3. Evaluate the expression if w = -4, x = 2, y = ½, and z = -6.
2. |6 + z| - |7| Original problem.
|6 + (-6)| - |7| Substitute values for the variables.
|0| - |7| Simplify inside each absolute value.
0 – 7 = Take absolute values and simplify.
Final Answer is -7.
6. |7 – x| + |3x| Original problem.
|7 – 2| + |3(2)| Substitute values for the variables.
|5| - |6| Simplify inside each absolute value.
= Take absolute values and simplify.
Final Answer is 11.

4. Evaluate the expression if w = -4, x = 2, y = ½, and z = -6.
|w| + 2|z – 3y| Original problem.
5|-4| + 2|-6 – 3(1/2)| Substitute values for the variables.
5|-4| + 2|-7 ½ | Simplify inside each absolute value.
5(4) + 2(-15/2) = 5 Take absolute values and simplify.
Final Answer is 5.
|wx| + (1/4)|4x + 8y| Original problem.
3|(-4)(2)| + (1/4)|4(2) + 8(1/2)| Substitute values for the variables.
3|-8| + (1/4)|12| Simplify inside each absolute value.
= 27 Take absolute values and simplify.
Final Answer is 27.

5. Evaluate the expression if w = -4, x = 2, y = ½, and z = -6.
18. |xyz| + |wxz| Original problem.
|(2)(1/2)(-6)| + |(-4)(2)(-6)| Substitute values for the variables.
|-6| + |48| Simplify inside each absolute value.
= Take absolute values and simplify.
Final Answer is 54.
22. |yz – 4w| - w Original problem.
|(1/2)(-6) – 4(-4)| - (-4) Substitute values for the variables.
|13| - (-4) Simplify inside each absolute value.
= 17 Take absolute values and simplify.
Final Answer is 17.

9. Solve the equation. Check your solutions.
|15 – 2k| = Original Equation.
Since Absolute Value bars are isolated, just set up two equations to be solved. One that is a positive answer the other a negative answer.
a. 15 – 2k = b. 15 – 2k = -45
15 – 2k = move 15.
-2k = move -2.
k = Solution.
15 – 2k = move 15.
-2k = move -2.
k = Solution.
Check the solutions k = -15 and 30 into original equation.
k = k = 30
|15 – 2k| = |15 – 2k| = 45
|15 – 2(-15)| = |15 – 2(30)| = 45
|45| = yes |-45| = yes
Final Answer is k = -15 and 30.

10. Solve the equation. Check your solutions.
8. |8 + 5a| = 14 – a Original Equation.
Since Absolute Value bars are isolated, just set up two equations to be solved. One that is a positive answer the other a negative answer.
a a = 14 – a b a = -(14 – a)
8 + 5a = 14 – 1a move -1a.
8 + 6a = move 8.
6a = move 6.
a = solution.
8 + 5a = a move 1a.
8 + 4a = move 8.
4a = move 4.
a = -22/4 = -11/2 solution.
Check the solutions a = 1 and -11/2 into original equation.
k = k = -11/2
|8 + 5a| = 14 – a |8 + 5a| = 14 – a
|8 + 5(1)| = 14 – |8 + 5(-11/2)| = 14 – (-11/2)
|13| = yes |-15/2| = 15 1/2 yes
Final Answer is a = 1 and -11/2.

11. Solve the equation. Check your solutions.
10. |3x – 1| = 2x Original Equation.
Since Absolute Value bars are isolated, just set up two equations to be solved. One that is a positive answer the other a negative answer.
a. 3x – 1 = 2x b. 3x – 1 = -(2x + 11)
3x – 1 = 2x + 11 move 2x.
x – 1 = move 1.
x = 12
x = solution.
3x – 1 = -2x – move -2x.
5x – 1 = move 1.
5x = move 5.
x = solution.
Check the solutions x = 12 and -2 into original equation.
k = k = -2
|3x – 1| = 2x |3x – 1| = 2x + 11
|3(12) – 1| = 2(12) |3(-2) – 1| = 2(-2) + 11
|35| = yes |-7| = yes
Final Answer is k = 12 and -2.

12. Solve the equation. Check your solutions.
– 4x = 2|3x – 10| Original Equation.
Since Absolute Value bars are not isolated, divide each side by 2 to isolate the bars. Then set up two equations to be solved.
a. 20 – 2x = 3x – b. 20 – 2x = -(3x – 10)
20 – 2x = 3x – 10 move 3x.
20 – 5x = move 20.
-5x = move -5.
x = solution.
20 – 2x = -3x move -3x.
20 + 1x = move 20.
1x = move 1.
x = solution.
Check the solutions x = 6 and -10 into original equation.
k = k = -10
40 – 4x = 2|3x – 10| – 4x = 2|3x – 10|
40 – 4(6) = 2|3(6) – 10| – 4(-10) = 2|3(-10) – 10|
16 = 2(8) yes = 2(40) yes
Final Answer is k = 6 and -10.

13. Solve the equation. Check your solutions.
14. |4b + 3| = 15 – 2b Original Equation.
Since Absolute Value bars are isolated, just set up two equations to be solved. One that is a positive answer the other a negative answer.
a. 4b + 3 = 15 – 2b b. 4b + 3 = -(15 – 2b)
4b + 3 = 15 – 2b move -2b.
6b + 3 = move 3.
6b = move 6.
b = solution.
4b + 3 = b move 2b.
2b + 3 = move 3.
2b = move 5.
b = solution.
Check the solutions x = 2 and -9 into original equation.
k = k = -9
|4b + 3| = 15 – 2b |4b + 3| = 15 – 2b
|4(2) + 3| = 15 – 2(2) |4(-9) + 3| = 15 – 2(-9)
|11| = yes |-33| = yes
Final Answer is k = 2 and -9.

14. Solve the equation. Check your solutions.
16. |16 – 3x| = 4x Original Equation.
Since Absolute Value bars are isolated, just set up two equations to be solved. One that is a positive answer the other a negative answer.
a. 16 – 3x = 4x – b. 16 – 3x = -(4x – 12)
16 – 3x = 4x – 12 move 4x.
16 – 7x = move 16.
-7x = move -7.
x = 4/ solution.
16 – 3x = -4x move -4x.
16 + 1x = move 16.
1x = move 1.
b = -1/4 solution.
Check the solutions x = 4/7 and -1/4 into original equation.
k = 4/ k = -1/4
|16 – 3x| = 4x – |16 – 3x| = 4x – 12
|16 – 3(4/7)| = 4(4/7) – 12 |16 – 3(-1/4)| = 4(-1/4) – 12
|100/7| = -68/ no |16 3/4| = no
Final Answer is No Solution.