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## Homework Statement

solve the equation: 3(x+1)-(x-1)/2 = -4

## Homework Equations

## The Attempt at a Solution

3(x+1)-(x-1)/2 = -4

= 3x+3-2(x-1)/2 = 2(-4)

= 3x+3-x+1 = -8

= 3x-x = -8-1-3

= 2x = -12

= x = -6

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- Thread starter tehmatriks
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- #1

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solve the equation: 3(x+1)-(x-1)/2 = -4

3(x+1)-(x-1)/2 = -4

= 3x+3-2(x-1)/2 = 2(-4)

= 3x+3-x+1 = -8

= 3x-x = -8-1-3

= 2x = -12

= x = -6

- #2

Mark44

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You can't do this (above). You can multiply both sides of an equation by a number, but you can't just pick and choose which terms on one side to multiply.## Homework Statement

solve the equation: 3(x+1)-(x-1)/2 = -4

## Homework Equations

## The Attempt at a Solution

3(x+1)-(x-1)/2 = -4

= 3x+3-2(x-1)/2 = 2(-4)

The first equation is equivalent to (NOT equal to)

3x + 3 -x/2 + 1/2 = -4

Combine the x terms and move the constant terms to the right side. If you like you can multiply both sides by 2 (all terms) first before moving the constant terms over to the right side.

Do not connect equations with =.

You should also get in the habit of checking your solutions. If you replace x in the first equation by -6, you= 3x+3-x+1 = -8

= 3x-x = -8-1-3

= 2x = -12

= x = -6

- #3

HallsofIvy

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- #4

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yo mark, sorry about the = thing.You can't do this (above). You can multiply both sides of an equation by a number, but you can't just pick and choose which terms on one side to multiply.

The first equation is equivalent to (NOT equal to)

3x + 3-x/2 + 1/2= -4

Combine the x terms and move the constant terms to the right side. If you like you can multiply both sides by 2 (all terms) first before moving the constant terms over to the right side.

Do not connect equations with =.

You should also get in the habit of checking your solutions. If you replace x in the first equation by -6, youshouldget a true statement. If you don't, you have made a mistake.

and yea i tried multiplying both sides completely[2(3x + 3 - x-1/2)/2 = 2(-4)

didn't work out, the answer in the back of the book is -3

anyway, what did you do with the bolded?

- #5

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canmultiply both sides by 2, but, as Mark44 said, you can't "pick and choose"- you have to multiplyeverythingby 2. Multiplying both sides of 3(x+ 1)- (x-1)/2= -4 by 2 you will get 6(x+ 1)- 2(x- 1)/2= -8. You did not multiply 3(x+ 1) by 2.

wait, so you dont have to divide the denominator by the LCM first? i've been doing all these questions like that and they have all been right up until now, how come you dont divide here? and you kept the fraction, my whole purpose was to remove the fraction

i really do hate fractions

- #6

Mark44

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No. And this doesn't even make any sense. The goal is to get rid of fractions completely, which you can do by multiplying both sides by 2.wait, so you dont have to divide the denominator by the LCM first?

Doing so in this problem gives you 6(x+ 1)- 2(x- 1)/2= -8, or 6x + 6 - x + 1 = -8.

The second term on the left side is the same as - (x - 1) after you cancel the 2 in the numerator and denominator.

Grouping the terms in x on the left and everything else on the right produces this equation:

5x = -15

<==> x = -3

I would have to see an example to see if you might have stumbled onto the correct answer by accident, which is not a good way to get the answer.i've been doing all these questions like that and they have all been right up until now,

how come you dont divide here? and you kept the fraction, my whole purpose was to remove the fraction

i really do hate fractions

- #7

eumyang

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("yo?")yo mark, sorry about the = thing.

and yea i tried multiplying both sides completely[2(3x + 3 - x-1/2)/2 = 2(-4)

didn't work out, the answer in the back of the book is -3

anyway, what did you do with the bolded?

The part I bolded is already wrong. It should be

2(3x + 3 - x/2 + 1/2) = 2(-4)

- #8

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an example on how the book has been teaching me how to do these types of question, and yes, it says multiply by both sides, but, just look down and check the example. also note out of 20 odd sums of these i've had them all right, till now ofcourse.

the books example:

consider the equation (2x-1)/5 = 3

to get rid of the fraction, we multiply both sides by 5.

5(2x-1)/5 = 3(5)

=> 2x - 1 = 15

=> 2x - 1 + 1 = 15 + 1

=> 2x = 16 => x = 8

and mark, i used the => because that's how it is in the book, there's one example.

here's another example:

**solve the equation 4x/5 - x/2 = 3/4**

The LCM of 5, 2 and 4 is 20.

We now multiply each term by 20.

20(4x)/5 - 20(x)/2 = 20(3)/4

=> 4(4x) - 10(x) = 5(3)

=> 16x - 10x = 15

=> 6x = 15

=> x = 15/6 = 5/2 = 2½

looks to me that the multiple or LCM is being divided by the denominator before the multiplying actually happens.

the books example:

consider the equation (2x-1)/5 = 3

to get rid of the fraction, we multiply both sides by 5.

5(2x-1)/5 = 3(5)

=> 2x - 1 = 15

=> 2x - 1 + 1 = 15 + 1

=> 2x = 16 => x = 8

and mark, i used the => because that's how it is in the book, there's one example.

here's another example:

The LCM of 5, 2 and 4 is 20.

We now multiply each term by 20.

20(4x)/5 - 20(x)/2 = 20(3)/4

=> 4(4x) - 10(x) = 5(3)

=> 16x - 10x = 15

=> 6x = 15

=> x = 15/6 = 5/2 = 2½

looks to me that the multiple or LCM is being divided by the denominator before the multiplying actually happens.

Last edited:

- #9

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yea, that way definitely worksNo. And this doesn't even make any sense. The goal is to get rid of fractions completely, which you can do by multiplying both sides by 2.

Doing so in this problem gives you 6(x+ 1)- 2(x- 1)/2= -8, or 6x + 6 - x + 1 = -8.

The second term on the left side is the same as - (x - 1) after you cancel the 2 in the numerator and denominator.

Grouping the terms in x on the left and everything else on the right produces this equation:

5x = -15

<==> x = -3

I would have to see an example to see if you might have stumbled onto the correct answer by accident, which is not a good way to get the answer.

anyway, you guys have been of great help and i actually understand it now, this forum is great, thanks again

and can i fire away with another equation(simultaneous) in this thread? i've done everything in the chapter now, just this simultaneous question

- #10

Mark44

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I suppose that's one way to look at it. Notice however that each term on both sides was multiplied by 20, which is the smallest number you can multiply the three fractions by to get a common (the same) denominator.an example on how the book has been teaching me how to do these types of question, and yes, it says multiply by both sides, but, just look down and check the example. also note out of 20 odd sums of these i've had them all right, till now ofcourse.

the books example:

consider the equation (2x-1)/5 = 3

to get rid of the fraction, we multiply both sides by 5.

5(2x-1)/5 = 3(5)

=> 2x - 1 = 15

=> 2x - 1 + 1 = 15 + 1

=> 2x = 16 => x = 8

and mark, i used the => because that's how it is in the book, there's one example.

here's another example:

solve the equation 4x/5 - x/2 = 3/4

The LCM of 5, 2 and 4 is 20.

We now multiply each term by 20.

20(4x)/5 - 20(x)/2 = 20(3)/4

=> 4(4x) - 10(x) = 5(3)

=> 16x - 10x = 15

=> 6x = 15

=> x = 15/6 = 5/2 = 2½

looks to me that the multiple or LCM is being divided by the denominator before the multiplying actually happens.

In the first term, you have 20/5, which is 4.

In the second term, you have -20/2, which is -10.

Finally, on the right side, you have 20/4, which is 5.

Now, take a look again at what you had in the first post.

3(x+1)-(x-1)/2 = -4

3x+3-2(x-1)/2 = 2(-4)

Multiplying both sides by 2 is the right thing to do, but you neglected to multiply the first term on the left side.3x+3-2(x-1)/2 = 2(-4)

- #11

Mark44

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Since it's a new problem, you should start a new thread.yea, that way definitely works

anyway, you guys have been of great help and i actually understand it now, this forum is great, thanks again

and can i fire away with another equation(simultaneous) in this thread? i've done everything in the chapter now, just this simultaneous question

- #12

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Since it's a new problem, you should start a new thread.

ok, i'll do it tommorrow then, thanks again, peace

- #13

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## The Attempt at a Solution

3(x+1)-(x-1)/2 = -4

= 3x+3-2(x-1)/2 = 2(-4)

= 3x+3-x+1 = -8

= 3x-x = -8-1-3

= 2x = -12

= x = -6

It looks like your error is where you multiplied the (x-1)/2 term by 2 without multiplying the rest of the left hand side of the equation by 2. To get rid of the fraction, you would have to multiply both sides of the equation by two as follows:

2[3(x+1)-(x-1)/2] = 2*[-4]

- #14

Mark44

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You're a day late and a dollar short. This was already noted in posts 2, 3, and 7.It looks like your error is where you multiplied the (x-1)/2 term by 2 without multiplying the rest of the left hand side of the equation by 2. To get rid of the fraction, you would have to multiply both sides of the equation by two as follows:

2[3(x+1)-(x-1)/2] = 2*[-4]

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