Solve the Simultaneous equation x^2+y^2=58 x+y=10.

Question

asked Aug 27, 2019 208k views

2 Answers

Concat · Answer 1 · 2019-08-28T22:01:13+0000

That's the intersection of a circle and a line.

y = 10 - x

x^2 + (10 - x)^2 = 58

x^2 + 100 -20 x + x^2 = 58

2x^2 -20 x + 42 = 0

x^2 -10 x + 21 = 0

(x-7)(x-3)=0

$x = 7 \quad \textrm{or} \quad x=3$

By the symmetry of the problem when x is one solution y is the other:

(x,y) = (7,3) or (3,7)

Check:

$7^2+3^2=49+9=58 \quad\checkmark$

Regis Frey · Answer 2 · 2019-08-30T19:50:46+0000

From the secon equation, derive

y = 10-x

Substitute in the first equation:

Expand and simplify:

x^2 - 10 x + 21 = 0</span>

Solve with the quadratic equation to get
x=3

or

Derive back the values for

:

$x=3 \implies y=7$

$x=7 \implies y=3$
which makes sense, since the problem is symmetric in

and