Question

X+y=10 y^2=8x x+y=10 x^2=4xy

Answer 1

1)
{x+y = 10    ⇒    y = 10 -  x
{y² = 8x
Метод подстановки:
(10 - x)²  = 8x
10²  - 2*10*x + x²  = 8x
x²  - 20x  + 100 - 8x = 0
x²  - 28x + 100 = 0
D = (-28)²  - 4*1*100 = 784 - 400 = 384= (8√6)²
D>0
x₁ = ( 28 - 8√6)/(2*1)  = 14 - 4√6
x₂ = (28 + 8√6)/(2*1)  = 14 + 4√6
y₁ = 10 - (14 - 4√6) = 10 - 14 + 4√6 = - 4 + 4√6
y₂ = 10 - (14  + 4√6) = 10 - 14 - 4√6 = - 4 - 4√6
ОТВЕТ:   ( (14 - 4√6)  ; ( - 4 + 4√6)  )  ;   
                ( (14 + 4√6) ;  (-4 - 4√6) ) .
2)
{ x + y  = 10    ⇒  y = 10 - x
{ x² = 4xy
x² = 4x(10-x)
x² = 40x - 4x²
x² - 40x + 4x²=0
5x²  - 40x = 0
5x(x - 8) = 0
5x = 0
x₁=0
x - 8 =0
x₂ = 8
y₁ =  10 - 0 = 10
y₂ =  10 - 8  = 2
ОТВЕТ :  ( 0; 10)  ;   
                ( 8;  2 ).

Answer 2

Oh, you've got a system of equations going on! Let's see what we can do with these.

The first two equations are:

1. $x + y = 10$
2. $y^2 = 8x$

And the next two are:

3. $x + y = 10$
4. $x^2 = 4xy$

It seems like the first and third equations are the same, so we essentially have three equations to work with. Let's focus on the first and second.

For the first pair (1 and 2), you could substitute the expression for $x + y$ from the first equation into the second one and see where that takes you.

For the second pair (3 and 4), try the same strategy. Substitute $x + y$ from the third equation into the fourth.

Give it a shot, and let me know what you find!