Questions on Algebra: Absolute value answered by real tutors!

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 = 6

To remove absolute value bars, you must make and solve two 
separate equations.

 = +6    and }} = -6

Multiply both sides of both equations by 10 to clear of fractions:

 8x-3 = 60         and    8x-3 = -60

Add 3 to both sides of both equations:

   8x = 63         and      8x = -57
    x =   and      x = 

Two solutions:   and  

Edwin

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|8x-3|=6
-----
10
---
If the fraction [(8x-3)/10] is in the absolute value sign
you get two equations:
(8x-3)/10 = 6 and (8x-3)/10 = -6
8x-3 = 60 or (8x-3)= -60
8x = 63 or 8x = -57
x = 63/8 or x = -57/8
------------------------------
If the 10 is not in the absolute value sign you get
|8x-3| = 60
Then you get:
8x-3 = 60 or 8x-3 = -60
8x = 63 or 8x = -57
8x = 63/8 or x = -57/8
================
Either way you get the same result.
===============
Cheers,
Stan H.
===============

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Hello,
First let's set this up as two equations:
3x+5=1 3x+5= -1
Now solve:
x=-4/3 x=-2
Make sense?
RJ
www.math-unlock.com

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what two numbers have an absolute of 10?
----
Ans: +10 and -10
=====================
Cheers,
Stan H.

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p^2+2p-35
factoring you get:
(p+7)(p-5)

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A better and more compact way is to write

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the basic definition of absolute value is:
if |x| = y, then:
x = y when x is positive.
x = -y when x is negative.
that basic definition applies also to:
if |x| = |y|, then:
x = |y| when x is positive.
x = -|y| when x is negative.
what this leads to is:
if you commute these equations, you get:
|y| = x when x is positive.
-|y| = x when x is negative.
the equation of -|y| = x when x is negative is equivalent to the equation of:
|y| = -x when x is negative.
if we look at the case when x is positive, then we get 2 more equations from:
|y| = x
we get:
y = x when y is positive.
y = -x when y is negative.
if we look at the case when x is negative, then we get 2 more equations from:
|y| = -x
we get:
y = -x when y is positive.
y = -(-x) when y is negative, which is equivalent to:
y = x when y is negative.
we wind up with 4 cases.
when x is positive, we get:
y = x when y is positive.
y = -x when y is negative.
when x is negative, we get:
y = -x when y is positive.
y = x when y is negative.
you will notice that we have really have only 2 equations.
if we are solving for y, they are:
y = x
y = -x
if we are solving for x, they are:
x = y
x = -y
note that we are talking about the expression within the absolute value signs, not the value of x or y. x or y can be part of the expression or the while expression. in our examples, they represented the whole expression.
let's see how this process works in solving one of your problems.
we'll take the first problem.
the first problem is:
|4-x| = |2x+1|
this leads to 2 equations:
(4-x) = (2x + 1)
and:
(4-x) = - (2x + 1)
the first equation of:
(4-x) = (2x + 1) is solved as follows:
remove parentheses to get:
4-x = 2x+1
subtract 2x from both sides of the equation and subtract 4 from both sides of the equation to get:
-x-2x = 1-4
simplify to get:
-3x = -3
divide both sides of the equation by -3 to get:
x=1.
the second equation of:
(4-x) = -(2x+1) is solved as follows:
remove parentheses to get:
4-x = -2x-1
add 2x to both sides of the equation and subtract 4 from both sides of the equation to get:
x = -5
you have 2 values for x that should satisfy the equation.
they are:
x = 1
your original equation is:
|4-x|=|2x+1|
when x = 1, this equation becomes:
|3| = |3| which is true.
when x = -5, this equation becomes:
|9| = |9| which is also true.
we can also graph your equations to see if these answers are accurate.
your equation is:
|4-x|=|2x+1|
subtract |2x+1| from both sides of the equation and set the equation equal to y to get:
y = |4-x| - |2x+1|
the graph of this equation looks like this:

the 2 equations are equal when the value of y is equal to 0.
that occurs when x = -5 and x = 1
this confirms those solutions are good.
we can apply the same logic to your other 2 equations and, if the method works on them as well, we should get the correct solutions that we can confirm graphically as well.
your second equation is:
|x+2|=|2-x|
this leads to:
(x+2) = (2-x)
and:
(x+2) = -(2-x)
solving for x in the first equation gets:
x = 0
solving for x in the second equation gets:
0 = 4
it appears the only solution for this equation is x = 0.
we'll graph to confirm.
we graph the equation of:
y = |x+2| - |2-x| to get:

the graph confirms that the only solution to this equation is x = 0.
your third equation is:
|x/5-21/10|=|x/2|
this leads to:
(x/5-21/10) = (x/2)
(x/5-21/10)= -(x/2)
we solve for x in the first equation to get:
x = -7
we solve for x in the second equation to get:
x = 3
we graph the equation of:
y = |x/5-21/10| - |x/2| to get:

the graph confirms those solutions are good.
this is the first time i've used this shortcut method.
it appears to work.
what this says is:
if you have an equation of:
|x| = y, then you solve it as:
x = y
x = -y
if you have an equation of:
|x| = |y|, then you solve it as:
x = y
x = -y
you use the same method regardless.
i won't say this method is bullet proof in all situations until i can conclusively prove that, but it looks like it will fit the bill and i'll use it in future problems until i see that it fails.
so far it hasn't.

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I just finished doing this one. See Question 557011




John

My calculator said it, I believe it, that settles it

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Either



or



Just solve for in both cases. Hint: The first step in each case should be to multiply both sides by 2. When you are done, remember to check both answers.

John

My calculator said it, I believe it, that settles it

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What is |y+5|>0
It's an inequality in y.