what is 3(y-2)=1-3y over 5?
回答 (7)
2009-05-16 4:28 pm
✔ 最佳答案
3(y - 2) = (1 - 3y) / 5Multiply both sides by 5.
15(y - 2) = 1 - 3y
Distribute.
15y - 30 = 1 - 3y
Isolate 'y'.
18y = 31
Divide.
y = 31 / 18
2009-05-16 11:46 pm
3(y - 2) = 1 - 3y
3*y - 3*2 = 1 - 3y
3y + 3y = 1 + 6
6y = 7
y = 7/6
3*y - 3*2 = 1 - 3y
3y + 3y = 1 + 6
6y = 7
y = 7/6
2009-05-16 11:35 pm
3(y-2)=1-3y/5
3y-6=1/5-3y/5 [Collect The Like Terms]
3y+3y/5=1/5+6 [Make Same Denominator Then Add]
18y/5=31/5 [Cross Multiplication]
90y=155 [Divide By 5]
18y=31 [Divide By 18]
y=31/18
3y-6=1/5-3y/5 [Collect The Like Terms]
3y+3y/5=1/5+6 [Make Same Denominator Then Add]
18y/5=31/5 [Cross Multiplication]
90y=155 [Divide By 5]
18y=31 [Divide By 18]
y=31/18
2009-05-16 11:32 pm
multiply by 5 to get rid of the fraction and get
15(y-2) = 1-3y
add 3y to gather the 'y's together
18y - 30 = 1
add 30 to isolate y
18y = 31
divide by 18 to solve for y
y = 31/18
done
15(y-2) = 1-3y
add 3y to gather the 'y's together
18y - 30 = 1
add 30 to isolate y
18y = 31
divide by 18 to solve for y
y = 31/18
done
2009-05-16 11:31 pm
3y - 6 = 1-3y/5
6y = 6 1/5
y= 1.03 (recurring) OR 1 1/6
3 (y-2) = 1-3y / 5
3y - 6 = 1-3y / 5
6y - 6 =1/5
6y = 6 + 1/5
y = 1.03 (recurring) OR 1 1/6
Or if you meant (3(y-2)=1-3y)/5 It is 31/18
6y = 6 1/5
y= 1.03 (recurring) OR 1 1/6
3 (y-2) = 1-3y / 5
3y - 6 = 1-3y / 5
6y - 6 =1/5
6y = 6 + 1/5
y = 1.03 (recurring) OR 1 1/6
Or if you meant (3(y-2)=1-3y)/5 It is 31/18
參考: So 1 1/6 (1.03 reccuring). OR 31/18
2009-05-16 11:29 pm
15y-30=1-3y
18y=31
y=31/18
18y=31
y=31/18
2009-05-16 11:30 pm
Use distributive property to solve the left side and you get 3y-6
3y-6=1-3y/5
Multiply both sides by 5 to clear out that 5 in the denom.
5(3y-6)=1-3y
15y-30=1-3y
18y=31
y=31/18 or 1 13/18
3y-6=1-3y/5
Multiply both sides by 5 to clear out that 5 in the denom.
5(3y-6)=1-3y
15y-30=1-3y
18y=31
y=31/18 or 1 13/18
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