Question

Which values of P and Q result in an equation with exactly one solution? Px+50=-2x+QPx+50=−2x+Q

Answer 1

Hello mate!!


First, TXH for being a thinking person and for asking this question multiple times.  Obviously, you want an answer you understand (not just an answer to turn in).

 

The words "under what conditions?" are the "if" part of an "if-then" statement.  An "if-then" phrase means that the result happens IF the conditions are met.  Sometimes, especially in math, it might be even more exclusive and be "if-and-only-if" conditions.  For example, "If you get an A on the next math test, then you get to go to on the trip to the Spy Museum (or something that you like so that it motivates you)."  Well, you can make up better examples than I can.  The "if" part are the conditions.

 

This problem gives three situations:

 

(i) an infinite number of solutions  - this happens if and only if (sometimes written 'iff') the two equations actually are the same line


(ii) no solution  - this happens iff the two lines are parallel (if they never intersect, there is no solution)


(iii) a unique solution, i.e. Only one solution  - this happens when two lines cross  (the solution is unique because the lines cross only once if they are straight lines)

 

So, what values of p and q make situation (i) true?

       px  - y=6
       8x -2y=q

  These equation are in Standard Form (leave them that way).  In order to get a common coefficient for y, multiply the first equation by 2:

       2px  - 2y = 12

        8x  -  2y = q

  For these equations to be exactly the same, 2p=8, or p=8, and q=12.

  For this system of equations to have infinite solutions (that is, be the same line), the conditions that both p=4 and q=12 must be met.

 

What situation (conditions) of p and q make (ii) true?

    Now, let's convert the equations to slope-intercept form:

         y = mx + b      {where m = slope and b = y-intercept)

         y = px - 6

         y = 4x -q/2

    The lines are parallel iff they have the same slope (that is, p=4) but they have a different y-intercept (that is, q≠12).

   For these equations to have no solutions (that is, be parallel), the conditions that both p=4 and q≠12 must be met.

 

And, what conditions of p and q make (iii) true?

   Now, some logic is important.  We can not allow p=4, ever!   Because if p=4 and q=12 it is case (i).  But, if p=4 and q≠12 it is case (ii)  --  that accounts for everything when p=4.  Thus, p can never be equal to 4.

   What can q be?    q can be any value (-∞ to +∞ as long as p≠4).

  For these equations to have exactly one solution (that is, intersect at a point), p≠4 and q can have any value.


Hope it helpful

Answer 2

Answer:

The values of P is -2 and Q is 50.

Step-by-step explanation:

Given : Equation $Px+50=-2x+Q$

To find : Which values of P and Q result in an equation with exactly one solution?

Solution :

Equation $Px+50=-2x+Q$

Subtract 50 both the sides,

$Px+50-50=-2x+Q-50$

$Px=-2x+Q-50$

Now, subtract Q from both the sides,

$Px-Q=-2x+Q-50-Q$

$Px-Q=-2x-50$

Now, we compare the coefficient of x and the constant term,  

P=-2 and Q=50

Therefore, The values of P is -2 and Q is 50.