+0

# Solve the system

0
503
4
+88

x^2+y^2=4

x+y=2

Type your answer as ordered pairs (x,y) with the order: from the smallest value of x to the largest, with comma and no space.

Mar 24, 2015

#3
+94558
+10

x^2+y^2=4

x+y=2

This is the intersecction of a line  and a circle

Using the second equation we can write y = 2 -  x

And substituting this into the first equation for y, we have

x^2 + (2 - x)^2 = 4   simplify

x^2 + x^2 - 4x + 4 = 4   subtract 4 from both sides  and simplify

2x^2 - 4x = 0  factor

2x(x - 2) = 0       setting each factor to 0, we have that x = 0, and x = 2

And using x + y = 2

When x = 0, y = 2

When x = 2, y =0  so the answers are (0,2) (2,0)

Here's a graph......https://www.desmos.com/calculator/eakj6g1q0t

Mar 24, 2015

#1
+95360
+5

$$\\x^2+y^2=4\qquad(1)\\\\ x+y=2 \;\;\rightarrow \;\;y=2-x\qquad(2)\\\\ sub\;\;2\;into\;1\\\\ x^2+(2-x)^2=4\\\\ x^2+4+x^2-4x=4\\\\ 2x^2-4x=0\\\\ 2(x-2)(x+2)=0\\\\ x=\pm 2$$

You can find the y values :)

Mar 24, 2015
#2
+890
+10

x = 2 implying y = 0 is ok. but x = -2 leads to a contradiction.

Also, the equations are symmetric in x and y, so it's possible to switch the x and y values to obtain a second solution corresponding to x=2, y=0.

An alternative method of solution would be to square the second equation and then combine that with the first leading to 2xy = 0, from which either x = 0, or y = 0.

Then substitute into the second equation, (not the first), to find the corresponding y or x value.

Mar 24, 2015
#3
+94558
+10

x^2+y^2=4

x+y=2

This is the intersecction of a line  and a circle

Using the second equation we can write y = 2 -  x

And substituting this into the first equation for y, we have

x^2 + (2 - x)^2 = 4   simplify

x^2 + x^2 - 4x + 4 = 4   subtract 4 from both sides  and simplify

2x^2 - 4x = 0  factor

2x(x - 2) = 0       setting each factor to 0, we have that x = 0, and x = 2

And using x + y = 2

When x = 0, y = 2

When x = 2, y =0  so the answers are (0,2) (2,0)

Here's a graph......https://www.desmos.com/calculator/eakj6g1q0t

CPhill Mar 24, 2015
#4
+95360
0

Thanks guys,

sometimes it pays to finish a question.

I suppose it also pays to look at a question sensibly before plowing headlong into the algebra   :)

Mar 25, 2015