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Algebra Level 3

Given that $x, y$ are real numbers such that $5x + 9y \neq 0$ , and

$x^{2}+xy=12 \text{ and } 2xy+3y^{2}+5=0,$

then what is the value of $x+4y$ ?

3 0 1 2

3 solutions

Pratik Goel
May 9, 2014

$5x^{2}+5xy=60,$ $24xy+36y^{2}=-60$ Add and factorise to get $(x+4y)(5x+9y)=0$

But you mentioned 5x+9y is not equal to 0. Also from your solution, I fail to understand why x+4y cannot be 0. Just substitute x=4 and y=-1, all conditions are satisfied and 5x+9y is 11 in that case.

Hari Krishna - 7 years, 1 month ago

$(-4,1)\rightarrow\boxed{-11}$ is a solution as well.

Trevor B. - 7 years, 1 month ago

Sorry, the question should have read "what is the value of $x+4y$ ". I introduced an error when I edited the problem.

Calvin Lin Staff - 7 years, 1 month ago

the answer is either +11 (x=4, y=-1) or -11 (x=-4, y=1). Both satisfy all the conditions given. none of the given options are correct.

anant rai - 7 years, 1 month ago

Pls see above comment. Sorry for inconvenience

Pratik Goel - 7 years, 1 month ago

A SIMPLE FACTORING...x^2 + xy - 12 = 0 can easily be factored into (x+6)(x-2), (x-6)(x+2),(x+4)(x-3),(x-4)(x+3), (x-12)(x+1), (x+12)(x-1) to find integer solutions for y. It can quickly be seen that x=-4 and y=1 or x=4 and y=-1 solves the second equation, while the other solutions do not, giving the answer (x+4y) = 0. Might save time on a test, if you assume x and y are integers, or might not. Another quick way to find an integer solution is to use the quadratic formula: (-y +- squareroot(y^2+48))/2. When y=-1, we get (1+7)/2 = 4. So again, x=4, y=-1 and (x+4y)=0.

Richard Levine - 7 years, 1 month ago

Apoorv Trivedi
Jun 14, 2014

Letz do it by crunch mathod.....put x=4 & y= -1..... It will satisfy both equation..... Putting the value in the equation... 4+ 4× (-1) = 0

Dheeman Kuaner
May 16, 2014

we see that 2xy + 3y^2+5 = 0, the last 2 terms of this problem are positive , so 2xy must be negative to give 0.So, one out of x and y must be positive and the other negative. 12 = 1 12,2 6,3*4.Factoring x^2 + xy = x(x+y) and substituting value and trying for both the equations, we get x = 4,y=-1 .Substituting this in x+4y, we get answer to be 0.

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