How to solve first-degree equation with two or more numbers to the right of the $=$? [closed]

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I can solve some equations of the first degree when it has a right number of the symbol $=$ like this one. For example,

$$
2x+2=12
$$

But when the equation has two or more numbers I cannot solve it. Look at this equation. For example,

$$
23x-16=14-17x
$$

I get confused when I try to solve an equation like that. How could I learn solve it?

Please excuse my English.

algebra-precalculus

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edited Jul 30 at 22:07

Math Lover

12.2k21132

asked Jul 30 at 21:44

gato

1064

closed as off-topic by José Carlos Santos, max_zorn, Xander Henderson, Isaac Browne, Rhys Steele Jul 31 at 10:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, max_zorn, Xander Henderson, Isaac Browne, Rhys Steele

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just move the constant terms on the right side, linear term on the left side
  – Mulliganaceous
  Jul 30 at 21:50
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If you have an equation, you may add the same thing to both sides and this will not change the solution to the equation. For $2x+2=12$ this is seen by "adding $-2$ to both sides" giving $2x+2=12implies 2x+2+colorred(-2)=12+colorred(-2)$ which simplifies to $2x=10$. The same technique is used in the second problem you mention, here adding $16$ to both sides and adding $17x$ to both sides, noting that what you add to both sides does not have to be limited to just constants, but you may add variables as well.
  – JMoravitz
  Jul 30 at 21:50
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

I can solve some equations of the first degree when it has a right number of the symbol $=$ like this one. For example,

$$
2x+2=12
$$

But when the equation has two or more numbers I cannot solve it. Look at this equation. For example,

$$
23x-16=14-17x
$$

I get confused when I try to solve an equation like that. How could I learn solve it?

Please excuse my English.

algebra-precalculus

share|cite|improve this question

edited Jul 30 at 22:07

Math Lover

12.2k21132

asked Jul 30 at 21:44

gato

1064

closed as off-topic by José Carlos Santos, max_zorn, Xander Henderson, Isaac Browne, Rhys Steele Jul 31 at 10:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, max_zorn, Xander Henderson, Isaac Browne, Rhys Steele

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just move the constant terms on the right side, linear term on the left side
  – Mulliganaceous
  Jul 30 at 21:50
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If you have an equation, you may add the same thing to both sides and this will not change the solution to the equation. For $2x+2=12$ this is seen by "adding $-2$ to both sides" giving $2x+2=12implies 2x+2+colorred(-2)=12+colorred(-2)$ which simplifies to $2x=10$. The same technique is used in the second problem you mention, here adding $16$ to both sides and adding $17x$ to both sides, noting that what you add to both sides does not have to be limited to just constants, but you may add variables as well.
  – JMoravitz
  Jul 30 at 21:50
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I can solve some equations of the first degree when it has a right number of the symbol $=$ like this one. For example,

$$
2x+2=12
$$

But when the equation has two or more numbers I cannot solve it. Look at this equation. For example,

$$
23x-16=14-17x
$$

I get confused when I try to solve an equation like that. How could I learn solve it?

Please excuse my English.

algebra-precalculus

share|cite|improve this question

edited Jul 30 at 22:07

Math Lover

12.2k21132

asked Jul 30 at 21:44

gato

1064

I can solve some equations of the first degree when it has a right number of the symbol $=$ like this one. For example,

$$
2x+2=12
$$

But when the equation has two or more numbers I cannot solve it. Look at this equation. For example,

$$
23x-16=14-17x
$$

I get confused when I try to solve an equation like that. How could I learn solve it?

Please excuse my English.

algebra-precalculus

share|cite|improve this question

edited Jul 30 at 22:07

Math Lover

12.2k21132

asked Jul 30 at 21:44

gato

1064

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 22:07

Math Lover

12.2k21132

edited Jul 30 at 22:07

Math Lover

12.2k21132

edited Jul 30 at 22:07

Math Lover

12.2k21132

12.2k21132

asked Jul 30 at 21:44

gato

1064

asked Jul 30 at 21:44

gato

1064

asked Jul 30 at 21:44

gato

1064

1064

closed as off-topic by José Carlos Santos, max_zorn, Xander Henderson, Isaac Browne, Rhys Steele Jul 31 at 10:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, max_zorn, Xander Henderson, Isaac Browne, Rhys Steele

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, max_zorn, Xander Henderson, Isaac Browne, Rhys Steele Jul 31 at 10:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, max_zorn, Xander Henderson, Isaac Browne, Rhys Steele

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just move the constant terms on the right side, linear term on the left side
  – Mulliganaceous
  Jul 30 at 21:50
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If you have an equation, you may add the same thing to both sides and this will not change the solution to the equation. For $2x+2=12$ this is seen by "adding $-2$ to both sides" giving $2x+2=12implies 2x+2+colorred(-2)=12+colorred(-2)$ which simplifies to $2x=10$. The same technique is used in the second problem you mention, here adding $16$ to both sides and adding $17x$ to both sides, noting that what you add to both sides does not have to be limited to just constants, but you may add variables as well.
  – JMoravitz
  Jul 30 at 21:50
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just move the constant terms on the right side, linear term on the left side
  – Mulliganaceous
  Jul 30 at 21:50
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If you have an equation, you may add the same thing to both sides and this will not change the solution to the equation. For $2x+2=12$ this is seen by "adding $-2$ to both sides" giving $2x+2=12implies 2x+2+colorred(-2)=12+colorred(-2)$ which simplifies to $2x=10$. The same technique is used in the second problem you mention, here adding $16$ to both sides and adding $17x$ to both sides, noting that what you add to both sides does not have to be limited to just constants, but you may add variables as well.
  – JMoravitz
  Jul 30 at 21:50
  
  

  
  
  
  

1

1

Just move the constant terms on the right side, linear term on the left side
– Mulliganaceous
Jul 30 at 21:50

Just move the constant terms on the right side, linear term on the left side
– Mulliganaceous
Jul 30 at 21:50

2

2

If you have an equation, you may add the same thing to both sides and this will not change the solution to the equation. For $2x+2=12$ this is seen by "adding $-2$ to both sides" giving $2x+2=12implies 2x+2+colorred(-2)=12+colorred(-2)$ which simplifies to $2x=10$. The same technique is used in the second problem you mention, here adding $16$ to both sides and adding $17x$ to both sides, noting that what you add to both sides does not have to be limited to just constants, but you may add variables as well.
– JMoravitz
Jul 30 at 21:50

If you have an equation, you may add the same thing to both sides and this will not change the solution to the equation. For $2x+2=12$ this is seen by "adding $-2$ to both sides" giving $2x+2=12implies 2x+2+colorred(-2)=12+colorred(-2)$ which simplifies to $2x=10$. The same technique is used in the second problem you mention, here adding $16$ to both sides and adding $17x$ to both sides, noting that what you add to both sides does not have to be limited to just constants, but you may add variables as well.
– JMoravitz
Jul 30 at 21:50

add a comment | 

5 Answers
5

active

oldest

votes

up vote
4
down vote



accepted

To solve this kind of equations you relay on two "tricks" that let's you go through various equivalent equations all the way to the solution. This two "tricks" are listed here.

Basically they say this (note that LHS stands for Left Hand Side and RHS for Right Hand Side)

1. Adding or subtracting a same quantity to the LHS and RHS of an equation, generates an equivalent equation.




2. Multiplying or dividing by the same quantity (different from zero) the LHS and RHS of an equation, generates an equivalent equation

By equivalent equation I mean an equation that has the same solution as the one you started with. Let's see how this principles are useful to solve your equation (I'll refer them as $(1)$ and $(2)$) $$beginalign&23x-16=14-17x \ &overset(1)longrightarrow 23xcolorred+17x-16=14-17xcolorred+17x \ &overset(1)longrightarrow 40x-16 colorred+16=14colorred+16 \&longrightarrow 40x = 30 \&overset(2)longrightarrowfrac40xcolorblue40 = frac30colorblue40 \&longrightarrow x = frac34endalign$$ so as you can see by only invoking this two principles you can go through many equivalent equations all the way to the simples equation of the form $x=b$ which is the solution to your problem!

You could ask why I choose to add, subtract, divide and multiplay by a specific number and the reason is to be found, to put it simple, in the fact that to solve an equation you want to have all the terms with the $x$ on one side and all the numbers on the other. So for example in the fist step I don't want that $+17x$ factor on the RHS so I try to remove it: what is a better way to remove it if not annihilate it by subtracting the same factor from both sides? By this means you get $+17x-17x=0$ on the RHS as we wanted.

In fact the same first rule that I told you before can be simplified in this manner

1. Moving a term from the RHS to the LHS, and vice versa, in an equation, changes the sing of that term.

share|cite|improve this answer

edited Jul 30 at 22:10

answered Jul 30 at 21:57

Davide Morgante

1,695220

add a comment | 

up vote
3
down vote

Hint: You can move some of the terms on the right to the left. Using your example,

$$23x-16=14-17ximplies 40x-16=14.$$

share|cite|improve this answer

answered Jul 30 at 21:47

Carl Schildkraut

8,23711238

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Note for OP: This is achieved here by adding $17x$ to both sides of the equation. Provided you do the same to both sides of the equation you can add or subtract any term you like. You can multiply or divide too - but you have to take care not to divide by zero, and multiplying by zero just gives the equation $0=0$.
  – Mark Bennet
  Jul 30 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the meaning of symbol $implies$?
  – gato
  Jul 30 at 22:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $Aimplies B$ is read as "A implies B." It means that if $A$ is true, then $B$ is true as well.
  – Carl Schildkraut
  Jul 30 at 22:20
  

  
  
  
  

add a comment | 

up vote
2
down vote

Hint:

Add $17x + 16$ to both sides of the equation (to move all quantities containing $x$ to one side and the constants to the other). Then divide by $40$ to get $x$.

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answered Jul 30 at 21:55

RayDansh

867113

add a comment | 

up vote
1
down vote

We have that

$$23x-16=14-17x$$

$$23x-16colorred+17x+16=14-17xcolorred+17x+16$$

$$40x=30$$

$$frac40xcolorred40=frac30colorred40$$

$$x=frac34$$

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answered Jul 30 at 21:54

gimusi

64.1k73480

add a comment | 

up vote
1
down vote

Golden Rule : Jump over the equal sign and change the sign.

Example $1$: $$3x+12=2x+20$$

$$3x-2x=20-12$$

$$x=8$$

Example $2$

$$ -3x+12-25=x-4+10$$

$$-3x-x=-4+10-12+25$$

$$-4x=19$$

$$x=-19/4$$

share|cite|improve this answer

answered Jul 30 at 22:28

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

To solve this kind of equations you relay on two "tricks" that let's you go through various equivalent equations all the way to the solution. This two "tricks" are listed here.

Basically they say this (note that LHS stands for Left Hand Side and RHS for Right Hand Side)

1. Adding or subtracting a same quantity to the LHS and RHS of an equation, generates an equivalent equation.




2. Multiplying or dividing by the same quantity (different from zero) the LHS and RHS of an equation, generates an equivalent equation

By equivalent equation I mean an equation that has the same solution as the one you started with. Let's see how this principles are useful to solve your equation (I'll refer them as $(1)$ and $(2)$) $$beginalign&23x-16=14-17x \ &overset(1)longrightarrow 23xcolorred+17x-16=14-17xcolorred+17x \ &overset(1)longrightarrow 40x-16 colorred+16=14colorred+16 \&longrightarrow 40x = 30 \&overset(2)longrightarrowfrac40xcolorblue40 = frac30colorblue40 \&longrightarrow x = frac34endalign$$ so as you can see by only invoking this two principles you can go through many equivalent equations all the way to the simples equation of the form $x=b$ which is the solution to your problem!

You could ask why I choose to add, subtract, divide and multiplay by a specific number and the reason is to be found, to put it simple, in the fact that to solve an equation you want to have all the terms with the $x$ on one side and all the numbers on the other. So for example in the fist step I don't want that $+17x$ factor on the RHS so I try to remove it: what is a better way to remove it if not annihilate it by subtracting the same factor from both sides? By this means you get $+17x-17x=0$ on the RHS as we wanted.

In fact the same first rule that I told you before can be simplified in this manner

1. Moving a term from the RHS to the LHS, and vice versa, in an equation, changes the sing of that term.

share|cite|improve this answer

edited Jul 30 at 22:10

answered Jul 30 at 21:57

Davide Morgante

1,695220

add a comment | 

up vote
4
down vote



accepted

To solve this kind of equations you relay on two "tricks" that let's you go through various equivalent equations all the way to the solution. This two "tricks" are listed here.

Basically they say this (note that LHS stands for Left Hand Side and RHS for Right Hand Side)

1. Adding or subtracting a same quantity to the LHS and RHS of an equation, generates an equivalent equation.




2. Multiplying or dividing by the same quantity (different from zero) the LHS and RHS of an equation, generates an equivalent equation

By equivalent equation I mean an equation that has the same solution as the one you started with. Let's see how this principles are useful to solve your equation (I'll refer them as $(1)$ and $(2)$) $$beginalign&23x-16=14-17x \ &overset(1)longrightarrow 23xcolorred+17x-16=14-17xcolorred+17x \ &overset(1)longrightarrow 40x-16 colorred+16=14colorred+16 \&longrightarrow 40x = 30 \&overset(2)longrightarrowfrac40xcolorblue40 = frac30colorblue40 \&longrightarrow x = frac34endalign$$ so as you can see by only invoking this two principles you can go through many equivalent equations all the way to the simples equation of the form $x=b$ which is the solution to your problem!

You could ask why I choose to add, subtract, divide and multiplay by a specific number and the reason is to be found, to put it simple, in the fact that to solve an equation you want to have all the terms with the $x$ on one side and all the numbers on the other. So for example in the fist step I don't want that $+17x$ factor on the RHS so I try to remove it: what is a better way to remove it if not annihilate it by subtracting the same factor from both sides? By this means you get $+17x-17x=0$ on the RHS as we wanted.

In fact the same first rule that I told you before can be simplified in this manner

1. Moving a term from the RHS to the LHS, and vice versa, in an equation, changes the sing of that term.

share|cite|improve this answer

edited Jul 30 at 22:10

answered Jul 30 at 21:57

Davide Morgante

1,695220

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

To solve this kind of equations you relay on two "tricks" that let's you go through various equivalent equations all the way to the solution. This two "tricks" are listed here.

Basically they say this (note that LHS stands for Left Hand Side and RHS for Right Hand Side)

1. Adding or subtracting a same quantity to the LHS and RHS of an equation, generates an equivalent equation.




2. Multiplying or dividing by the same quantity (different from zero) the LHS and RHS of an equation, generates an equivalent equation

By equivalent equation I mean an equation that has the same solution as the one you started with. Let's see how this principles are useful to solve your equation (I'll refer them as $(1)$ and $(2)$) $$beginalign&23x-16=14-17x \ &overset(1)longrightarrow 23xcolorred+17x-16=14-17xcolorred+17x \ &overset(1)longrightarrow 40x-16 colorred+16=14colorred+16 \&longrightarrow 40x = 30 \&overset(2)longrightarrowfrac40xcolorblue40 = frac30colorblue40 \&longrightarrow x = frac34endalign$$ so as you can see by only invoking this two principles you can go through many equivalent equations all the way to the simples equation of the form $x=b$ which is the solution to your problem!

You could ask why I choose to add, subtract, divide and multiplay by a specific number and the reason is to be found, to put it simple, in the fact that to solve an equation you want to have all the terms with the $x$ on one side and all the numbers on the other. So for example in the fist step I don't want that $+17x$ factor on the RHS so I try to remove it: what is a better way to remove it if not annihilate it by subtracting the same factor from both sides? By this means you get $+17x-17x=0$ on the RHS as we wanted.

In fact the same first rule that I told you before can be simplified in this manner

1. Moving a term from the RHS to the LHS, and vice versa, in an equation, changes the sing of that term.

share|cite|improve this answer

edited Jul 30 at 22:10

answered Jul 30 at 21:57

Davide Morgante

1,695220

To solve this kind of equations you relay on two "tricks" that let's you go through various equivalent equations all the way to the solution. This two "tricks" are listed here.

Basically they say this (note that LHS stands for Left Hand Side and RHS for Right Hand Side)

1. Adding or subtracting a same quantity to the LHS and RHS of an equation, generates an equivalent equation.




2. Multiplying or dividing by the same quantity (different from zero) the LHS and RHS of an equation, generates an equivalent equation

By equivalent equation I mean an equation that has the same solution as the one you started with. Let's see how this principles are useful to solve your equation (I'll refer them as $(1)$ and $(2)$) $$beginalign&23x-16=14-17x \ &overset(1)longrightarrow 23xcolorred+17x-16=14-17xcolorred+17x \ &overset(1)longrightarrow 40x-16 colorred+16=14colorred+16 \&longrightarrow 40x = 30 \&overset(2)longrightarrowfrac40xcolorblue40 = frac30colorblue40 \&longrightarrow x = frac34endalign$$ so as you can see by only invoking this two principles you can go through many equivalent equations all the way to the simples equation of the form $x=b$ which is the solution to your problem!

You could ask why I choose to add, subtract, divide and multiplay by a specific number and the reason is to be found, to put it simple, in the fact that to solve an equation you want to have all the terms with the $x$ on one side and all the numbers on the other. So for example in the fist step I don't want that $+17x$ factor on the RHS so I try to remove it: what is a better way to remove it if not annihilate it by subtracting the same factor from both sides? By this means you get $+17x-17x=0$ on the RHS as we wanted.

In fact the same first rule that I told you before can be simplified in this manner

1. Moving a term from the RHS to the LHS, and vice versa, in an equation, changes the sing of that term.

share|cite|improve this answer

edited Jul 30 at 22:10

answered Jul 30 at 21:57

Davide Morgante

1,695220

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 22:10

edited Jul 30 at 22:10

edited Jul 30 at 22:10

answered Jul 30 at 21:57

Davide Morgante

1,695220

answered Jul 30 at 21:57

Davide Morgante

1,695220

answered Jul 30 at 21:57

Davide Morgante

1,695220

1,695220

add a comment | 

add a comment | 

up vote
3
down vote

Hint: You can move some of the terms on the right to the left. Using your example,

$$23x-16=14-17ximplies 40x-16=14.$$

share|cite|improve this answer

answered Jul 30 at 21:47

Carl Schildkraut

8,23711238

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Note for OP: This is achieved here by adding $17x$ to both sides of the equation. Provided you do the same to both sides of the equation you can add or subtract any term you like. You can multiply or divide too - but you have to take care not to divide by zero, and multiplying by zero just gives the equation $0=0$.
  – Mark Bennet
  Jul 30 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the meaning of symbol $implies$?
  – gato
  Jul 30 at 22:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $Aimplies B$ is read as "A implies B." It means that if $A$ is true, then $B$ is true as well.
  – Carl Schildkraut
  Jul 30 at 22:20
  

  
  
  
  

add a comment | 

up vote
3
down vote

Hint: You can move some of the terms on the right to the left. Using your example,

$$23x-16=14-17ximplies 40x-16=14.$$

share|cite|improve this answer

answered Jul 30 at 21:47

Carl Schildkraut

8,23711238

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Note for OP: This is achieved here by adding $17x$ to both sides of the equation. Provided you do the same to both sides of the equation you can add or subtract any term you like. You can multiply or divide too - but you have to take care not to divide by zero, and multiplying by zero just gives the equation $0=0$.
  – Mark Bennet
  Jul 30 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the meaning of symbol $implies$?
  – gato
  Jul 30 at 22:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $Aimplies B$ is read as "A implies B." It means that if $A$ is true, then $B$ is true as well.
  – Carl Schildkraut
  Jul 30 at 22:20
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Hint: You can move some of the terms on the right to the left. Using your example,

$$23x-16=14-17ximplies 40x-16=14.$$

share|cite|improve this answer

answered Jul 30 at 21:47

Carl Schildkraut

8,23711238

Hint: You can move some of the terms on the right to the left. Using your example,

$$23x-16=14-17ximplies 40x-16=14.$$

share|cite|improve this answer

answered Jul 30 at 21:47

Carl Schildkraut

8,23711238

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 21:47

Carl Schildkraut

8,23711238

answered Jul 30 at 21:47

Carl Schildkraut

8,23711238

answered Jul 30 at 21:47

Carl Schildkraut

8,23711238

8,23711238

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Note for OP: This is achieved here by adding $17x$ to both sides of the equation. Provided you do the same to both sides of the equation you can add or subtract any term you like. You can multiply or divide too - but you have to take care not to divide by zero, and multiplying by zero just gives the equation $0=0$.
  – Mark Bennet
  Jul 30 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the meaning of symbol $implies$?
  – gato
  Jul 30 at 22:19
  

  
  
  
  


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  $Aimplies B$ is read as "A implies B." It means that if $A$ is true, then $B$ is true as well.
  – Carl Schildkraut
  Jul 30 at 22:20
  

  
  
  
  

add a comment | 

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  Note for OP: This is achieved here by adding $17x$ to both sides of the equation. Provided you do the same to both sides of the equation you can add or subtract any term you like. You can multiply or divide too - but you have to take care not to divide by zero, and multiplying by zero just gives the equation $0=0$.
  – Mark Bennet
  Jul 30 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the meaning of symbol $implies$?
  – gato
  Jul 30 at 22:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $Aimplies B$ is read as "A implies B." It means that if $A$ is true, then $B$ is true as well.
  – Carl Schildkraut
  Jul 30 at 22:20
  

  
  
  
  

4

4

Note for OP: This is achieved here by adding $17x$ to both sides of the equation. Provided you do the same to both sides of the equation you can add or subtract any term you like. You can multiply or divide too - but you have to take care not to divide by zero, and multiplying by zero just gives the equation $0=0$.
– Mark Bennet
Jul 30 at 21:53

Note for OP: This is achieved here by adding $17x$ to both sides of the equation. Provided you do the same to both sides of the equation you can add or subtract any term you like. You can multiply or divide too - but you have to take care not to divide by zero, and multiplying by zero just gives the equation $0=0$.
– Mark Bennet
Jul 30 at 21:53

What is the meaning of symbol $implies$?
– gato
Jul 30 at 22:19

What is the meaning of symbol $implies$?
– gato
Jul 30 at 22:19

2

2

$Aimplies B$ is read as "A implies B." It means that if $A$ is true, then $B$ is true as well.
– Carl Schildkraut
Jul 30 at 22:20

$Aimplies B$ is read as "A implies B." It means that if $A$ is true, then $B$ is true as well.
– Carl Schildkraut
Jul 30 at 22:20

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up vote
2
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Hint:

Add $17x + 16$ to both sides of the equation (to move all quantities containing $x$ to one side and the constants to the other). Then divide by $40$ to get $x$.

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answered Jul 30 at 21:55

RayDansh

867113

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up vote
2
down vote

Hint:

Add $17x + 16$ to both sides of the equation (to move all quantities containing $x$ to one side and the constants to the other). Then divide by $40$ to get $x$.

share|cite|improve this answer

answered Jul 30 at 21:55

RayDansh

867113

add a comment | 

up vote
2
down vote

up vote
2
down vote

Hint:

Add $17x + 16$ to both sides of the equation (to move all quantities containing $x$ to one side and the constants to the other). Then divide by $40$ to get $x$.

share|cite|improve this answer

answered Jul 30 at 21:55

RayDansh

867113

Hint:

Add $17x + 16$ to both sides of the equation (to move all quantities containing $x$ to one side and the constants to the other). Then divide by $40$ to get $x$.

share|cite|improve this answer

answered Jul 30 at 21:55

RayDansh

867113

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 21:55

RayDansh

867113

answered Jul 30 at 21:55

RayDansh

867113

answered Jul 30 at 21:55

RayDansh

867113

867113

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up vote
1
down vote

We have that

$$23x-16=14-17x$$

$$23x-16colorred+17x+16=14-17xcolorred+17x+16$$

$$40x=30$$

$$frac40xcolorred40=frac30colorred40$$

$$x=frac34$$

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answered Jul 30 at 21:54

gimusi

64.1k73480

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up vote
1
down vote

We have that

$$23x-16=14-17x$$

$$23x-16colorred+17x+16=14-17xcolorred+17x+16$$

$$40x=30$$

$$frac40xcolorred40=frac30colorred40$$

$$x=frac34$$

share|cite|improve this answer

answered Jul 30 at 21:54

gimusi

64.1k73480

add a comment | 

up vote
1
down vote

up vote
1
down vote

We have that

$$23x-16=14-17x$$

$$23x-16colorred+17x+16=14-17xcolorred+17x+16$$

$$40x=30$$

$$frac40xcolorred40=frac30colorred40$$

$$x=frac34$$

share|cite|improve this answer

answered Jul 30 at 21:54

gimusi

64.1k73480

We have that

$$23x-16=14-17x$$

$$23x-16colorred+17x+16=14-17xcolorred+17x+16$$

$$40x=30$$

$$frac40xcolorred40=frac30colorred40$$

$$x=frac34$$

share|cite|improve this answer

answered Jul 30 at 21:54

gimusi

64.1k73480

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 21:54

gimusi

64.1k73480

answered Jul 30 at 21:54

gimusi

64.1k73480

answered Jul 30 at 21:54

gimusi

64.1k73480

64.1k73480

add a comment | 

add a comment | 

up vote
1
down vote

Golden Rule : Jump over the equal sign and change the sign.

Example $1$: $$3x+12=2x+20$$

$$3x-2x=20-12$$

$$x=8$$

Example $2$

$$ -3x+12-25=x-4+10$$

$$-3x-x=-4+10-12+25$$

$$-4x=19$$

$$x=-19/4$$

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answered Jul 30 at 22:28

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
1
down vote

Golden Rule : Jump over the equal sign and change the sign.

Example $1$: $$3x+12=2x+20$$

$$3x-2x=20-12$$

$$x=8$$

Example $2$

$$ -3x+12-25=x-4+10$$

$$-3x-x=-4+10-12+25$$

$$-4x=19$$

$$x=-19/4$$

share|cite|improve this answer

answered Jul 30 at 22:28

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
1
down vote

up vote
1
down vote

Golden Rule : Jump over the equal sign and change the sign.

Example $1$: $$3x+12=2x+20$$

$$3x-2x=20-12$$

$$x=8$$

Example $2$

$$ -3x+12-25=x-4+10$$

$$-3x-x=-4+10-12+25$$

$$-4x=19$$

$$x=-19/4$$

share|cite|improve this answer

answered Jul 30 at 22:28

Mohammad Riazi-Kermani

27.3k41851

Golden Rule : Jump over the equal sign and change the sign.

Example $1$: $$3x+12=2x+20$$

$$3x-2x=20-12$$

$$x=8$$

Example $2$

$$ -3x+12-25=x-4+10$$

$$-3x-x=-4+10-12+25$$

$$-4x=19$$

$$x=-19/4$$

share|cite|improve this answer

answered Jul 30 at 22:28

Mohammad Riazi-Kermani

27.3k41851

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 22:28

Mohammad Riazi-Kermani

27.3k41851

answered Jul 30 at 22:28

Mohammad Riazi-Kermani

27.3k41851

answered Jul 30 at 22:28

Mohammad Riazi-Kermani

27.3k41851

27.3k41851

add a comment | 

add a comment |