Question

Решите системы уравнений: 1) x+y=8, 3x-y=0 2)4х-3у=0, х²+у²=25 3)у=х², у=х+6.

Answer 1

1.  $\left \{ {{x+y=8} \atop {3x-y=0}} \right. x=8-y 3(8-y)-y=0 24-3y-y=0 24-4y=0 -4y=-24 y=5 x+5=8 x=8-5 x=3$

$\left \{ {{y=x^2} \atop {y=x+6}} \right. x^2=x+6 x=3 x=-2 y=3+6 y=-2+6 y=9 y=4 (3 9) (-2 4)$

$\left \{ {{y= \frac{4}{3}x } \atop {x^2+y^2=25}} \right. x^2+( \frac{4}{3}x )^2 = 25 x=3 x=-3$

$y= \frac{4}{3} *3$

$y= \frac{4}{3} *(-3)$

ОТВЕТ  (3 4)   ( -3  -4 )

Answer 2

1. x+y=8, 3x-y=0:

From the first equation, we get:

y = 8 - x

Substituting into the second equation, we get:

3x - (8 - x) = 0

Simplifying this equation, we get:

4x - 8 = 0

Solving for x, we get:

x = 2

Substituting x = 2 into the first equation, we get:

2 + y = 8

Solving for y, we get:

y = 6

Therefore, the solution to this system of equations is x = 2 and y = 6.

2. 4х-3у=0, х²+у²=25:

From the first equation, we get:

y = (4/3)x

Substituting this into the second equation, we get:

x² + (4/3)x² = 25

Simplifying this equation, we get:

7x²/3 = 25

Solving for x, we get:

x² = 75/7

x = ±(5/√7)

Substituting x into the equation y = (4/3)x, we get:

y = ±(20/3√7)

Therefore, the solutions to this system of equations are (5/√7, 20/3√7) and (-5/√7, -20/3√7).

3. у=х², у=х+6:

Substituting у = х² into у = х+6, we get:

х² = х + 6

Simplifying this equation, we get:

х² - х - 6 = 0

Factoring this equation, we get:

(х - 3)(х + 2) = 0

Solving for x, we get:

x = 3 or x = -2

Substituting x = 3 into y = x², we get:

y = 9

Substituting x = -2 into y = x², we get:

y = 4

Therefore, the solutions to this system of equations are (3, 9) and (-2, 4).