Linear Equations
Absolute Value
PEMDAS
Inverse Functions
Complex numbers
100

4t/t2−25=1/5−t

t = -1

100

6u = |1 + 3u|

u = 1/3

100

7×(2+5)

49

100

Given f(x)= ( x - 2)3 + 1, find f-1 (x)

f-1(x) = (x−1)1/3 + 2

100

(-3 - i) - (6 - 7i)

-9 + 6i

200

5x/3x-3+6/x+2= 5/3

x=8/23

200

|1/2z + 4| = |4z - 6|

z = 4/9 & 20/7

200

 7/(5×2−(6/3)

0.875

200

Given h (x) = 5 - 9x find h -1 (x)

x = 5-x/9

200

(2 + 7t) (8 + 3t)

−5 + 62t

300

2 ( w+3) - 10 = 6 (32-3w)

w = 196/20 or 49/5

300

|2 - 4x| = 1

x= 1 & -1

300

17 + 77 / 11 + 22 - 8 x 5

6

300

Given g (x) = 1/2x + 7 find g-1 (x)

g-1 (x) = 2x - 14

300

(4 − 5a) (12 + 11a)

103 − 16a

400

4x - 7 ( 2 - x) = 3x + 2

x = 2

400

|x2+2x|=15

x= -5 & 3

400

11−((6/2)×((7/5)+8))

-17.2

400

Given j (x) = 1 + 2x/7 + x find h-1 (x)

h-1 (x) = 7x - 1/2 - x

400

(1 + 4z)−(−16 + 9z)

17−5z