SOLUTION: Please help me solve this system:{{{y=1/3x+10}}} {{{x=3y+6}}}
I have tried this so far:
y=1/3(3y+6)
y=y+2
I think this is the wrong way to solve this system by substitutin
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-> SOLUTION: Please help me solve this system:{{{y=1/3x+10}}} {{{x=3y+6}}}
I have tried this so far:
y=1/3(3y+6)
y=y+2
I think this is the wrong way to solve this system by substitutin
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Question 65928: Please help me solve this system:
I have tried this so far:
y=1/3(3y+6)
y=y+2
I think this is the wrong way to solve this system by substituting. Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Please help me solve this system:
I have tried this so far:
y=1/3(3y+6)
y=y+2
I think this is the wrong way to solve this system by substituting.
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Substituting will work for this problem and you are sort of on the right track. At least you get a gold star for trying. When you substituted x=3y+6 into the other equation, you should have gotten y=(1/3(3y+6))+10. I'm not sure where the y=y+2 comes from. Anyway,try following my approach:
Generally, I like to get rid of the fractions first.
(1) y=(1/3x)+10
(2) x=3y+6
First, we'll get rid of the fraction in(1) by multiplying every term by (3x). This gives us:
(1) 3xy=1+30x
(2) x=3y+6
Now, we can substitute x=3y+6 into (1) and we have:
3y(3y+6)=1+30(3y+6) simplifying we get:
9y^2+18y=1+90y+180 collecting like terms, we have:
9y^2-72y-181=0
We can solve using the quadratic formula:
y=(-b+or-sqrt(b^2-4ac))/2a and we get:
y=(+72+or-sqrt(72^2-(4)(9)(-181))/18 simplifying we have:
y=(72+or-108.166)/18 so we have for y:
y=(72+108.166)/18=180.166/18=10.009
y=(72-108.166)/18=-36.166/18=-2.009
Next, substitute these values for y in eq(2)
For y=10.009
x=3y+6
x=3(10.009)+6
x=30.027+6=36.027
