Answer:

Step-by-step explanation:
Given
--- proper representation of the question
Required
Evaluate
Convert numbers to improper fractions

Take LCM


Convert to improper fraction

Step-by-step explanation:

3(p2q + pq2 - pq) - 6pq + 3pq2 =
3[-20 × 2 + (-20) × 2 - (-20)] - 6 × (-20) + 3 × 2 × (-20) =
3[-40 + (-40) + 20] - 6 × (-20) + 3 × 2 × (-20) =
3(-40 - 40 + 20) - 6 × (-20) + 3 × 2 × (-20) =
3(-80 + 20) - 6 × (-20) + 3 × 2 × (-20) =
3 × (-60) - 6 × (-20) + 3 × 2 × (-20) =
-180 + 120 + 6 × (-20) =
-180 + 120 + (-120) =
-180 + 120 - 120 =
-60 - 120 = -180
1) 7-5 = 2
2)-7+ 5 = -2
3) -7-5 = -12
4)7 - (-5) = 12
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Let "a" and "b" be the two numbers
So a*b=-8 and a+b=-7
Therefore S=-7 and P=-8
Using (x^2)-S+P=0 => (x^2)-(-7)+(-8)=0 => (x^2)+7-8=0
By factorizing we get (x+8)(x-1)=0
So a can be either -8 or 1 and b can be 1 or -8
Meaning that the two numbers are -8 and 1.