Question

If 3x^2+xy-2y^2=0 and 2x^2-3xy+y^2=-1 then (x,y)

Answer 1

Answer:

3x + 2y - 1

=(3) 2 + (2) -2 - 1

=6 - 4 - 1

=2 - 1

= 1

Step-by-step explanation:

Answer 2

Answer:

If $3x^2+xy-2y^2=0$ and $2x^2-3xy+y^2=-1$, then (x,y) is (2,3) and (-2, -3).

Step-by-step explanation:

Given:

Equation of line 1 : $3x^2+xy-2y^2=0$

Equation of line 2 : $2x^2-3xy+y^2=-1$

To find:

Points of intersection (x,y)

Step 1:

Let us first solve the equation of line 1. We will do it by the factorizing method.

$3x^2+xy-2y^2=0$

$3x^2+3xy-2xy-2y^2=0$

$3x(x+y)-2y(x+y) = 0$

$(x+y)(3x-2y)=0$

$x+y=0$ OR $3x-2y = 0$

x = -y OR x = 2y/3

Step 2:

Substituting the obtained values of x in the equation of line 2 to find y.

Let us first substitute x = -y

$2x^2-3xy+y^2=-1$

$2(-y)^2-3(-y)y+y^2=-1$

$2y^2+3y^2+y^2=-1$

$6y^2=-1$

$y^2=\frac{-1}{6}$

$y=\sqrt{\frac{-1}{6} }$

x = -y gives us imaginary values. Therefore, we neglect x = -y.

Step 3:

Now, let us substitute x = 2y/3

$2x^2-3xy+y^2=-1$

$2(\frac{2y}{3})^2-3( \frac{2y}{3})y+y^2=-1$

$\frac{8y^2}{9} - 2y^2 +y^2=-1$

$8y^2-18y^2+9y^2=-9$

$-y^2=-9$

$y^2=9$

$y=\sqrt{9}$

y = ±3

Step 4:

When y = 3,

x = 2y/3

x = 2×3/3

x = 2

When y = -3,

x = 2y/3

x = 2×(-3)/3

x = -2

Therefore, the points of intersection of the given lines are (2., 3) and (-2, -3).

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