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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
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)
You should also get in the habit of checking your solutions. If you replace x in the first equation by -6, you should get a true statement. If you don't, you have made a mistake.= 3x+3-x+1 = -8
= 3x-x = -8-1-3
= 2x = -12
= x = -6
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, you should get a true statement. If you don't, you have made a mistake.
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 fractionConversely, if you really don't like fractions, you can multiply both sides by 2, but, as Mark44 said, you can't "pick and choose"- you have to multiply everything by 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.
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?
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
("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?
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.
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.
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
ok, i'll do it tommorrow then, thanks again, peaceSince it's a new problem, you should start a new thread.
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: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
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]