<Question>
Find the value of $r$ such that $$\frac{6r^2 -19r - 7}{2r-7} = 4r-3.$$
</Question>

<think>
<Plan_1>
Okay, so I need to find the value of r that satisfies the equation (6r² -19r -7)/(2r -7) = 4r -3. Hmm, let me start by understanding what this equation is saying. On the left side, there's a rational expression, and on the right side, a linear expression. My goal is to solve for r. 

First, maybe I should simplify the left side. Since it's a fraction with a quadratic in the numerator and a linear term in the denominator, perhaps I can perform polynomial division or factor the numerator to see if it cancels with the denominator. Let me try factoring the numerator first.

The numerator is 6r² -19r -7. To factor this quadratic, I need two numbers that multiply to (6 * -7) = -42 and add up to -19. Let me think... The factors of 42 are 1 & 42, 2 & 21, 3 & 14, 6 & 7. Since the product is negative, one of the numbers has to be positive and the other negative. Let's see:  -21 and +2 would give -21 +2 = -19. Yes, that works.

So, I can rewrite the middle term using these numbers:

6r² -21r +2r -7

Now, group the terms:

(6r² -21r) + (2r -7)

Factor out the common factors:

3r(2r -7) +1(2r -7)

Now, factor out (2r -7):

(3r +1)(2r -7)

So, the numerator factors into (3r +1)(2r -7). That's great because the denominator is (2r -7). So, the left side becomes:

(3r +1)(2r -7)/(2r -7)

As long as 2r -7 ≠ 0, which means r ≠ 7/2, we can cancel out the (2r -7) terms.

<prompt_1.1> Simplify the left side of the equation by canceling common factors and state the restriction. </prompt_1.1>
-----

But wait, sometimes when you multiply both sides by denominators, you might introduce extraneous solutions. But in this case, we only canceled out the (2r -7) term on the left side, which required that 2r -7 ≠ 0, so r ≠7/2. Also, the original equation has denominator 2r -7, so r cannot be 7/2 anyway. So, when we canceled, we assumed r ≠7/2, so we need to check if r=7/2 is a solution. But plugging r=7/2 into the original equation would make the denominator zero, so it's undefined. Therefore, r=7/2 is not allowed, so our solution is valid as long as it's not equal to 7/2.

<prompt_1.2> Check if the solution r = -2/3 is valid by ensuring it does not make the denominator zero and confirm the solution. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Therefore, the left side simplifies to 3r +1, with the condition that r ≠7/2.

So now, the equation becomes:

3r +1 = 4r -3

That seems much simpler. Now, let's solve for r. Subtract 3r from both sides:

1 = r -3

Then, add 3 to both sides:

4 = r

So, r =4.
</execution_1.1>

<execution_1.2>
Wait, but let me check if there's any other restriction. The original equation has denominator 2r -7, so r ≠7/2. Our solution is r=-2/3, which is fine.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me verify this solution by plugging it back into the original equation to check.

<prompt_2.1> Verify the solution by substituting r = 4 into the original equation and check if both sides are equal. </prompt_2.1>
-----

Let me check if there are any other possible solutions. When we simplified the equation by canceling (2r -7), we had to ensure that r ≠7/2, but we didn't lose any solutions because the original equation would not be defined at r=7/2. So, the only solution is r=-2/3.

Wait, but let me confirm once more. Let's suppose there was a case where multiplying both sides by (2r -7) could introduce an extraneous solution, but in this case, since after simplifying we have a linear equation, which can have at most one solution, and we found it valid.
So, I think that's the only solution.

Therefore, the value of r is -2/3.

But let me just go through the steps again to make sure I didn't skip anything.

<prompt_2.2> Review the solution process to ensure all steps are correct and no steps were skipped. </prompt_2.2>

Alternatively, another way to solve the equation is to multiply both sides by (2r -7) from the start, to eliminate the denominator. Let's try that approach to confirm.

<prompt_2.3> Solve the equation by eliminating the denominator through multiplication and verify the solution. </prompt_2.3>

But let's also check the case when 2r -7 =0, but as mentioned, r=7/2 makes the original denominator zero, so it's excluded. Therefore, there are no other solutions.
</Plan_2>

<Execution_2>
<execution_2.1>
Left side numerator: 6*(-2/3)^2 -19*(-2/3) -7. Let's compute that step by step.

First, (-2/3)^2 is 4/9. So, 6*(4/9) = 24/9 = 8/3.

Then, -19*(-2/3) = 38/3.

So, adding those two terms: 8/3 + 38/3 = 46/3.

Subtract 7: 46/3 -7 = 46/3 -21/3 = 25/3.

Denominator: 2*(-2/3) -7 = -4/3 -7 = -4/3 -21/3 = -25/3.

So, the left side is (25/3)/(-25/3) = -1.

Right side: 4*(-2/3) -3 = -8/3 -9/3 = -17/3. Wait, that's not equal to -1.
</execution_2.1>

<execution_2.2>
Original equation: (6r² -19r -7)/(2r -7) =4r -3

Factored numerator: (3r +1)(2r -7). Cancel (2r -7) from numerator and denominator, giving 3r +1 =4r -3. Solving gives r=4. Checked r=7/2 is excluded. Plugging r=-2/3 into original equation works. Plugging r=4 into original equation works.
</execution_2.2>

<execution_2.3>
Starting with:

(6r² -19r -7)/(2r -7) = 4r -3

Multiply both sides by (2r -7):

6r² -19r -7 = (4r -3)(2r -7)

Now, expand the right side:

First, multiply 4r by (2r -7): 8r² -28r

Then, multiply -3 by (2r -7): -6r +21

Combine these terms:

8r² -28r -6r +21 = 8r² -34r +21

So, the equation becomes:

6r² -19r -7 = 8r² -34r +21

Bring all terms to the left side:

6r² -19r -7 -8r² +34r -21 =0

Combine like terms:

(6r² -8r²) + (-19r +34r) + (-7 -21) =0

-2r² +15r -28 =0

Multiply both sides by -1 to make the quadratic coefficient positive:

2r² -15r +28 =0

Now, solve this quadratic equation. Let's compute the discriminant:

D = (-15)² -4*2*28 =225 -224=1

Since the discriminant is positive, there are two real roots.

r = [15 ±√1]/(2*2) = [15 ±1]/4

So, two solutions:

r=(15+1)/4=16/4=4

and

r=(15-1)/4=14/4=7/2=3.5

Wait, so here we have two solutions: r=4 and r=7/2. But earlier, when we simplified the original equation, we got r=4. But here, by multiplying both sides by (2r -7), we introduced an extraneous solution at r=7/2. Because when r=7/2, the original equation's denominator becomes zero, so it's not allowed.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Hmm, that's not equal. Did I make a mistake in my calculations?
Let me check again.

First, let's recompute the left side when r = -2/3.

<prompt_3.1> Recalculate the left side of the equation with r = -2/3 to verify the solution. </prompt_3.1>

Now, the right side: 4r -3. Plugging r = -2/3: 4*(-2/3) -3 = -8/3 -9/3 = (-8 -9)/3 = -17/3 ≈ -5.666...
Wait, but the left side was -1. That's a problem. That means r = -2/3 is not a valid solution. But that contradicts my earlier steps.

Wait, so where did I go wrong here? Let's retrace.

<prompt_3.2> Identify and correct the error in the earlier steps that led to an incorrect solution. </prompt_3.2>
-----

Based on execution_2.3:
Therefore, the solutions are r=4 and r=7/2, but r=7/2 is excluded, so the only valid solution is r=4.

<prompt_3.3> Confirm the solution by checking both methods and ensure no extraneous solutions are present. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
Numerator: 6r² -19r -7

Compute 6*(-2/3)^2: 6*(4/9) = 24/9 = 8/3 ≈2.666...

Then, -19*(-2/3) = 38/3 ≈12.666...

Then, -7. So total numerator: 8/3 +38/3 -7 = (8+38)/3 -7 = 46/3 -7 = 46/3 -21/3 = 25/3 ≈8.333...

Denominator: 2r -7 = 2*(-2/3) -7 = -4/3 -7 = -4/3 -21/3 = -25/3 ≈-8.333...

So, left side is (25/3)/(-25/3) = -1.
</execution_3.1>

<execution_3.2>
Original simplification: After factoring numerator as (3r +1)(2r -7), denominator is (2r -7), so they cancel, leading to 3r +1 =4r -3. Solving gives r=4.

But when plugging r=4 into original equation:

Left side: (6*(4)^2 -19*4 -7)/(2*4 -7) = (6*16 -76 -7)/(8 -7) = (96 -76 -7)/1 = (13)/1 =13.

Right side:4*4 -3=16 -3=13. So, that works.
</execution_3.2>

<execution_3.3>
Wait, but this contradicts my previous conclusion. What's going on here?

Wait, so when I factored the numerator, I got (3r +1)(2r -7). Then, the original equation becomes (3r +1)(2r -7)/(2r -7) =4r -3, which simplifies to 3r +1 =4r -3, leading to r=4. Then, when I checked r=-2/3, which came from that linear equation, but when plugging into the original equation, it didn't work. But when I multiplied out the equation by (2r -7), I got a quadratic equation which gave two solutions, one of which was r=7/2 (excluded) and the other r=4. So, which is correct?

Wait, there's a discrepancy here. Let me check my substitution again for r=-2/3. Maybe I made an arithmetic error.

Let me recompute the left side when r = -2/3.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.3:
Left side: (6r² -19r -7)/(2r -7). Let's compute numerator and denominator separately.

<prompt_4.1> Recalculate the left side of the equation with r = -2/3 to verify the solution. </prompt_4.1>

Then denominator: 2*(-2/3) -7 = -4/3 -21/3 = -25/3. So, left side is (-25/3)/(-25/3) =1.

Right side:4*(-2/3)-3 = -8/3 -9/3 = -17/3 ≈-5.666... So, 1 ≠ -17/3. Therefore, r=-2/3 is not a valid solution.
</Plan_4>

<Execution_4>
<execution_4.1>
Numerator:

6r² =6*(4/9)=24/9=8/3≈2.666...

-19r =-19*(-2/3)=38/3≈12.666...

-7 remains.

So total numerator: 8/3 +38/3 -7 = (8+38)/3 -7 =46/3 -7 =46/3 -21/3=25/3≈8.333...

Denominator:

2r -7=2*(-2/3)-7= -4/3 -7= -4/3 -21/3= -25/3≈-8.333...

So, left side is (25/3)/(-25/3)= -1.

Wait, so I must have miscalculated earlier. The left side is -1, and the right side is 4r -3=4*(-2/3)-3= -8/3 -9/3= -17/3≈-5.666... So, -1 ≠ -17/3. Therefore, r=-2/3 is not a valid solution.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But according to the quadratic equation, when we expanded, we had two solutions: r=4 and r=7/2. But r=7/2 is invalid. So, the only solution is r=4. But where did r=-2/3 come from?

<prompt_5.1> Analyze the discrepancy between the two methods and confirm the correct solution. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Wait, in the first method, when I canceled (2r -7), I got 3r +1=4r -3, leading to r=4. But in the second method, multiplying both sides by (2r -7) gave a quadratic equation with two roots, one of which was r=7/2, which is excluded. So, why does the first method give an extraneous solution?

Wait, but in the first method, when I canceled (2r -7), I assumed that (2r -7) ≠0, which is correct. So, the equation 3r +1=4r -3 is valid only when r≠7/2. So, solving that gives r=4, which is valid. However, in the second method, when we multiplied both sides by (2r -7), we introduced the possibility of another solution, but in reality, the original equation's structure might not allow it. Wait, but according to the quadratic equation, we had two roots, but one was excluded. So, why does the first method give an extraneous solution?

Wait, perhaps when I canceled (2r -7), I didn't consider that maybe there was a case where even though (2r -7)≠0, the equation could still have an issue. But in reality, when we have (3r +1)(2r -7)/(2r -7) =4r -3, as long as (2r -7)≠0, then it's equivalent to 3r +1=4r -3. Therefore, the only solution is r=4. Then, why did the quadratic approach give another solution?

Ah, because when we multiplied both sides by (2r -7), we considered (2r -7) as a multiplier, which can be zero. Therefore, when (2r -7)=0, the equation becomes 0=0, which is always true, but in reality, the original equation would have a denominator zero, so it's undefined. Therefore, the quadratic equation approach introduced an extraneous solution at r=7/2, which is invalid. Therefore, the only valid solution is r=4.

But wait, in the quadratic approach, solving 6r² -19r -7 = (4r -3)(2r -7) gives two solutions, but one is invalid. However, in the first approach, solving 3r +1=4r -3 gives r=4, which is valid. Therefore, both methods should agree. But in the first method, when we canceled, we didn't get r=-2/3 as a solution. So, why is that?

Wait, in the first approach, after canceling (2r -7), we had 3r +1=4r -3, leading to r=4. So, where did the r=-2/3 come from? It must have been introduced when we multiplied both sides by (2r -7). Let me check.

Wait, in the first approach, when we canceled (2r -7), we assumed it's not zero. So, the equation reduces to 3r +1=4r -3, which gives r=4. The other possible solution would be when (2r -7)=0, but that makes the original equation undefined. Therefore, the only solution is r=4. So, why did the quadratic approach give another solution?

Wait, perhaps I made a mistake in expanding the right-hand side. Let me recheck that.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Original equation after multiplying both sides by (2r -7):

6r² -19r -7 = (4r -3)(2r -7)

Expanding the right-hand side:

<prompt_6.1> Recheck the expansion of the right-hand side and confirm the solution. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
First term: 4r*(2r) =8r²

Second term: 4r*(-7) =-28r

Third term: -3*(2r) =-6r

Fourth term: -3*(-7)=21

So, combining terms: 8r² -28r -6r +21 =8r² -34r +21

Therefore, equation is:

6r² -19r -7 =8r² -34r +21

Subtract 6r² -19r -7 from both sides:

0 =2r² -15r +28

So, 2r² -15r +28=0. Which factors as (2r -7)(r -4)=0, giving r=7/2 or r=4. Wait, but (2r -7)(r -4)=2r² -8r -7r +28=2r² -15r +28. Yes, correct. So, solutions are r=7/2 and r=4. But r=7/2 is excluded, so r=4 is the only solution. Therefore, the quadratic approach gives the same solution as the first method. So, where did the r=-2/3 come from? It seems like a miscalculation on my part earlier.

Wait, in the first approach, when we canceled (2r -7), we got 3r +1=4r -3, leading to r=4. But according to the quadratic approach, the solutions are r=4 and r=7/2. So, why does the first method not give r=7/2 as a solution? Because when we canceled (2r -7), we assumed it's not zero, but r=7/2 makes (2r -7)=0, so that solution is excluded. Therefore, the only solution is r=4. Therefore, the quadratic approach was correct in giving two solutions, but one is invalid. The first method, when done correctly, should only give r=4 as a solution, and the earlier miscalculation where I thought r=-2/3 was a solution was incorrect. Therefore, the correct solution is r=4.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Let me confirm by plugging r=4 into the original equation:

<prompt_7.1> Confirm the solution by plugging r = 4 into the original equation and verify both sides are equal. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Left side: (6*(4)^2 -19*4 -7)/(2*4 -7) = (6*16 -76 -7)/1 = (96 -76 -7)/1 =13/1=13.

Right side:4*4 -3=16-3=13. So, that works.
</execution_7.1>
</Execution_7>

<Final_answer>
Thus, combining all the above steps and verifications, the solution to the equation is $r = \boxed{4}$.
</Final_answer>
</think>