Mixing
Understanding transitive property of equality
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Suppose we have two equations such that neither left side equals zero. Is the quotient of the two left sides equal to the quotient of the two right sides? In other words, if $a=b$ and $c=d$ and neither $a$ nor $c$ is 0, then is $a/c$ = $b/d$?
I don't quite get what the question is asking here. What concept is it trying to illustrate over here?
I interpret it as: if $a$ is 1 and $c$ is 2, since $a = b$ and $c = d$, I should get the equation $1/2$ = $1/2$, and therefore $a/c$ = $b/d$ are equal.
algebra-precalculus
asked Jan 26 at 15:23
ilovetolearnilovetolearn
63021021
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add a comment |
$begingroup$
Suppose we have two equations such that neither left side equals zero. Is the quotient of the two left sides equal to the quotient of the two right sides? In other words, if $a=b$ and $c=d$ and neither $a$ nor $c$ is 0, then is $a/c$ = $b/d$?
I don't quite get what the question is asking here. What concept is it trying to illustrate over here?
I interpret it as: if $a$ is 1 and $c$ is 2, since $a = b$ and $c = d$, I should get the equation $1/2$ = $1/2$, and therefore $a/c$ = $b/d$ are equal.
algebra-precalculus
asked Jan 26 at 15:23
ilovetolearnilovetolearn
63021021
$endgroup$
add a comment |
$begingroup$
Suppose we have two equations such that neither left side equals zero. Is the quotient of the two left sides equal to the quotient of the two right sides? In other words, if $a=b$ and $c=d$ and neither $a$ nor $c$ is 0, then is $a/c$ = $b/d$?
I don't quite get what the question is asking here. What concept is it trying to illustrate over here?
I interpret it as: if $a$ is 1 and $c$ is 2, since $a = b$ and $c = d$, I should get the equation $1/2$ = $1/2$, and therefore $a/c$ = $b/d$ are equal.
algebra-precalculus
asked Jan 26 at 15:23
ilovetolearnilovetolearn
63021021
$endgroup$
Suppose we have two equations such that neither left side equals zero. Is the quotient of the two left sides equal to the quotient of the two right sides? In other words, if $a=b$ and $c=d$ and neither $a$ nor $c$ is 0, then is $a/c$ = $b/d$?
I don't quite get what the question is asking here. What concept is it trying to illustrate over here?
I interpret it as: if $a$ is 1 and $c$ is 2, since $a = b$ and $c = d$, I should get the equation $1/2$ = $1/2$, and therefore $a/c$ = $b/d$ are equal.
algebra-precalculus
algebra-precalculus
asked Jan 26 at 15:23
ilovetolearnilovetolearn
63021021
asked Jan 26 at 15:23
ilovetolearnilovetolearn
63021021
edited Jan 26 at 15:28
ilovetolearn
asked Jan 26 at 15:23
ilovetolearnilovetolearn
63021021
asked Jan 26 at 15:23
ilovetolearnilovetolearn
63021021
asked Jan 26 at 15:23
ilovetolearnilovetolearn
63021021
63021021
add a comment |
add a comment |
1 Answer
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We can use transitivity to prove this, viz. $a/c=b/c=b/d$.
answered Jan 26 at 15:31
J.G.J.G.
31.3k23149
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add a comment |
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$begingroup$
We can use transitivity to prove this, viz. $a/c=b/c=b/d$.
answered Jan 26 at 15:31
J.G.J.G.
31.3k23149
$endgroup$
add a comment |
$begingroup$
We can use transitivity to prove this, viz. $a/c=b/c=b/d$.
answered Jan 26 at 15:31
J.G.J.G.
31.3k23149
$endgroup$
add a comment |
$begingroup$
We can use transitivity to prove this, viz. $a/c=b/c=b/d$.
answered Jan 26 at 15:31
J.G.J.G.
31.3k23149
$endgroup$
We can use transitivity to prove this, viz. $a/c=b/c=b/d$.
answered Jan 26 at 15:31
J.G.J.G.
31.3k23149
answered Jan 26 at 15:31
J.G.J.G.
31.3k23149
answered Jan 26 at 15:31
J.G.J.G.
31.3k23149
answered Jan 26 at 15:31
J.G.J.G.
31.3k23149
31.3k23149
add a comment |
add a comment |
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