Simplfying expression to get $b-a$
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$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
I know that $$left( frac{2a}{3}+frac{b}{3} right)^0 = 1$$
and $$left( frac{b-a}{4} right)b^0=frac{b-a}{4}$$
but then how do I simplify the rest to get the solution which Is $b-a$?
algebra-precalculus
edited Nov 23 at 3:26
Joey Kilpatrick
1,183422
asked Nov 23 at 2:41
mt12345
958
add a comment |
up vote
0
down vote
favorite
$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
I know that $$left( frac{2a}{3}+frac{b}{3} right)^0 = 1$$
and $$left( frac{b-a}{4} right)b^0=frac{b-a}{4}$$
but then how do I simplify the rest to get the solution which Is $b-a$?
algebra-precalculus
edited Nov 23 at 3:26
Joey Kilpatrick
1,183422
asked Nov 23 at 2:41
mt12345
958
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
I know that $$left( frac{2a}{3}+frac{b}{3} right)^0 = 1$$
and $$left( frac{b-a}{4} right)b^0=frac{b-a}{4}$$
but then how do I simplify the rest to get the solution which Is $b-a$?
algebra-precalculus
edited Nov 23 at 3:26
Joey Kilpatrick
1,183422
asked Nov 23 at 2:41
mt12345
958
$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
I know that $$left( frac{2a}{3}+frac{b}{3} right)^0 = 1$$
and $$left( frac{b-a}{4} right)b^0=frac{b-a}{4}$$
but then how do I simplify the rest to get the solution which Is $b-a$?
algebra-precalculus
algebra-precalculus
edited Nov 23 at 3:26
Joey Kilpatrick
1,183422
asked Nov 23 at 2:41
mt12345
958
edited Nov 23 at 3:26
Joey Kilpatrick
1,183422
asked Nov 23 at 2:41
mt12345
958
edited Nov 23 at 3:26
Joey Kilpatrick
1,183422
edited Nov 23 at 3:26
Joey Kilpatrick
1,183422
edited Nov 23 at 3:26
Joey Kilpatrick
1,183422
1,183422
asked Nov 23 at 2:41
mt12345
958
asked Nov 23 at 2:41
mt12345
958
asked Nov 23 at 2:41
mt12345
958
958
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?
answered Nov 23 at 2:52
Rhys Hughes
4,6751327
add a comment |
up vote
0
down vote
$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
$$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
$$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
$$dfrac{4b-4a}{4}=b-a$$
$$b-a=b-a$$
answered Nov 23 at 2:44
Key Flex
7,44941232
sorry that was a typo in my question
– mt12345
Nov 23 at 2:45
add a comment |
up vote
0
down vote
The equation becomes $$
frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
b-a=b-a
$$
answered Nov 23 at 2:45
Joey Kilpatrick
1,183422
sorry that was a typo in my question
– mt12345
Nov 23 at 2:46
I don't think the typo has been fixed.
– Joey Kilpatrick
Nov 23 at 2:47
fixed, thanks!!
– mt12345
Nov 23 at 2:48
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?
answered Nov 23 at 2:52
Rhys Hughes
4,6751327
add a comment |
up vote
1
down vote
Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?
answered Nov 23 at 2:52
Rhys Hughes
4,6751327
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?
answered Nov 23 at 2:52
Rhys Hughes
4,6751327
Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?
answered Nov 23 at 2:52
Rhys Hughes
4,6751327
answered Nov 23 at 2:52
Rhys Hughes
4,6751327
answered Nov 23 at 2:52
Rhys Hughes
4,6751327
answered Nov 23 at 2:52
Rhys Hughes
4,6751327
4,6751327
add a comment |
add a comment |
up vote
0
down vote
$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
$$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
$$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
$$dfrac{4b-4a}{4}=b-a$$
$$b-a=b-a$$
answered Nov 23 at 2:44
Key Flex
7,44941232
sorry that was a typo in my question
– mt12345
Nov 23 at 2:45
add a comment |
up vote
0
down vote
$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
$$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
$$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
$$dfrac{4b-4a}{4}=b-a$$
$$b-a=b-a$$
answered Nov 23 at 2:44
Key Flex
7,44941232
sorry that was a typo in my question
– mt12345
Nov 23 at 2:45
add a comment |
up vote
0
down vote
up vote
0
down vote
$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
$$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
$$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
$$dfrac{4b-4a}{4}=b-a$$
$$b-a=b-a$$
answered Nov 23 at 2:44
Key Flex
7,44941232
$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
$$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
$$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
$$dfrac{4b-4a}{4}=b-a$$
$$b-a=b-a$$
answered Nov 23 at 2:44
Key Flex
7,44941232
edited Nov 23 at 2:46
answered Nov 23 at 2:44
Key Flex
7,44941232
answered Nov 23 at 2:44
Key Flex
7,44941232
answered Nov 23 at 2:44
Key Flex
7,44941232
7,44941232
sorry that was a typo in my question
– mt12345
Nov 23 at 2:45
add a comment |
sorry that was a typo in my question
– mt12345
Nov 23 at 2:45
sorry that was a typo in my question
– mt12345
Nov 23 at 2:45
sorry that was a typo in my question
– mt12345
Nov 23 at 2:45
add a comment |
up vote
0
down vote
The equation becomes $$
frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
b-a=b-a
$$
answered Nov 23 at 2:45
Joey Kilpatrick
1,183422
sorry that was a typo in my question
– mt12345
Nov 23 at 2:46
I don't think the typo has been fixed.
– Joey Kilpatrick
Nov 23 at 2:47
fixed, thanks!!
– mt12345
Nov 23 at 2:48
add a comment |
up vote
0
down vote
The equation becomes $$
frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
b-a=b-a
$$
answered Nov 23 at 2:45
Joey Kilpatrick
1,183422
sorry that was a typo in my question
– mt12345
Nov 23 at 2:46
I don't think the typo has been fixed.
– Joey Kilpatrick
Nov 23 at 2:47
fixed, thanks!!
– mt12345
Nov 23 at 2:48
add a comment |
up vote
0
down vote
up vote
0
down vote
The equation becomes $$
frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
b-a=b-a
$$
answered Nov 23 at 2:45
Joey Kilpatrick
1,183422
The equation becomes $$
frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
b-a=b-a
$$
answered Nov 23 at 2:45
Joey Kilpatrick
1,183422
edited Nov 23 at 3:29
answered Nov 23 at 2:45
Joey Kilpatrick
1,183422
answered Nov 23 at 2:45
Joey Kilpatrick
1,183422
answered Nov 23 at 2:45
Joey Kilpatrick
1,183422
1,183422
sorry that was a typo in my question
– mt12345
Nov 23 at 2:46
I don't think the typo has been fixed.
– Joey Kilpatrick
Nov 23 at 2:47
fixed, thanks!!
– mt12345
Nov 23 at 2:48
add a comment |
sorry that was a typo in my question
– mt12345
Nov 23 at 2:46
I don't think the typo has been fixed.
– Joey Kilpatrick
Nov 23 at 2:47
fixed, thanks!!
– mt12345
Nov 23 at 2:48
sorry that was a typo in my question
– mt12345
Nov 23 at 2:46
sorry that was a typo in my question
– mt12345
Nov 23 at 2:46
I don't think the typo has been fixed.
– Joey Kilpatrick
Nov 23 at 2:47
I don't think the typo has been fixed.
– Joey Kilpatrick
Nov 23 at 2:47
fixed, thanks!!
– mt12345
Nov 23 at 2:48
fixed, thanks!!
– mt12345
Nov 23 at 2:48
add a comment |
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