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We should set the two equations for y equal and algebraically solve |x – 1| = 3x + 3 for x.

This requires two solutions: one for the case that x – 1 is positive, the other for the case that x –1 is negative:

x – 1 is positive: (x – 1) = 3x + 3 --> x = -2. INCORRECT: x – 1 = (–2) – 1 = –3 Therefore, the original assumption that x – 1 is positive does not hold.

x – 1 is negative: –(x – 1) = 3x + 3 --> x = -1/2. CORRECT: Therefore, the original assumption that x – 1 is negatives holds.

Thus, there is only one solution for x and y, which is y = 3/2 and x = -1/2.

If y = |x – 1| and y = 3x + 3, then x must be between which of the [#permalink]

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25 Jul 2016, 05:43

5

This post received KUDOS

Since we're finding the intersection points of two relatively easy functions to think about visually, the best way to approach this problem is with a graph.

The first equation should be a simple absolute value function with the bend at x = 1.

The second is a basic line with y intercept at y = 3 and x intercept at x = -1.

Based on these two functions, we see that they have to intersect somewhere between - 1 and 0 .

We should set the two equations for y equal and algebraically solve |x – 1| = 3x + 3 for x.

This requires two solutions: one for the case that x – 1 is positive, the other for the case that x –1 is negative:

x – 1 is positive: (x – 1) = 3x + 3 --> x = -2. INCORRECT: x – 1 = (–2) – 1 = –3 Therefore, the original assumption that x – 1 is positive does not hold.

x – 1 is negative: –(x – 1) = 3x + 3 --> x = -1/2. CORRECT: Therefore, the original assumption that x – 1 is negatives holds.

Thus, there is only one solution for x and y, which is y = 3/2 and x = -1/2.

The correct answer is D.

So I solved for y initially to find the distance. One answer was -3 and the second for 3/2. Since distance from 1 cannot be negative, only 3/2 is valid.

But why then can it not be between 2 and 3 as well? Is that because it x as 5/2 needs to satisfy y=3x+3? That is the only reason I could think of. The possible answer I solved for was -1/2 and 5/2 (distance of 3/2 from 1 but only one of them is correct).

We should set the two equations for y equal and algebraically solve |x – 1| = 3x + 3 for x.

This requires two solutions: one for the case that x – 1 is positive, the other for the case that x –1 is negative:

x – 1 is positive: (x – 1) = 3x + 3 --> x = -2. INCORRECT: x – 1 = (–2) – 1 = –3 Therefore, the original assumption that x – 1 is positive does not hold.

x – 1 is negative: –(x – 1) = 3x + 3 --> x = -1/2. CORRECT: Therefore, the original assumption that x – 1 is negatives holds.

Thus, there is only one solution for x and y, which is y = 3/2 and x = -1/2.

The correct answer is D.

Bunuel, this is was the same thing I did, but I just had one question. Isn't it more accurate to say for the case where x>1, that since it leads to x=-2, this is an invalid solution because -2 is not >1, rather than saying -2 is invalid because -2 <0? Like when are testing solutions to be valid we are making sure that fall within the range, not just if they are positive or negative right?

We should set the two equations for y equal and algebraically solve |x – 1| = 3x + 3 for x.

This requires two solutions: one for the case that x – 1 is positive, the other for the case that x –1 is negative:

x – 1 is positive: (x – 1) = 3x + 3 --> x = -2. INCORRECT: x – 1 = (–2) – 1 = –3 Therefore, the original assumption that x – 1 is positive does not hold.

x – 1 is negative: –(x – 1) = 3x + 3 --> x = -1/2. CORRECT: Therefore, the original assumption that x – 1 is negatives holds.

Thus, there is only one solution for x and y, which is y = 3/2 and x = -1/2.

The correct answer is D.

Bunuel, this is was the same thing I did, but I just had one question. Isn't it more accurate to say for the case where x>1, that since it leads to x=-2, this is an invalid solution because -2 is not >1, rather than saying -2 is invalid because -2 <0? Like when are testing solutions to be valid we are making sure that fall within the range, not just if they are positive or negative right?

Here is another way, which might help:

\(|x – 1| = 3x + 3\)

When \(x - 1 < 0\), so when \(x < 1\) we'll get \(-(x - 1) = 3x + 3\) --> \(x = -\frac{1}{2}\): keep because \(-\frac{1}{2}\) IS in the range \(x < 1\).

When \(x - 1 \geq 0\), so when \(x \geq 1\) we'll get \(x - 1 = 3x + 3\) --> \(x = -2\): discard because -2 is NOT in the range \(x \geq 1\)).

So, we got that \(|x – 1| = 3x + 3\) has only one solutions \(x = -\frac{1}{2}\).

Re: If y = |x – 1| and y = 3x + 3, then x must be between which of the [#permalink]

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02 Oct 2017, 01:21

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This post received KUDOS

Bunuel ... this is how I approached this question. is it ok to do this way?

y=|x-1| so y=(x-1) or y=-(x-1)

First consider y=(x-1) since y=3x+3 we have an equation 3x+3=x-1, solving this equation gives x=-2 which when plugged in the given equations does not hold. Plugging x=-2 in y=|x-1| gives y=3 and Plugging x=-2 in y=3x+3 gives y=-3 so x=-2 is not the desired solution.

Now consider y=-(x-1) Using the same approach, we get x=-1/2 which when plugged in the given equations gives y=3/2 in both cases. Plugging x=-1/2 in y=|x-1| gives y=3/2 and Plugging x=-1/2 in y=3x+3 gives y=3/2 so x=-1/2 is the desired solution. It lies within -1 and 0 so D is the right answer.

Bunuel ... this is how I approached this question. is it ok to do this way?

y=|x-1| so y=(x-1) or y=-(x-1)

First consider y=(x-1) since y=3x+3 we have an equation 3x+3=x-1, solving this equation gives x=-2 which when plugged in the given equations does not hold. Plugging x=-2 in y=|x-1| gives y=3 and Plugging x=-2 in y=3x+3 gives y=-3 so x=-2 is not the desired solution.

Now consider y=-(x-1) Using the same approach, we get x=-1/2 which when plugged in the given equations gives y=3/2 in both cases. Plugging x=-1/2 in y=|x-1| gives y=3/2 and Plugging x=-1/2 in y=3x+3 gives y=3/2 so x=-1/2 is the desired solution. It lies within -1 and 0 so D is the right answer.

Yes, you can expand absolute value with positive sign and negative sing, solve and substitute back to check the validity of the roots.
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Re: If y = |x – 1| and y = 3x + 3, then x must be between which of the [#permalink]

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13 Nov 2017, 14:03

eaze wrote:

Since we're finding the intersection points of two relatively easy functions to think about visually, the best way to approach this problem is with a graph.

The first equation should be a simple absolute value function with the bend at x = 1.

The second is a basic line with y intercept at y = 3 and x intercept at x = -1.

Based on these two functions, we see that they have to intersect somewhere between - 1 and 0 .

Answer: D

A graph is the "best way to approach this problem" ? Not really.

Equating |x-1| to 3x+3 is quick and painless.
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Kudosity killed the cat but your kudos can save it.