SOLUTION: Solve: {{{ 6x+8y=14}}},
{{{ xy=-12}}}
Algebra.Com
Question 601880: Solve: ,
Found 2 solutions by mananth, unlockmath:
Answer by mananth(16949) (Show Source): You can put this solution on YOUR website!
6x+8y=14
xy=-12
x=-12/y
substitute x
6*(-12/y) +8y=14
multiply the equation by y
-72 +8y^2=14y
8y^2-14y-72=0
/2
4y^2-7y-36=0
4y^2-7y-36=0
You have to factorize . convert the trinomial into a a polynomial with 4 terms by splitting the middle term.
compare with equation ax^2 +bx+c
co-efficient of 1st term - a=4
co-efficient of second term = b= -7
co-eficient of third term = -36
The product ac= 4*-36 = -144
factorize -144 .
group the factors into parts, such that the sum of I part & II part = -7
-144= -2*2*2*2* 3*3
-16 & 9 are the factors add them you get -7
multiply them you get -144
4y^2-16y+9y-36=0
49(y-4)+9(y-4)=0
(y-4)(4y+9)=0
y = 4 Or -9/4
plug y in xy =-12 to get x= 3 Or x= 16/3
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
Let's do substitution with these.
6x+8y=14
xy=-12
Rewrite the second equation as:
x=-12/y
Plug this into the first:
6(-12/y)+8y=14
Multiply by y:
-72+8y^2=14y
Rearrange:
8y^2-14y-72=0
Factor this as:
2(4y+9)(y-4)=0
Solve for y:
y=4
y=-9/4
Plug 4 into the original equation:
x=-3
The other is:
x=16/3
Make sense?
RJ
www.math-unlock.com
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