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# Algebra/absolute value

Question
please provide the solutions to the following. Please show me step by step.

1) Solve /-2x-28/=/8x-8/

2) Solve /2x-15/=8x-75/

3) Solve /-10x+75/=/-8x+87/

The / stands forabsolut value.

Thank you.

All we've got to do is find the range where each of the functions inside the absolute values is positive and negative.

1) -2x-28 = 0; add 28 to both sides, giving -2x = 28; divide by -2, giving x = -14.  This is the break point.  For x=-15, -2x-28 = 30-28 = 2, which is positive.  Since we are using a number slightly to the left of the break point, -2x-28 is positive on the left side of the break point.  The function is then negative on the other side of the break point, so it is
–(2x-28) and for x >= -14, which is = 2x+28.

Note that both are the same when x=-14, for they both are 0.  This means the equal sign can be applied on either side of the break point.  This property will be used throughtout the rest of this answer for the other equations as well.

For 8x-8=0, by adding 8 to both sides we get 8x=8.
This is the same as x=1 (divide both sides by 8).
For x=0, the function 8x-8 = -8, and 0 is to the left of 1,
so the function is -8x+8 for x<=1.  This means it is 8x-8 for x>=1.

Putting both conditions together, the edges will be at x=-14 and x=1.  They will be
A) x<=-14 gives -2x-28 = -8x+8;
B) -14<=x<=1 gives 2x+28 = -8x+8; and
C) 1<=x gives 2x+28 = 8x-8.

Looking at (A), add 8x+28 to both sides gives 6x = 36, or x=6.
Since 6 is not less than of equal to -14, there is no solution for (A).

Looking at (B), add 8x-28 to both sides, giving 10x = -20, so x = -2.
Now -2 does satisfy -14<=-2<=1, so that is a solution.

Looking at (C), add -8x-28 to both sides, giving -6x = -36; divide by -6 and get x=6.
Since 1<=6, this is also a solution.  {note that (C) is the same as –(A)}

Both (B) and (C) gave a solution, so there are two solutions: x=-2 or x=6.

Checking x=-2 gives -4+28 = 24 on the left and 16+8 = 24 on the right, so that's good.
Checking x=6 gives 12+28=40 on the left and 48-8=40 on the right, so that's good as well.

2) Solve /2x-15/=/8x-75/
The boundaries are at 2x=15, so x=7.5 and at 8x=75, so x=9.375.
At x=0, 2x-15<0 and 8x-75<0, so flip the signs on the left side on both.

The equations are then
for x<=7.5, -2x+15 = -8x+75;
for 7.5<=x<=9.375, 2x-15 = -8x+75; and
for 9.375<x, 2x-15 = 8x-75.

Note that the 2nd equation can be solved, but only the 1st or 3rd equation has a solution in the right limits since they both have the same solution but different limits.

3) Solve /-10x+75/=/-8x+87/
The points to check are where 10x=75 and 8x=87; that is, x=7.5 and x=10.875.
The 1st equation is 10x-75 for x>=7.5, which makes it -10x+75 for x<=7.5.
The 2nd equation is 8x-87 for -8x+87 for x<=10.875 and 8x-87 for x>=10.875.
This means the ranges to check are x<=7.5, 7.5<=x<=10.875, and 10.75<=x.

This gives
(A)-10x+75 = -8x+87 for x<=7.5,
(B) 10x-75 = -8x+87 for 7.5<=x<=10.875, and
(C) 10x-75 =  8x-87 for 10.875<=x.

Solving these will once again give an answer for (B), but for (A) and (C) {since (A) is the negative of the (C) and vice versa}, only one of them meets the right range for the solution.

Algebra

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#### Scott A Wilson

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