variable manipulation. what is the value of 2x+y if 4x^+6xy+2y^=32 and x+y=4

Question

what is the value of 2x+y if 4x^+6xy+2y^=32 and x+y=4

what is the value of 2x+y if 4x^+6xy+2y^=32 and x+y=4

 

 

 

Answer 1

This is often an SAT type problem. Firstly, I am assuming that you mean 4x^2 + 6xy + 2y^2 = 32.

 

The thing you need to realize with these types of problems is that you usually just need to multiply the expression by a number, or square the expression then subtract or substitute. There is no set rules, just a lot of trial and error.

 

Since one expression has squares in it while the other doesn't, I am going to start by squaring the expression. So

x+y = 4     square both sides

x^2 + 2xy + y^2 = 16      Now, if I multirply by 2 and subtract, the y^2 term will drop out.

So we get

2x^2 + 4xy + 2y^2 = 32.    Now subtract this from the other expression and we get

2x^2 + 2xy = 0  We can the factor since it's a quadratic equation to get

2x(x + y) = 0

So x=0 or x=-y

 

Since x+y = 4

if x=0, then y=4 so substitute back into 2x+y and get 4.

if x=-y, then -y+y=4, which has no solutions.

 

So your answer is 4