How do you solve the system $x^2+y^2=25, x+2y=10$ ?

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Answer 1 · 2018-03-28T21:48:47

$(x, y) = (0, 5)" "$ or $" "(x, y) = (4, 3)$

Given:

${ (x^2+y^2=25), (x+2y=10) :}$

From the second equation, we have $x = 10-2y$ .

Substituting this in the first equation, we find:

$25 = x^2+y^2$

$color(white)(25) = (10-2y)^2+y^2$

$color(white)(25) = 5y^2-40y+100$

Divide both ends by $5$ to get:

$5 = y^2-8y+20$

Subtract $5$ from both sides to get:

$0 = y^2-8y+15 = (y-3)(y-5)$

So $y = 3$ or $y = 5$

If $y=3$ then:

$x = 10-2y = 10-6 = 4$

If $y = 5$ then:

$x = 10-2y = 10-10 = 0$

So:

$(x, y) = (0, 5)" "$ or $" "(x, y) = (4, 3)$

How do you solve the system x^2+y^2=25, x+2y=10x^2+y^2=25, x+2y=10?