How do I solve these questions for x?

Question

6|x+4| = 8|x-2|

7|x+4| = 8|x-2|

2|x-6| + 8 = 4|x+2|-10

Answer 1

its simple..
like in the first example, solve for a few cases differently
case 1: x<-4
eq : -6(4+x)=-8(x-2)
-24-6x=-8x+16
2x=40
x=20 (but x should be < -4)
reject this solution

case 2: -4<=x<2
eq: 6(4+x)=-8(x-2)

case 3: x>=2
eq: 6(4+x)=8(x-2)

finally join all the answers
likewise you can try the others!

Answer 2

You may have to solve for 4 cases (plus and minus for each side).
6|x+4| = 8|x-2|

Case A (+ +)
6x + 24 = 8x - 16
40 = 2x
20 = x

Case B (- +)
6*(-1)(x+4) = 8x - 16
-6x -24 = 8x - 16
-8 = 14 x
-8/14 = x = -4/7

Case C (+ -)
6x + 24 =8*(-1)(x-2) = -8x + 16
14 x = -8
x = -8/14 = -4/7

Case D (- -)
-6x - 24 = -8x + 16
2x = 40
x = 20

---

Here, because each side is a single, first degree function inside the absolute value, there can only be a total of 2 solutions.

This can be shown geometrically by using two functions
y = 6|x+4|
y = 8|x-2|

each one, when plotted, will have a V shape, with a minimum of 0 when x is the additive inverse of the other operand (i.e., when x = -4 in the first one).

If you plot two V's on a graph, they can only intersect twice (in other words, there can only be a maximum of two values of x where y is equal for both equations). If you have more complicated functions, then you may have to check out more cases.

Answer 3

assuming ur only problem is with the absolute value signs, | |,

absolute values are related to inequalities

so u need to solve two equations for each one, because the absolute value signs CAN hav 2 different values and complete the equality, some hav ONE, others NONE

so the first one,
6|x+4| = 8|x-2|
goes to,
6(x+4) = 8(x-2) AND 6(x+4) = -8(x-2)
this is because of the properties of the absolute value aspect, so now, solve
x= 20 AND x= - 4/7

Answer 4

i think that u would do.... for number one x-2 times 8 equals 4+x times 6.
i cant figure it out because i might give you a wrong answer. I hope that is right and it is helpful.
srry if it isnt
bye
JJ

Answer 5

6|x+4| = 8|x-2|
6x+24=8x-16
6x-8x=-16-24
-2x= -40 therefore x= 20

7|x+4| = 8|x-2|
7x+28=8x -16
7x - 8x = -16 -28
-x= -44 or x=44

2|x-6| + 8 = 4|x+2|-10
2x -12 +8=4x+8 -10
2x -4= 4x -2
2x -4x = 4 -2
-2x=2 or x= -1

Answer 6

6(x+4) = 8(x-2)
6x+24-6x= 8x-16-6x
24=2x-16
+16 +16
40=2x
40/2=2x/2
20=x

i think after this, u can figure the rest out

Answer 7

so first you would multiply x and 4 by 6 and on the other side you would multiply x and -2 by 8.(first problem)
so you would get 6x+24=8x+-16
then u have to get the variables to one side so u subtract 6x from 8x
24=2x+-16
then u add 16 to both sides
40=2x
then u need to get x by itself so u divide both sides by 2
so the final answer is
20=x
so first do distributive property, then get variables to one side, then add/subtract, finally either multiply/divide
i hope i helped
sorry if its wrong, im only in 8th grade

Answer 8

I'm not sure if I'm suppose to help you but they have to be equal on both sides and the x's have to be on 1 side and the numbers are suppose to be on the other

Answer 9

treat the absolute value symbols as if they were parenthesis and solve as another equation (combine like terms, get "x" by itself, and reduce franction), just make sure to solve for both positive and negative "x"

i.e. 6|x+4|=8|x-2| would be
6(x+4)=8(x-2) ...... and .......... 6(-x+4)=8(-x-2)
6x+24=8x-16 .................. -6x+24=-8x-16
-2x=40........................................................................ 2x=40
x= -20......................................................................... x=20

Answer 10

i usually just set one side to 0 and graph it on my graphing calculator, then go into table and find when y=0, then the x value is the answer