:20 PM
mr fantastic
Quote:

Originally Posted by skeeter
most probably, the longest thread I've seen over a linear inequality. (Wondering)

And it ain't gettin' any longer.