The answer would be 25/9 as a improper fraction
Answer:
b = 0
Step-by-step explanation:
x -3 (2x-3) + 5x = bx + 9
x - 6x +9 + 5x = bx + 9
-9 -9
x - 6x + 5x = bx
-5x + 5x = bx
0 = bx
b/x 0/x
b = 0
Answer:
The answer is 660.4 millimeters in diameter.
Step-by-step explanation:
The wheel is 26 inches in diameter. An inch is 25.4 millimeters. 26 * 25.4 = 660.4
1.) Solve for x:
5 x + 7 = 3 x + 21
Subtract 3 x from both sides:
(5 x - 3 x) + 7 = (3 x - 3 x) + 21
5 x - 3 x = 2 x:
2 x + 7 = (3 x - 3 x) + 21
3 x - 3 x = 0:
2 x + 7 = 21
Subtract 7 from both sides:
2 x + (7 - 7) = 21 - 7
7 - 7 = 0:
2 x = 21 - 7
21 - 7 = 14:
2 x = 14
Divide both sides of 2 x = 14 by 2:
(2 x)/2 = 14/2
2/2 = 1:
x = 14/2
The gcd of 14 and 2 is 2, so 14/2 = (2×7)/(2×1) = 2/2×7 = 7:
Answer: x = 7
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2.) Solve for x:
3 x - 2 (5 - x) = 3 x - 3 (x - 10)
-2 (5 - x) = 2 x - 10:
2 x - 10 + 3 x = 3 x - 3 (x - 10)
Grouping like terms, 3 x + 2 x - 10 = (3 x + 2 x) - 10:
(3 x + 2 x) - 10 = 3 x - 3 (x - 10)
3 x + 2 x = 5 x:
5 x - 10 = 3 x - 3 (x - 10)
-3 (x - 10) = 30 - 3 x:
5 x - 10 = 30 - 3 x + 3 x
3 x - 3 x = 0:
5 x - 10 = 30
Add 10 to both sides:
5 x + (10 - 10) = 10 + 30
10 - 10 = 0:
5 x = 30 + 10
30 + 10 = 40:
5 x = 40
Divide both sides of 5 x = 40 by 5:
(5 x)/5 = 40/5
5/5 = 1:
x = 40/5
The gcd of 40 and 5 is 5, so 40/5 = (5×8)/(5×1) = 5/5×8 = 8:
<span>Answer: x = 8
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3.) Solve for x:</span>
5 (x + 1) = 3 (2 x + 3) + 5
3 (2 x + 3) = 6 x + 9:
5 (x + 1) = 6 x + 9 + 5
Grouping like terms, 6 x + 5 + 9 = 6 x + (9 + 5):
5 (x + 1) = 6 x + (9 + 5)
9 + 5 = 14:
5 (x + 1) = 6 x + 14
Expand out terms of the left hand side:
5 x + 5 = 6 x + 14
Subtract 6 x from both sides:
(5 x - 6 x) + 5 = (6 x - 6 x) + 14
5 x - 6 x = -x:
-x + 5 = (6 x - 6 x) + 14
6 x - 6 x = 0:
5 - x = 14
Subtract 5 from both sides:
(5 - 5) - x = 14 - 5
5 - 5 = 0:
-x = 14 - 5
14 - 5 = 9:
-x = 9
Multiply both sides of -x = 9 by -1:
(-x)/(-1) = -9
(-1)/(-1) = 1:
<span>Answer: x = -9</span>
Answer: See below
Step-by-step explanation:
27. -(a-3)
28. (b-1)(b+3)
29. (c+4)(c+5)
30. d(d+5)
31. -(3/4)(2e-5)
Sorry - I don't have time to enter the details. Look for areas where the expressions can be factored in a manner that forms as many equivalent expressions in both the numerator and denominator.
For example: In problem 30:
(5d-20)/(d^2+d-20) * [??]/20d = 1/4
Factor:
<u>(5(d-4))</u> <u>d(d+5)</u> = 1/4
(d-4)(d+5<u>)</u> 20d
The (d-4), d+5, and d terms cancel, leaving
5/20 = 1/4
