Question 206077: What is the lesser value of x out of the two solutions for x and y in this equation: 4x^2-3xy+9y^2=15
2x+3y=5
Thank you!!
Answer by mickclns(59)   (Show Source):
You can put this solution on YOUR website! First solve the linear equation for 3y (so you can substitute it into the other equation and solve for x -- both terms containing a y contain a 3y) by subtracting 2x from both sides:
2x + 3y = 5 --> 3y = 5 - 2x
The first equation can be rewritten 4x^2 - (3y)x + (3y)^2 = 15. Substitution 5-2x for 3y, we get
4x^2 - ( 5-2x)x + ( 5-2x)^2 = 15.
Multiplying these out we get 4x^2 - 5x + 2x^2 + 25 - 20x + 4x^2 = 15.
Subtracting 15 from both sides and combining like terms, we get 10x^2 -25x + 10 = 0
Dividing both sides by 5, we get 2x^2 - 5x + 2 = 0. There are various ways of factoring a quadratic (if it factors, which this does). I like this way: First multiply the coefficient of the x^2 term, 2 by the constant term, also 2, to get 4. Now you want to find two numbers that multiply to this number, 4, and that add to the coefficient of x, in this case -5. The two numbers are -1 and -4.
Now you put those two numbers into the blanks in 2x^2 + _x + _x + 2.
Case 1: 2x^2 + -4x + -1x + 2. Now factor out whatever you can out of the first two terms and factor out whatever you can out of the last two terms:
2x(x - 2) + -1(x - 2). Notice both terms have a factor of x-2 which you can factor out:
(2x - 1) (x-2)
Case 2: 2x^2 + -1x + -4x + 2. Now factor out whatever you can out of the first two terms and factor out whatever you can out of the last two terms:
x(2x - 1) + -2(2x - 1). Again, both terms have a factor, 2x - 1 this time, which you can factor out:
(x-2) (2x - 1)
In both cases the factors are the same except in opposite order, which makes not a whit of difference.
Now we use the zero product property which says that if a product of a bunch of numbers is 0 then one of the numbers must be 0, which is what allows us to set each of the factors to 0 (and which is why, far above, we subtracted 15 from each side).
So, (2x - 1) (x-2) = 0 --> 2x-1 = 0 or x-2 = 0, so x = 1/2 or x = 2.
Going back to the original question (always remember to do that for any problem, especially a word problem), we have, the lesser value of x out of the two solutions for x and y is ***** 1/2 *****. And notice that in order to answer the question we didn't even have to find the values for y because we substituted an expression of x for an equivalent expression of y in the fourth or fifth line. We could have, by plugging in 1/2 for x in 2x + 3y = 5 and solving for y, then plugging in 2 for x and solving for y.
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