Step-by-step explanation:
I assume
SR = 2x + 23
RQ = x + 21
if that is true, then the situation is completely simple :
14 = (2x + 23) + (x + 21) = 3x + 44
3x = -30
x = -10
SR = 2×-10 + 23 = -20 + 23 = 3
RQ = -10 + 21 = 11
50% chance they will land on the same number and there’s also a 50% chance they won’t.
Answer:
4c^2+7c-5=0
4c^2+7c=5
4c^2+7c=5
4c^2+7c-5=0\quad :\quad c=\frac{-7+\sqrt{129}}{8},\:c=\frac{-7-\sqrt{129}}{8}\quad \left(\mathrm{Decimal}:\quad c=0.54472\dots ,\:c=-2.29472\dots \right)
Hope This Helps!!!
Let x represent the larger number.
Let y represent the smaller number.
x-y=4 Given
3x=5y-2 Given
Now we can just substitute; let x=4+y
Substitute 4+y for x in the second equation:
3(4+y)=5y-2
12+3y=5y-2
-2y=-14
y=7
Substitute back (into BOTH equations to double check work).
x, the larger number, is 11