Question

Question #f5d8c

Answer 1

Solution : `p=-6 , r=12.5 and p=3 , r=-1`

Explanation:

`3p+2r=7(1) and p^2-2r=11(2)`. From equation(1)

we get `2r= 7-3p` Putting `2r= 7-3p` in equaition (2)

we get `p^2-(7-3p)=11or p^2+3p-7=11` or

`p^2+3p-7-11=0 or p^2+3p-18=0` or

`p^2+6p-3p-18=0 or p(p+6) -3(p+6)=0` or

`(p+6)(p-3)=0 :. p= -6 or p=3`

When `p=-6 ; 2r = 7- 3*(-6)=7+18 or 2r=25` or

`r=12.5 :.` Solution: `p=-6 , r=12.5`

When `p=3 ; 2r = 7- 3*3=7-9 or 2r=-2` or

`r=-1 :.` Solution: `p=3 , r=-1`

Solution : `p=-6 , r=12.5 and p=3 , r=-1` [Ans]

Answer 2

`3p + 2r = 7................[1]`

and

`p^2 - 2r =11..............[2]`

Adding [1] and [2] we get

`p^2 +3p-18=0`

`=> p^2 +6p-3p-18=0`

`=> p(p +6)-3(p+6)=0`

`=> (p +6)(p-3)=0`

So `p= - 6 and 3`

Inserting these in [1] we get

`3(-6)+2r=7`

`=>r=25/2=12.5`

and

`3xx3+2r=7`

`r=-2/2=-1`

So solutions are
`p=-6 and r= 12.5`

or
`p=3and r= -1`

Answer 3

`(p,r)to(-6,25/2)" or "(3,-1)`

Explanation:

`3p+2r=7to(1)`

`p^2-2r=11to(2)`

`"from equation "(1)" we can express 2r as"`

`2r=7-3pto(3)`

`color(blue)"substitute in "(2)`

`p^2-(7-3p)=11`

`rArrp^2-7+3p-11=0`

`rArrp^2+3p-18=0`

`"the factors of - 18 which sum to + 3 are + 6 and - 3"`

`rArr(p+6)(p-3)=0`

`"equate each factor to zero and solve for p"`

`p+6=0rArrp=-6`

`p-3=0rArrp=3`

`"substitute each value in "(3)" and solve for r"`

`2r=7-3prArrr=1/2(7-3p)`

`p=-6tor=1/2(7+18)=25/2rArr(-6,25/2)`

`p=3tor=1/2(7-9)=-1rArr(3,-1)`

Answer 4

p=3 r=-1

Explanation:

In order to do this problem with simultaneous equations, we must first isolate -2r from the first equation.
`3p+2r=7`
`3p-7=-2r`
Now that we have negative 2r, we can substitute it into the second equation.
`p^2-2r=11`
`p^2+(3p-7)=11`
Now, let's bring all the variables and constants to one side and solve the equation by factorization.
`p^2+3p-18=0`
`(p+6)(p-3)`
`p=-6,3`
With both of these solutions, plug them into the original equations and solve for r. The only solution that works in this circumstance is if p=3 and r=-1.