Negative 10 and positive 3
Wiki User
∙ 2011-02-22 22:58:006 and 1
-12 and +5
a+b=-7 rearrange: b= -a-7=-(a+7) substitute into: ab=30 -a(a+7)=30 -a^2-7a-30=0 a^2+7a+30=0 x = complex number, therefore there are no two Real numbers that exist that satisfy these two conditions.
29
y*7+25=165 y*7+25-25 = 165-25 y*7+ 0 =140 y*7/7 = 140/7 y=20
Any number can multiply 144 and be added to 7!
-713
7
-25
151
There is no answer, it is not possible
6 and 1
6*1
-31
7
-151
-6 and -1.