hello i need help please

x²+y²=25 x+7y=25   x=? y=?

$$\\\boxed{x^2+y^2=25} \small{\text{ is a circle }}\\ \boxed{x+7y=25} \small{\text{ is a line}}\\$$

$$\small{\text{The line cut the circle in 2 Points.}} \\ \small{\text{Point 1 is ( x=4, y=3) }}\\ \small{\text{Point 2 is ( x=-3, y=4) }}\\$$

The first equation can be written as x² + y² = 5²

There is a well-known Pythagorean triple: 3² + 4² = 5²

so it is possible that x and y are 3 and 4.

If x = 3 and y = 4 then the second equation, x + 7y = 25 is not true.  However, if x = 4 and y = 3 then it is true that x + 7y = 25.  Therefore 

x = 4

y = 3

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If you didn't spot the Pythagorean triple, then, more generally, you would write the 2nd equation as x = 25 - 7y, substitute this into the first equation and solve the resulting quadratic equation in y.

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yes but thats what i wrote and i dont know how to solve it

x=25-7y

(25-7y)²+y²=25

625-350+49y²+y²=25

50y²=-250

y²=-5

You have left out a "y" that multiplies the 350.

(25 - 7y)² +y² = 25

25² - 2*25*7*y + 7²y² + y² = 25

625 - 350y + 50y² = 25

50y² - 350y + 600 = 0

y² - 7y + 12 = 0

(y - 3)(y - 4) = 0

y  = 3 or y = 4

so x = 25 - 7*3 or x = 4  when y = 3

and x = 25 -7*4 or x = -3  when y = 4 (as Heureka illustrates).

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x=25-7y

(25-7y)²+y²=25  okay

$$\underbrace{(25-7y)^2}_{=25^2-2*25*7*y+49y^2}+y^2=25 \\\\ 25^2-2*25*7*y+49y^2+y^2=25 \\\\ 50y^2-50*7*y + 625 - 25 = 0 \\ \\ 50y^2-50*7*y + 600= 0 \quad | \quad : 50\\ \\ y^2-7*y + 12= 0 \\ \\ y_{1,2}= \frac{7\pm\sqrt{49-4*12} }{2} \\ \\ y_{1,2}= \frac{7\pm\sqrt{49-48} }{2} \\ \\ y_{1,2}= \frac{7\pm 1}{2} \\ \\ y_1 = \frac{7 +1}{2} = \frac{8}{2} = 4 \qquad x_1 = 25 - 7*4 = -3\\ \\ y_2 = \frac{7 -1}{2} = \frac{6}{2} = 3 \qquad x_2 = 25 - 7*3 = 4\\ \\$$