Integers
BEDMAS
Substituting the variable
Algebra: collecting like terms
100

3 - 4

-1

100

4(2 − 7)(1 − 5)

4(2 - 7) (1 - 5) = 4 x (-5) x (-4) 

= -20 x -4

= 80

100

(x= 4) x - 5 + 2x

4 - 5 + 2(4)

4 - 5 + 8

-1 + 8

7

100

3x + 2y - x + 4y

4x + 6y

200

-5 - 6

-11

200

2(8 − 3) + (9 − 12)

2 x 5 + (-3)

= 10 + (-3)

= 7

200

(x= 7) 7(x + 4) - (x1)

7(7 + 4) - (7 x 1)

7 x 11 - 7

77 - 7

70

200

5x² + 8 - 2x + 3 - x²

6x² + 11 - 2x

300

5 (-3)

-15

300

−6(8 − 9)³(6 − 7)

-6 x (-1)³(-1)

-6 x -1 x -1

-6

300

 (x= 7, y= 2) -x + 5y - (x - 6)

-7 + 5 x 2 - 7 - 6

-7 + 10 - 7 - 6

3 - 7 - 6

-4 - 6

-10

300

4(5x + 3) + 2(7x - 1)

10 x 12x

400

−6 + (−1) − (−8)

1

400

5² + 2⁴− 6(2)(8)(−4)

25 + 16 - 6 x 2 x 8 x (-4)

25 + 16 - (-384) 

41 - (-384)

425

400

(x= 8, y= 5) -x2 x y - (3x + 2)y

-8 x 2 x 5 - (3 x 8 + 2)5

-8 x 2 x 5 - 26 + 5

-8 x 2 x 5 - 26 + 5

-16 x 2 x 5 - 26 + 5

-32 x 5 - 26 + 5

-160 - 26 + 5

-186 + 5

-181

400

6x(2x + 3) - 7x (1x - 1)

(16x)4 

500

(-7) - 2 + (-5) + (-8)

-22

500

5√2(4) + 7(4) − 1

7 x 4 + 7 x 4 - 1

28 + 28 - 1

56 - 1

55

500

 (x= 3, y= 6, z= 2) -5 + x (y - 5z) - 3

-5 + 3 (6 - 5 x 2) - 3

-5 + 3 x 2 - 3

-5 + 6 - 3

1 - 3

-2

500

6(x + 3) - 5x = 8

x = -10 

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