Mixing
Algebra: Can you think of each side of the equation as a term?
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For example with the equation
$$5x-4 = 2x +5$$
Is the accepted theory that you think of this equation in terms of:
$(5x-4) = (2x+5)$ when you are doing an operation to both sides.
Lets say I wanted to multiply by $3$, I would do:
$$3(5x-4) = 3(2x+5)$$
Is there ever a scenario that would break the rule of thinking of equations in a matter of each side being a whole term? Is this the accepted way of solving equations?
algebra-precalculus
edited 15 mins ago
N. F. Taussig
40.3k93253
asked 4 hours ago
Marlon Henderson
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up vote
3
down vote
favorite
For example with the equation
$$5x-4 = 2x +5$$
Is the accepted theory that you think of this equation in terms of:
$(5x-4) = (2x+5)$ when you are doing an operation to both sides.
Lets say I wanted to multiply by $3$, I would do:
$$3(5x-4) = 3(2x+5)$$
Is there ever a scenario that would break the rule of thinking of equations in a matter of each side being a whole term? Is this the accepted way of solving equations?
algebra-precalculus
edited 15 mins ago
N. F. Taussig
40.3k93253
asked 4 hours ago
Marlon Henderson
342
New contributor
Marlon Henderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
The name to use is expression rather than term. One expression is equal to another. Yes, if you wish to multiply both sides by $3$ you would do so to each expression. To answer your question of if you would ever not think of each side as a larger piece, sure... consider situations where you want to multiply by $1$ or add $0$. Take the expression $fracn-1n+1$. You could rewrite this as $fracn-1 +0n+1=fracn-1+2-2n+1=fracn+1-2n+1=fracn+1n+1+frac-2n+1=1-frac2n+1$. If I happened to have that expression equal to something else, I only need to modify first
– JMoravitz
4 hours ago
your idea is correct, when we apply such rule we think that x is some fixed number thus 5x-4 and 2x+5 are also some fixed numbers thus what you did here you say : "Suppose 5x - 4 = 2x+5 for some real number x, then 3*(5x-4) = 3*(2x+5)" here you apply a theorem that says : "for any real numbers a,b if a = b then ac = bc for any real number c" thus having 3(5x-4)=3(2x+5)
– famesyasd
4 hours ago
This is essentially the same question you posted a few hours ago. You could have just edited that one, instead.
– dxiv
3 hours ago
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
10 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For example with the equation
$$5x-4 = 2x +5$$
Is the accepted theory that you think of this equation in terms of:
$(5x-4) = (2x+5)$ when you are doing an operation to both sides.
Lets say I wanted to multiply by $3$, I would do:
$$3(5x-4) = 3(2x+5)$$
Is there ever a scenario that would break the rule of thinking of equations in a matter of each side being a whole term? Is this the accepted way of solving equations?
algebra-precalculus
edited 15 mins ago
N. F. Taussig
40.3k93253
asked 4 hours ago
Marlon Henderson
342
New contributor
Marlon Henderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For example with the equation
$$5x-4 = 2x +5$$
Is the accepted theory that you think of this equation in terms of:
$(5x-4) = (2x+5)$ when you are doing an operation to both sides.
Lets say I wanted to multiply by $3$, I would do:
$$3(5x-4) = 3(2x+5)$$
Is there ever a scenario that would break the rule of thinking of equations in a matter of each side being a whole term? Is this the accepted way of solving equations?
algebra-precalculus
algebra-precalculus
edited 15 mins ago
N. F. Taussig
40.3k93253
asked 4 hours ago
Marlon Henderson
342
New contributor
Marlon Henderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 15 mins ago
N. F. Taussig
40.3k93253
asked 4 hours ago
Marlon Henderson
342
New contributor
Marlon Henderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 15 mins ago
N. F. Taussig
40.3k93253
edited 15 mins ago
N. F. Taussig
40.3k93253
edited 15 mins ago
N. F. Taussig
40.3k93253
40.3k93253
asked 4 hours ago
Marlon Henderson
342
New contributor
Marlon Henderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Marlon Henderson
342
asked 4 hours ago
Marlon Henderson
342
342
New contributor
Marlon Henderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Marlon Henderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Marlon Henderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
The name to use is expression rather than term. One expression is equal to another. Yes, if you wish to multiply both sides by $3$ you would do so to each expression. To answer your question of if you would ever not think of each side as a larger piece, sure... consider situations where you want to multiply by $1$ or add $0$. Take the expression $fracn-1n+1$. You could rewrite this as $fracn-1 +0n+1=fracn-1+2-2n+1=fracn+1-2n+1=fracn+1n+1+frac-2n+1=1-frac2n+1$. If I happened to have that expression equal to something else, I only need to modify first
– JMoravitz
4 hours ago
your idea is correct, when we apply such rule we think that x is some fixed number thus 5x-4 and 2x+5 are also some fixed numbers thus what you did here you say : "Suppose 5x - 4 = 2x+5 for some real number x, then 3*(5x-4) = 3*(2x+5)" here you apply a theorem that says : "for any real numbers a,b if a = b then ac = bc for any real number c" thus having 3(5x-4)=3(2x+5)
– famesyasd
4 hours ago
This is essentially the same question you posted a few hours ago. You could have just edited that one, instead.
– dxiv
3 hours ago
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
10 mins ago
add a comment |Â
2
The name to use is expression rather than term. One expression is equal to another. Yes, if you wish to multiply both sides by $3$ you would do so to each expression. To answer your question of if you would ever not think of each side as a larger piece, sure... consider situations where you want to multiply by $1$ or add $0$. Take the expression $fracn-1n+1$. You could rewrite this as $fracn-1 +0n+1=fracn-1+2-2n+1=fracn+1-2n+1=fracn+1n+1+frac-2n+1=1-frac2n+1$. If I happened to have that expression equal to something else, I only need to modify first
– JMoravitz
4 hours ago
your idea is correct, when we apply such rule we think that x is some fixed number thus 5x-4 and 2x+5 are also some fixed numbers thus what you did here you say : "Suppose 5x - 4 = 2x+5 for some real number x, then 3*(5x-4) = 3*(2x+5)" here you apply a theorem that says : "for any real numbers a,b if a = b then ac = bc for any real number c" thus having 3(5x-4)=3(2x+5)
– famesyasd
4 hours ago
This is essentially the same question you posted a few hours ago. You could have just edited that one, instead.
– dxiv
3 hours ago
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
10 mins ago
2
2
The name to use is expression rather than term. One expression is equal to another. Yes, if you wish to multiply both sides by $3$ you would do so to each expression. To answer your question of if you would ever not think of each side as a larger piece, sure... consider situations where you want to multiply by $1$ or add $0$. Take the expression $fracn-1n+1$. You could rewrite this as $fracn-1 +0n+1=fracn-1+2-2n+1=fracn+1-2n+1=fracn+1n+1+frac-2n+1=1-frac2n+1$. If I happened to have that expression equal to something else, I only need to modify first
– JMoravitz
4 hours ago
The name to use is expression rather than term. One expression is equal to another. Yes, if you wish to multiply both sides by $3$ you would do so to each expression. To answer your question of if you would ever not think of each side as a larger piece, sure... consider situations where you want to multiply by $1$ or add $0$. Take the expression $fracn-1n+1$. You could rewrite this as $fracn-1 +0n+1=fracn-1+2-2n+1=fracn+1-2n+1=fracn+1n+1+frac-2n+1=1-frac2n+1$. If I happened to have that expression equal to something else, I only need to modify first
– JMoravitz
4 hours ago
your idea is correct, when we apply such rule we think that x is some fixed number thus 5x-4 and 2x+5 are also some fixed numbers thus what you did here you say : "Suppose 5x - 4 = 2x+5 for some real number x, then 3*(5x-4) = 3*(2x+5)" here you apply a theorem that says : "for any real numbers a,b if a = b then ac = bc for any real number c" thus having 3(5x-4)=3(2x+5)
– famesyasd
4 hours ago
your idea is correct, when we apply such rule we think that x is some fixed number thus 5x-4 and 2x+5 are also some fixed numbers thus what you did here you say : "Suppose 5x - 4 = 2x+5 for some real number x, then 3*(5x-4) = 3*(2x+5)" here you apply a theorem that says : "for any real numbers a,b if a = b then ac = bc for any real number c" thus having 3(5x-4)=3(2x+5)
– famesyasd
4 hours ago
This is essentially the same question you posted a few hours ago. You could have just edited that one, instead.
– dxiv
3 hours ago
This is essentially the same question you posted a few hours ago. You could have just edited that one, instead.
– dxiv
3 hours ago
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
10 mins ago
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
10 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
A deeper perspective on what you're missing is the notion of substitution.
Here is a toy example.
Suppose that
$$a+colorblueb=ctag1$$
And suppose as well that
$$colorblueb=colorblued+etag2$$
Then, because $b$ equals $d+e$, we may substitute $d+e$ for $b$ in the first equation (indeed, in any equation containing $b$), giving
$$a+underbracecolorblued+e_b=ctag3$$
Thus, informally speaking, you might say that in moving from (1) to (3) we have "manipulated one term in the left side of equation (1) without manipulating the whole left side, and without manipulating the right side at all." This is correct, and a perfectly valid way of reasoning.
On the other hand, apparently you have heard it said there is a "rule" that you must "always do the same thing to both sides of an equation." This is both imprecise and wrong. As we saw above, it is logically valid to "do something" to one side of an equation -- even part of one side of an equation -- without doing something to the other side.
What the rule means to say is much wordier: if you apply an arithmetic operation (addition, subtraction, multiplication, division, exponentiation, substitution into a trig function, whatever) to the entire expression on one side of an equation, you must perform that operation on the entire expression on the other side, too, in order to preserve the equality.
Even more precisely: if $a=b$, then $f(a)=f(b)$, no matter the function $f$.
In this case, we are substituting $a$ and $b$ as inputs into some function and deducing that the outputs of the function are equal. For more on this perspective as it relates to equation-solving, see my answer here.
A postscript: one way you can tell the "rule" is wrong is that it says to always do something, period.
But mathematical theory doesn't dictate behavior ex cathedra: it doesn't issue statements of the form "do this." That's what pseudocode for a computer program says; it's not what mathematics says.
Rather, mathematics says do this if you want to be consistent, or draw a certain conclusion, or whatever. What you do in mathematics depends on your goal. In order to obtain such-and-such a goal, you should do such-and-such.
You will be a better mathematical thinker if you constantly scrutinize procedural or imperative commands from your teachers -- do such-and-such -- for their real mathematical meaning. Procedures are only important to the extent they help us achieve a certain goal.
answered 3 hours ago
symplectomorphic
11.7k21838
add a comment |Â
up vote
0
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Yes, this is the most common way of thinking of the two sides of an expression. There is no case when you can have an equality, in your case $5x-4=2x+5$, and multiply only one part of each side by a constant. As for a reason why this is impossible, if one does this it will change the solution set of the expression, which is essentially another way to identify the expression.
answered 4 hours ago
Tyler6
575210
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
A deeper perspective on what you're missing is the notion of substitution.
Here is a toy example.
Suppose that
$$a+colorblueb=ctag1$$
And suppose as well that
$$colorblueb=colorblued+etag2$$
Then, because $b$ equals $d+e$, we may substitute $d+e$ for $b$ in the first equation (indeed, in any equation containing $b$), giving
$$a+underbracecolorblued+e_b=ctag3$$
Thus, informally speaking, you might say that in moving from (1) to (3) we have "manipulated one term in the left side of equation (1) without manipulating the whole left side, and without manipulating the right side at all." This is correct, and a perfectly valid way of reasoning.
On the other hand, apparently you have heard it said there is a "rule" that you must "always do the same thing to both sides of an equation." This is both imprecise and wrong. As we saw above, it is logically valid to "do something" to one side of an equation -- even part of one side of an equation -- without doing something to the other side.
What the rule means to say is much wordier: if you apply an arithmetic operation (addition, subtraction, multiplication, division, exponentiation, substitution into a trig function, whatever) to the entire expression on one side of an equation, you must perform that operation on the entire expression on the other side, too, in order to preserve the equality.
Even more precisely: if $a=b$, then $f(a)=f(b)$, no matter the function $f$.
In this case, we are substituting $a$ and $b$ as inputs into some function and deducing that the outputs of the function are equal. For more on this perspective as it relates to equation-solving, see my answer here.
A postscript: one way you can tell the "rule" is wrong is that it says to always do something, period.
But mathematical theory doesn't dictate behavior ex cathedra: it doesn't issue statements of the form "do this." That's what pseudocode for a computer program says; it's not what mathematics says.
Rather, mathematics says do this if you want to be consistent, or draw a certain conclusion, or whatever. What you do in mathematics depends on your goal. In order to obtain such-and-such a goal, you should do such-and-such.
You will be a better mathematical thinker if you constantly scrutinize procedural or imperative commands from your teachers -- do such-and-such -- for their real mathematical meaning. Procedures are only important to the extent they help us achieve a certain goal.
answered 3 hours ago
symplectomorphic
11.7k21838
add a comment |Â
up vote
5
down vote
A deeper perspective on what you're missing is the notion of substitution.
Here is a toy example.
Suppose that
$$a+colorblueb=ctag1$$
And suppose as well that
$$colorblueb=colorblued+etag2$$
Then, because $b$ equals $d+e$, we may substitute $d+e$ for $b$ in the first equation (indeed, in any equation containing $b$), giving
$$a+underbracecolorblued+e_b=ctag3$$
Thus, informally speaking, you might say that in moving from (1) to (3) we have "manipulated one term in the left side of equation (1) without manipulating the whole left side, and without manipulating the right side at all." This is correct, and a perfectly valid way of reasoning.
On the other hand, apparently you have heard it said there is a "rule" that you must "always do the same thing to both sides of an equation." This is both imprecise and wrong. As we saw above, it is logically valid to "do something" to one side of an equation -- even part of one side of an equation -- without doing something to the other side.
What the rule means to say is much wordier: if you apply an arithmetic operation (addition, subtraction, multiplication, division, exponentiation, substitution into a trig function, whatever) to the entire expression on one side of an equation, you must perform that operation on the entire expression on the other side, too, in order to preserve the equality.
Even more precisely: if $a=b$, then $f(a)=f(b)$, no matter the function $f$.
In this case, we are substituting $a$ and $b$ as inputs into some function and deducing that the outputs of the function are equal. For more on this perspective as it relates to equation-solving, see my answer here.
A postscript: one way you can tell the "rule" is wrong is that it says to always do something, period.
But mathematical theory doesn't dictate behavior ex cathedra: it doesn't issue statements of the form "do this." That's what pseudocode for a computer program says; it's not what mathematics says.
Rather, mathematics says do this if you want to be consistent, or draw a certain conclusion, or whatever. What you do in mathematics depends on your goal. In order to obtain such-and-such a goal, you should do such-and-such.
You will be a better mathematical thinker if you constantly scrutinize procedural or imperative commands from your teachers -- do such-and-such -- for their real mathematical meaning. Procedures are only important to the extent they help us achieve a certain goal.
answered 3 hours ago
symplectomorphic
11.7k21838
add a comment |Â
up vote
5
down vote
up vote
5
down vote
A deeper perspective on what you're missing is the notion of substitution.
Here is a toy example.
Suppose that
$$a+colorblueb=ctag1$$
And suppose as well that
$$colorblueb=colorblued+etag2$$
Then, because $b$ equals $d+e$, we may substitute $d+e$ for $b$ in the first equation (indeed, in any equation containing $b$), giving
$$a+underbracecolorblued+e_b=ctag3$$
Thus, informally speaking, you might say that in moving from (1) to (3) we have "manipulated one term in the left side of equation (1) without manipulating the whole left side, and without manipulating the right side at all." This is correct, and a perfectly valid way of reasoning.
On the other hand, apparently you have heard it said there is a "rule" that you must "always do the same thing to both sides of an equation." This is both imprecise and wrong. As we saw above, it is logically valid to "do something" to one side of an equation -- even part of one side of an equation -- without doing something to the other side.
What the rule means to say is much wordier: if you apply an arithmetic operation (addition, subtraction, multiplication, division, exponentiation, substitution into a trig function, whatever) to the entire expression on one side of an equation, you must perform that operation on the entire expression on the other side, too, in order to preserve the equality.
Even more precisely: if $a=b$, then $f(a)=f(b)$, no matter the function $f$.
In this case, we are substituting $a$ and $b$ as inputs into some function and deducing that the outputs of the function are equal. For more on this perspective as it relates to equation-solving, see my answer here.
A postscript: one way you can tell the "rule" is wrong is that it says to always do something, period.
But mathematical theory doesn't dictate behavior ex cathedra: it doesn't issue statements of the form "do this." That's what pseudocode for a computer program says; it's not what mathematics says.
Rather, mathematics says do this if you want to be consistent, or draw a certain conclusion, or whatever. What you do in mathematics depends on your goal. In order to obtain such-and-such a goal, you should do such-and-such.
You will be a better mathematical thinker if you constantly scrutinize procedural or imperative commands from your teachers -- do such-and-such -- for their real mathematical meaning. Procedures are only important to the extent they help us achieve a certain goal.
answered 3 hours ago
symplectomorphic
11.7k21838
A deeper perspective on what you're missing is the notion of substitution.
Here is a toy example.
Suppose that
$$a+colorblueb=ctag1$$
And suppose as well that
$$colorblueb=colorblued+etag2$$
Then, because $b$ equals $d+e$, we may substitute $d+e$ for $b$ in the first equation (indeed, in any equation containing $b$), giving
$$a+underbracecolorblued+e_b=ctag3$$
Thus, informally speaking, you might say that in moving from (1) to (3) we have "manipulated one term in the left side of equation (1) without manipulating the whole left side, and without manipulating the right side at all." This is correct, and a perfectly valid way of reasoning.
On the other hand, apparently you have heard it said there is a "rule" that you must "always do the same thing to both sides of an equation." This is both imprecise and wrong. As we saw above, it is logically valid to "do something" to one side of an equation -- even part of one side of an equation -- without doing something to the other side.
What the rule means to say is much wordier: if you apply an arithmetic operation (addition, subtraction, multiplication, division, exponentiation, substitution into a trig function, whatever) to the entire expression on one side of an equation, you must perform that operation on the entire expression on the other side, too, in order to preserve the equality.
Even more precisely: if $a=b$, then $f(a)=f(b)$, no matter the function $f$.
In this case, we are substituting $a$ and $b$ as inputs into some function and deducing that the outputs of the function are equal. For more on this perspective as it relates to equation-solving, see my answer here.
A postscript: one way you can tell the "rule" is wrong is that it says to always do something, period.
But mathematical theory doesn't dictate behavior ex cathedra: it doesn't issue statements of the form "do this." That's what pseudocode for a computer program says; it's not what mathematics says.
Rather, mathematics says do this if you want to be consistent, or draw a certain conclusion, or whatever. What you do in mathematics depends on your goal. In order to obtain such-and-such a goal, you should do such-and-such.
You will be a better mathematical thinker if you constantly scrutinize procedural or imperative commands from your teachers -- do such-and-such -- for their real mathematical meaning. Procedures are only important to the extent they help us achieve a certain goal.
answered 3 hours ago
symplectomorphic
11.7k21838
edited 3 hours ago
answered 3 hours ago
symplectomorphic
11.7k21838
answered 3 hours ago
symplectomorphic
11.7k21838
answered 3 hours ago
symplectomorphic
11.7k21838
11.7k21838
add a comment |Â
add a comment |Â
up vote
0
down vote
Yes, this is the most common way of thinking of the two sides of an expression. There is no case when you can have an equality, in your case $5x-4=2x+5$, and multiply only one part of each side by a constant. As for a reason why this is impossible, if one does this it will change the solution set of the expression, which is essentially another way to identify the expression.
answered 4 hours ago
Tyler6
575210
add a comment |Â
up vote
0
down vote
Yes, this is the most common way of thinking of the two sides of an expression. There is no case when you can have an equality, in your case $5x-4=2x+5$, and multiply only one part of each side by a constant. As for a reason why this is impossible, if one does this it will change the solution set of the expression, which is essentially another way to identify the expression.
answered 4 hours ago
Tyler6
575210
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes, this is the most common way of thinking of the two sides of an expression. There is no case when you can have an equality, in your case $5x-4=2x+5$, and multiply only one part of each side by a constant. As for a reason why this is impossible, if one does this it will change the solution set of the expression, which is essentially another way to identify the expression.
answered 4 hours ago
Tyler6
575210
Yes, this is the most common way of thinking of the two sides of an expression. There is no case when you can have an equality, in your case $5x-4=2x+5$, and multiply only one part of each side by a constant. As for a reason why this is impossible, if one does this it will change the solution set of the expression, which is essentially another way to identify the expression.
answered 4 hours ago
Tyler6
575210
answered 4 hours ago
Tyler6
575210
answered 4 hours ago
Tyler6
575210
answered 4 hours ago
Tyler6
575210
575210
add a comment |Â
add a comment |Â
Marlon Henderson is a new contributor. Be nice, and check out our Code of Conduct.
Marlon Henderson is a new contributor. Be nice, and check out our Code of Conduct.
Marlon Henderson is a new contributor. Be nice, and check out our Code of Conduct.
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2
The name to use is expression rather than term. One expression is equal to another. Yes, if you wish to multiply both sides by $3$ you would do so to each expression. To answer your question of if you would ever not think of each side as a larger piece, sure... consider situations where you want to multiply by $1$ or add $0$. Take the expression $fracn-1n+1$. You could rewrite this as $fracn-1 +0n+1=fracn-1+2-2n+1=fracn+1-2n+1=fracn+1n+1+frac-2n+1=1-frac2n+1$. If I happened to have that expression equal to something else, I only need to modify first
– JMoravitz
4 hours ago
your idea is correct, when we apply such rule we think that x is some fixed number thus 5x-4 and 2x+5 are also some fixed numbers thus what you did here you say : "Suppose 5x - 4 = 2x+5 for some real number x, then 3*(5x-4) = 3*(2x+5)" here you apply a theorem that says : "for any real numbers a,b if a = b then ac = bc for any real number c" thus having 3(5x-4)=3(2x+5)
– famesyasd
4 hours ago
This is essentially the same question you posted a few hours ago. You could have just edited that one, instead.
– dxiv
3 hours ago
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
10 mins ago