Thank you for the opportunity to help you with your question!
You've got the right first step, which is to cross multiply!
The 9 that I'm seeing in your first step is actually a q (lowercase Q) in the original problem, which means it's a variable.
So we have x^2 - qx - px + pq = x^2 -3x + 2 .
If you subtract x^2 from both sides, then the x^2's go away altogether:
-qx - px + pq = -3x + 2
You want the two sides to sort of look like each other so that you can figure out what p and q to be, so I would try to get the left side to say ___x + ____ so that it matches the right side: -3x + 2 .
This means you need to factor x out of the first half of the left side of the equation:
x(-q - p) + pq = -3x + 2
Based on the second sentence in the directions of the problem, you know that the coefficients of x and the constants need to equal each other:
-q - p = -3 and pq = 2
At this point, you would need to choose an equation and solve for a variable and then substitute what you get into the other equation. I would choose the first equation and solve for p:
-q - p = -3
-p = -3 + q
p = 3 - q
Substitute for p into the second equation:
pq = 2
(3-q)q = 2
3q - q^2 = 2
Solve this quadratic equation:
-q^2 - 3q - 2 = 0 or
q^2 + 3q + 2 = 0
You could use the quadratic formula here or factor. I'll factor.
(q+1)(q+2) = 0
q = -1 and q = -2
Now we know what numbers q can be, so you should plug each one separately into one of your equations (like p = 3 - q) Remember that each specific q value goes with a specific p value.
p = 3 - q
p = 3 - (-1)
p = 4
So p = 4 and q = -1 is a solution to the original equation.
Also, we have
p = 3 - q
p = 3 - (-2)
p = 5
So p = 5 and q = -2 is also a solution to the original equation.
Those are the values for p and q that would make the original rational equation a tautology.I hope this helps you move forward on the problem! Let me know if you need any more clarification!
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