Which mathematical law is used in $ab+ac-(b+c)=(a-1)(b+c)$
I just stumbled upon a question to figure out how to simplify
J = (ab)+(ac)-(b+c)
My steps:
<=> a*(b+c)-b-c
<=> a*(b+c) -1*(b+c)
But that was not one of the solutions. One of these was, as mentioned above,
(a-1)*(b+c).
As I saw this I somehow knew it is correct, calculated it and yes it is. But my math is a bit outdated and I cannot remember the law to see this. I do know it is correct, but by the love of god, I still don't know how to pull it of.
algebra-precalculus simple-functions
edited Nov 26 at 19:07
A.Γ.
21.8k22455
asked Nov 26 at 18:39
InDaPond
32
add a comment |
I just stumbled upon a question to figure out how to simplify
J = (ab)+(ac)-(b+c)
My steps:
<=> a*(b+c)-b-c
<=> a*(b+c) -1*(b+c)
But that was not one of the solutions. One of these was, as mentioned above,
(a-1)*(b+c).
As I saw this I somehow knew it is correct, calculated it and yes it is. But my math is a bit outdated and I cannot remember the law to see this. I do know it is correct, but by the love of god, I still don't know how to pull it of.
algebra-precalculus simple-functions
edited Nov 26 at 19:07
A.Γ.
21.8k22455
asked Nov 26 at 18:39
InDaPond
32
You are almost done. $a(b+c) -1(b+c)$ can be factored again (b+c) and will yield: $(a-1)(b+c)$
– Dashi
Nov 26 at 18:42
Your title has $+$ instead of $*$
– Ross Millikan
Nov 26 at 18:54
aaah, now I see it. It is not any fancy rule, it is just another factorization. Something tree, something forest.. Thank you!!
– InDaPond
Nov 26 at 18:57
add a comment |
I just stumbled upon a question to figure out how to simplify
J = (ab)+(ac)-(b+c)
My steps:
<=> a*(b+c)-b-c
<=> a*(b+c) -1*(b+c)
But that was not one of the solutions. One of these was, as mentioned above,
(a-1)*(b+c).
As I saw this I somehow knew it is correct, calculated it and yes it is. But my math is a bit outdated and I cannot remember the law to see this. I do know it is correct, but by the love of god, I still don't know how to pull it of.
algebra-precalculus simple-functions
edited Nov 26 at 19:07
A.Γ.
21.8k22455
asked Nov 26 at 18:39
InDaPond
32
I just stumbled upon a question to figure out how to simplify
J = (ab)+(ac)-(b+c)
My steps:
<=> a*(b+c)-b-c
<=> a*(b+c) -1*(b+c)
But that was not one of the solutions. One of these was, as mentioned above,
(a-1)*(b+c).
As I saw this I somehow knew it is correct, calculated it and yes it is. But my math is a bit outdated and I cannot remember the law to see this. I do know it is correct, but by the love of god, I still don't know how to pull it of.
algebra-precalculus simple-functions
algebra-precalculus simple-functions
edited Nov 26 at 19:07
A.Γ.
21.8k22455
asked Nov 26 at 18:39
InDaPond
32
edited Nov 26 at 19:07
A.Γ.
21.8k22455
asked Nov 26 at 18:39
InDaPond
32
edited Nov 26 at 19:07
A.Γ.
21.8k22455
edited Nov 26 at 19:07
A.Γ.
21.8k22455
edited Nov 26 at 19:07
A.Γ.
21.8k22455
21.8k22455
asked Nov 26 at 18:39
InDaPond
32
asked Nov 26 at 18:39
InDaPond
32
asked Nov 26 at 18:39
InDaPond
32
32
You are almost done. $a(b+c) -1(b+c)$ can be factored again (b+c) and will yield: $(a-1)(b+c)$
– Dashi
Nov 26 at 18:42
Your title has $+$ instead of $*$
– Ross Millikan
Nov 26 at 18:54
aaah, now I see it. It is not any fancy rule, it is just another factorization. Something tree, something forest.. Thank you!!
– InDaPond
Nov 26 at 18:57
add a comment |
You are almost done. $a(b+c) -1(b+c)$ can be factored again (b+c) and will yield: $(a-1)(b+c)$
– Dashi
Nov 26 at 18:42
Your title has $+$ instead of $*$
– Ross Millikan
Nov 26 at 18:54
aaah, now I see it. It is not any fancy rule, it is just another factorization. Something tree, something forest.. Thank you!!
– InDaPond
Nov 26 at 18:57
You are almost done. $a(b+c) -1(b+c)$ can be factored again (b+c) and will yield: $(a-1)(b+c)$
– Dashi
Nov 26 at 18:42
You are almost done. $a(b+c) -1(b+c)$ can be factored again (b+c) and will yield: $(a-1)(b+c)$
– Dashi
Nov 26 at 18:42
Your title has $+$ instead of $*$
– Ross Millikan
Nov 26 at 18:54
Your title has $+$ instead of $*$
– Ross Millikan
Nov 26 at 18:54
aaah, now I see it. It is not any fancy rule, it is just another factorization. Something tree, something forest.. Thank you!!
– InDaPond
Nov 26 at 18:57
aaah, now I see it. It is not any fancy rule, it is just another factorization. Something tree, something forest.. Thank you!!
– InDaPond
Nov 26 at 18:57
add a comment |
2 Answers
2
active
oldest
votes
It is called the distributive law
$$
(x+y)z=xz+yz.
$$
You apply it backwards in
$$
(a-1)underbrace{(b+c)}_{z}=aunderbrace{(b+c)}_{z}-1underbrace{(b+c)}_{z}.
$$
answered Nov 26 at 19:13
A.Γ.
21.8k22455
add a comment |
The first two terms are $a(b+c)$ by distributivity of multiplication over addition. A reuse of this rule, together with $x=1x$ with $x=b+c$, completes the proof.
answered Nov 26 at 19:05
J.G.
22.3k22035
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is called the distributive law
$$
(x+y)z=xz+yz.
$$
You apply it backwards in
$$
(a-1)underbrace{(b+c)}_{z}=aunderbrace{(b+c)}_{z}-1underbrace{(b+c)}_{z}.
$$
answered Nov 26 at 19:13
A.Γ.
21.8k22455
add a comment |
It is called the distributive law
$$
(x+y)z=xz+yz.
$$
You apply it backwards in
$$
(a-1)underbrace{(b+c)}_{z}=aunderbrace{(b+c)}_{z}-1underbrace{(b+c)}_{z}.
$$
answered Nov 26 at 19:13
A.Γ.
21.8k22455
add a comment |
It is called the distributive law
$$
(x+y)z=xz+yz.
$$
You apply it backwards in
$$
(a-1)underbrace{(b+c)}_{z}=aunderbrace{(b+c)}_{z}-1underbrace{(b+c)}_{z}.
$$
answered Nov 26 at 19:13
A.Γ.
21.8k22455
It is called the distributive law
$$
(x+y)z=xz+yz.
$$
You apply it backwards in
$$
(a-1)underbrace{(b+c)}_{z}=aunderbrace{(b+c)}_{z}-1underbrace{(b+c)}_{z}.
$$
answered Nov 26 at 19:13
A.Γ.
21.8k22455
answered Nov 26 at 19:13
A.Γ.
21.8k22455
answered Nov 26 at 19:13
A.Γ.
21.8k22455
answered Nov 26 at 19:13
A.Γ.
21.8k22455
21.8k22455
add a comment |
add a comment |
The first two terms are $a(b+c)$ by distributivity of multiplication over addition. A reuse of this rule, together with $x=1x$ with $x=b+c$, completes the proof.
answered Nov 26 at 19:05
J.G.
22.3k22035
add a comment |
The first two terms are $a(b+c)$ by distributivity of multiplication over addition. A reuse of this rule, together with $x=1x$ with $x=b+c$, completes the proof.
answered Nov 26 at 19:05
J.G.
22.3k22035
add a comment |
The first two terms are $a(b+c)$ by distributivity of multiplication over addition. A reuse of this rule, together with $x=1x$ with $x=b+c$, completes the proof.
answered Nov 26 at 19:05
J.G.
22.3k22035
The first two terms are $a(b+c)$ by distributivity of multiplication over addition. A reuse of this rule, together with $x=1x$ with $x=b+c$, completes the proof.
answered Nov 26 at 19:05
J.G.
22.3k22035
answered Nov 26 at 19:05
J.G.
22.3k22035
answered Nov 26 at 19:05
J.G.
22.3k22035
answered Nov 26 at 19:05
J.G.
22.3k22035
22.3k22035
add a comment |
add a comment |
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You are almost done. $a(b+c) -1(b+c)$ can be factored again (b+c) and will yield: $(a-1)(b+c)$
– Dashi
Nov 26 at 18:42
Your title has $+$ instead of $*$
– Ross Millikan
Nov 26 at 18:54
aaah, now I see it. It is not any fancy rule, it is just another factorization. Something tree, something forest.. Thank you!!
– InDaPond
Nov 26 at 18:57