Question
If 2x^2+3xy-2y^2-5(2x-y)=0 and x^2-2xy-3y^2+15=0
Find the value of x,y
Answer
Factor the expression
2x^2+3xy-2y^2-5(2x-y)
=(2x-y)(x+2y)-5(2x-y)
=(2x-y)(x+2y-5)
Then,
(2x-y)(x+2y-5)=0
We get,
y=2x
(1)
and
y = \dfrac{5-x}{2}
(2)
Substitute (1) to the second equation x^2-2xy-3y^2+15=0
x^2-4x^2-12x^2+15=0
Then,
x^2=1
(3)
x=\pm1 and y = \pm2
Substitute (2) to the second equation
x^2-2x\cdot \dfrac{5-x}{2}-3\Big( \dfrac{5-x}{2}\Big) ^2+15=0
x^2-5x+x^2-\dfrac{3}{4}(25-10x+x^2)+15=0
Multiply 4 on both sides and Collect like terms
8x^2-20x-75+30x-3x^2+60=0
5x^2+10x-15=0
x^2+2x-3=0
x=1, x=-3
Using the formula (2)
y = 2, y =4
Therefore, the solution set for x,y is \{(1,2),(-1,-2),(-3,4)\}