Using Substitution Method

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TokioHotelRox

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Joined
Dec 4, 2008
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I was wondering how to do this problem.
y=0.7(100+x)
x=0.4(100+y)
I am supposed to use the substitution method to solve it, and this is what I have came up with so far.
y=0.7(100+(0.4(100+y)
y=0.7(100+40+0.4y)
y=0.7(140+0.4y)
y=98+0.28y
-0.28y=98
y=-350
Am I working on this problem in the correct way? If not, could you walk me through it, step by step?
Thanks.
 
You're halfway there. Now that you know what "y" is, substitute it back into the second one and simplify the right-hand side to find "x". And then you're done, assuming the arithmetic is correct, which I didn't check.
 
So I substituted it into the next equation
x=.4(100+y)
x=.4(100+-350)
x=.4(-250)
x=-100
So I check it, and
y=.7(100+x)
y=.7(100-100)
y=.7
Y isn't equal to 0.7. At least, my first equation doesn't say that. What did I do wrong?
 
TokioHotelRox said:
I was wondering how to do this problem.
y=0.7(100+x)
x=0.4(100+y)
I am supposed to use the substitution method to solve it, and this is what I have came up with so far.
y=0.7(100+(0.4(100+y)
y=0.7(100+40+0.4y)
y=0.7(140+0.4y)
y=98+0.28y

y - .28 y = 98

.72 y = 98

y = 98/.72 = 49*25/9

Now continue.....
-0.28y=98
y=-350
Am I working on this problem in the correct way? If not, could you walk me through it, step by step?
Thanks.
Click to expand...
 
y - .28 y = 98
.72y=98
.72 y /.72= 98/.72
y = 136.1111
Now I substitute in to get x:
x=.4(100+y)
x=.4(100+136.111)
x=94.44444
Is this correct? *Edit!*
Just checked my work, here it is.
y=.7(00+94.4444)
y=.7(194.444)
y=136.111
Thank You so much! :D
 
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