document.write( "Question 1092649: I need a little help on my homework. I tried them all and I know some of them are right i just have a question on a few because if i can understand those problems the ones similar I can do. So we are learning about how to solve inequalities and graphing them on a number line. I started this one but need a little help finishing it.
\n" ); document.write( "(//= absolute value bars) \r
\n" ); document.write( "\n" ); document.write( "/x/>x-1, and i need 2 inequalities. i have the first which is x>x-1 but i don't have the other one and is not quite such how to solve either one and my other one is
\n" ); document.write( "7-/2x/>5, I have both the inequalities and answers (x>-6; x<6) i just don't know how to put on the third number line because my teacher said something about the union?
\n" ); document.write( "Can you help? That would be appreciative? \n" ); document.write( "
Algebra.Com's Answer #707281 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!

For the first one, yes, you need two inequalities. But one of them is not just
\n" ); document.write( "\"x%3Ex-1\"

\n" ); document.write( "That inequality without absolute value symbols is equivalent to the original inequality only if |x| = x, which is only true if x is 0 or positive.

\n" ); document.write( "So you can't just write two inequalities; you need to state when each of the inequalities is true.

\n" ); document.write( "If x is 0 or positive, then |x| = x. If x is negative, then |x| = -x. So the correct statement of your two inequalities needs to be
\n" ); document.write( "x > x-1, if x >= 0;
\n" ); document.write( "-x > x-1, if x < 0

\n" ); document.write( "Solving the first inequality gives 0 < 1, which is always true. That means that, in the interval x >=0 where this inequality is applicable, every number is a solution. So the first part of your solution set in interval notation is [0,infinity).

\n" ); document.write( "Solving the second inequality leads to 2x < 1, or x < 1/2. This means that any value of x less than 1/2 in the interval where this inequality is applicable is a solution to the original inequality. But every number in this interval (x < 0) is less than 1/2. So every number in the interval where this inequality is applicable is also a solution to the original inequality. So the second part of your solution set in interval notation is (-infinity,0).

\n" ); document.write( "Then the complete solution to the original inequality is the union of (-infinity,0) and [0,infinity} -- and in this case that union is all real numbers.

\n" ); document.write( "For your second problem, I don't know what two inequalities you have; but the answers are not (x>-6; x<6) -- at least for the original inequality as you show it. Since the variable only appears once in the inequality, there is no need for two equivalent inequalities. The usual solution to you second example would be
\n" ); document.write( "\"7-abs%282x%29%3E5\"
\n" ); document.write( "\"2+%3E+abs%282x%29\"
\n" ); document.write( "\"abs%28x%29+%3C+1\"

\n" ); document.write( "You COULD use two inequalities to separate this into two cases, where x is either negative or positive. However, when the inequality is in this form, it is easiest just to jump directly to the solution,
\n" ); document.write( "\"-1+%3C+x+%3C+1\" or, in interval notation, (-1,1). \n" ); document.write( "
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