To solve an absolute value equation of the form |X| = k, where X is an expression with a variable and k is a non-negative number or an expression, you must solve the compound equation X = k or X = -k.
The first step is to divide both sides by 9 to get the absolute value of an expression equal to a non-negative number.
9|9 – 8x| = 2x + 3
Divide both sides by 9.
[tex] |9 - 8x| = \dfrac{2x + 3}{9} [/tex]
Now we eliminate the absolute value by rewriting this as a compound equation.
[tex] 9 - 8x = \dfrac{2x + 3}{9} ~~~or~~~ 9 - 8x = -\dfrac{2x + 3}{9} [/tex]
We must solve both equations.
Multiply both sides by 9.
[tex] 9(9 - 8x) = 2x + 3 ~~~or~~~ 9(9 - 8x) = -(2x + 3) [/tex]
[tex] 81 - 72x = 2x + 3 ~~~or~~~ 81 - 72x = -2x - 3 [/tex]
[tex] -74x = -78 ~~~or~~~ -70x = -84 [/tex]
[tex] x = \dfrac{-78}{-74} ~~~or~~~ x = \dfrac{-84}{-70} [/tex]
[tex] x = \dfrac{39}{37} ~~~or~~~ x = \dfrac{6}{5} [/tex]
Answer: x = 39/37 or x = 6/5
To check for extraneous solutions, we plug in each solution into the original equation and check if it works.
Check 39/37:
9|9 – 8(39/37)| = 2(39/37) + 3
|81 - 72(39/37)| = 78/37 + 3
|2997/37 - 2808/37| = 78/37 + 111/37
|189/37| = 189/37
189/37 = 189/37
The solution 39/37 is valid.
Check x = 6/5.
9|9 – 8x| = 2x + 3
9|9 – 8(6/5)| = 2(6/5) + 3
|81 - 72(6/5)| = 12/5 + 15/5
|405/5 - 432/5| = 27/5
|-27/5| = 27/5
27/5 = 27/5
The solution x = 6/5 is also valid.
There are no extraneous solutions.
