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
x²+y²=25 x+7y=25 x=? y=?
$$\\\boxed{x^2+y^2=25} \small{\text{ is a circle }}\\
\boxed{x+7y=25} \small{\text{ is a line}}\\$$
$$\small{\text{The line cut the circle in 2 Points.}} \\
\small{\text{Point 1 is ( x=4, y=3) }}\\
\small{\text{Point 2 is ( x=-3, y=4) }}\\$$
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
$$\underbrace{(25-7y)^2}_{=25^2-2*25*7*y+49y^2}+y^2=25 \\\\
25^2-2*25*7*y+49y^2+y^2=25 \\\\
50y^2-50*7*y + 625 - 25 = 0 \\ \\
50y^2-50*7*y + 600= 0 \quad | \quad : 50\\ \\
y^2-7*y + 12= 0 \\ \\
y_{1,2}= \frac{7\pm\sqrt{49-4*12} }{2} \\ \\
y_{1,2}= \frac{7\pm\sqrt{49-48} }{2} \\ \\
y_{1,2}= \frac{7\pm 1}{2} \\ \\
y_1 = \frac{7 +1}{2} = \frac{8}{2} = 4 \qquad x_1 = 25 - 7*4 = -3\\ \\
y_2 = \frac{7 -1}{2} = \frac{6}{2} = 3 \qquad x_2 = 25 - 7*3 = 4\\ \\$$