SOLUTION: half of the sum of two numbers is 6 less than the first number.one-third of their difference is one less than the second number. Find the numbers
Question 649437: half of the sum of two numbers is 6 less than the first number.one-third of their difference is one less than the second number. Find the numbers Answer by shweta(56) (Show Source):
You can put this solution on YOUR website! Let the two numbers be x and y
Half of the sum of the numbers= {1*(x+y)}/2=(x+y)/2
(x+y)/2=x-6 .......(1)
One-third of their difference={1*(x-y)}/3=(x-y)/3
(x-y)/3=y-1 ........(2)
x-y=3*(y-1)
x-y=3y-3
x-y-3y=-3
x-4y=-3 ........(3)
Now consider equation (1)
(x+y)=2*(x-6) (on cross multiplication)
x+y=2x-12
x-2x+y=-12
-x+y=-12 .......(4)
Now we have got two linear equations .
x-4y=-3
-x+ y=-12
we need to find the value of each variable .We will start by cancelling one variable so that we can get the value of other variable
Add the above equation, we get
-4y+y=-3-12 (x-x=0)
-3y= -15
Negative sign on both the sides get cancelled
3y=15
y=15/3
y= 5
Now substitute the value of y in equation(4)
y-x=-12
5-x=-12
-x= -12-5
-x=-17
x=17
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