Question

1.х^2 - у^2 - 5ху=-62. х+у=3

Answer 1

$\left \{ {{x^{2} - y^{2} -5xy=-6} \atop {x+y=3}} \right. \left \{ {{(3-y)^{2}-y^{2}-5y \cdot (3-y) +6=0} \atop {x=3-y}} \right. \\ \left \{ {{{9 -6y+y^{2}-y^{2}-15y+5y^{2}+6=0} \atop {x=3-y}} \right. \left \{ {{5y^{2}-21y+15=0} \atop {x=3-y}} \right. \\ 5y^{2}-21y+15=0 \\ y_{1,2}= \frac{21 \pm \sqrt{441 - 300} }{10}= \frac{21 \pm \sqrt{141} }{10} \; \left {{y_{1}=\frac{21 + \sqrt{141} }{10}} \atop {y_{2}=\frac{21 - \sqrt{141} }{10}}} \right.$
$x_{1}=3 - \frac{21 + \sqrt{141} }{10}}, \; x_{2}=3 - \frac{21 - \sqrt{141} }{10}}} \\ x_{1}= \frac{9 - \sqrt{141} }{10}}, \; x_{2}=\frac{9 + \sqrt{141} }{10}}}$

Answer 2

To solve the system of equations:

х^2 - у^2 - 5ху = -62 --- Equation (1) х + у = 3 --- Equation (2)

We can use substitution or elimination method. Let's use substitution method:

From equation (2), we can express y in terms of x as:

y = 3 - x

Substitute y = 3 - x in equation (1) to get:

x^2 - (3 - x)^2 - 5x(3 - x) = -62

Simplifying and solving for x:

x^2 - (9 - 6x + x^2) - 15x + 5x^2 = -62

6x^2 - 6x - 71 = 0

Using the quadratic formula:

x = (6 ± √(6^2 - 4(6)(-71))) / (2*6) = (3 ± √607) / 3

So the two solutions for x are:

x1 = (3 + √607) / 3 x2 = (3 - √607) / 3

Substitute each value of x in equation (2) to find the corresponding value of y:

For x1: y1 = 3 - x1 = 3 - (3 + √607) / 3 = (6 - √607) / 3 For x2: y2 = 3 - x2 = 3 - (3 - √607) / 3 = (6 + √607) / 3

Therefore, the solutions to the system of equations are:

(x1, y1) = ((3 + √607) / 3, (6 - √607) / 3) (x2, y2) = ((3 - √607) / 3, (6 + √607) / 3)