When the difference between one and two numbers is two, the two numbers are -10 and -12.
Given that,
The difference between one and two numbers is two. Twelve more than six times the second is four times the first.
We have to find the number.
We know that,
The system of equation we get
x=y-2 ----->equation(1)
4x=12+6y ----->equation(2)
Substitute for x in the equation(2)
4(y-2)=12+6y
4y-8=12+6y
4y-6y=12+8
-2y=20
y=-10
Substitute y=-10 in equation(1)
x=-10-2
x=-12
Therefore, The two numbers area -10 and -12 when the difference between one and two numbers is two.
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Answer:
AB = x+ 8
Step-by-step explanation:
Given : QS=6x and CS=5x-8
From the given figure,
QS = QC + CS
We are given with QS and CS
Replace it the above equation
6x = QC + 5x -8
Solve for QC
Subtract 5x on both sides
x = QC + -8
Add 8 on both sodes
QC= x + 8
QABC is a rectangle
QC = AB
So AB = x+ 8
Line segment of length k is divided into 3 equal parts.
so first segment is 0-k/3 and third segment is 2/3k-k
so mid-pt of 1st = k/6 and 3rd = 5/6k
so the distance in between = 5/6k-k/6 = 4/6k = 2/3k
Answer:
Let the original number be x
Successor is defined as the number which comes immediately after a particular number.
also, the successor of a whole number is the number obtained by adding 1 to that number.
Then, the successor of a number x is, x+1
As per the given condition :
we have;
Using distributive property on LHS (i.e, 
)
Then, we have
5x+5+x=83
Combine like terms;
6x+5=83
Subtract 5 from both the sides we get;
6x+5-5=83-5
Simplify:
6x=78
Divide both side by 6,
Simplify:
x =13
Therefore, the original number x is, 13
