Using algebra to slove problems

5. LET X, X+1, X+2 BE THE THREE NUMBERS.
X + (X+1) + (X+2) = 63
3X + 3 = 63
3X = 63 - 3
3X = 60
X = 20
THEREFORE, THE THREE NUMBERS ARE 20, 21 AND 22.

2006-10-17 23:01:11 補充：
7. LET X BE THE SMALLER SUMBER AND Y BE THE LARGER NUMBER. Y - X = 10-----------(A) 12 X - Y = 45---------(B)FROM (A), WE HAVE Y - X = 10 Y = 10 + X------(C)

2006-10-17 23:01:32 補充：
INSERT (C) INTO (B),12X - Y = 4512X - (10+X) = 4512X - 10 - X = 4511X = 55X = 5 THEREFORE, THE SMALLER NUMBER IS 5.

2006-10-17 23:06:15 補充：
8. 4(y+6)=56 4y + 24 = 56 4y = 56 - 24 4y = 32 y = 89. 2c-9/6-1/2=0 (2c-9/6-1/2)*6 = 0 x 6 12c - 9 - 3 = 0 12c = 0 + 3 + 9 12c = 12 c = 1

2006-10-17 23:06:32 補充：
12. x/4-x+1/8=2 (x/4-x+1/8) * 8 = 2 * 8 2x - 8x + 8 = 16 -6x = 16 - 8 -6x = 8 x = - 4/3

5.let a, a+1,a+2 be the numbersa
a+a+1+a+2=63
a=20

so the numbers are 20,21 &22

7.let x &x-10 be the numbers

12(x-10)-x=45
x=15

so the numbers are15&10

8.4(y+6)=56
y+6=14
y=8

9.2c-9/6-i/2=0
2c=2
c=1

12.x/4-x+1/8=2
-3x/4=15/8
-24x=60
x=-2.5

2006-10-17 22:17:44 補充：
7.the numbers should be 15 &5(the smaller 1 is 5)

5. let the first number be a
then a+1 and a+2

a + (a+1) + (a+2) = 63
3a + 3 = 63
3a = 60
a = 20

so the three consecutive number is 20,21,22

7. let a and b be two numbers
a - b = 10 (1)
12b - a = 45 (2)

from (1), a = 10 + b
add into (2), 12b - (10+b) = 45
11b = 55
b = 5
smaller number is b, so is 5

8. y = 10
9. don't understand
12. x = -7/6

5. 20,21,22
7. Let x be the smaller number.
x+10-12x=45
8. y=8
9. c=0.5
12. 5/2