# Points of intersection of x^2 + y^2 = 25 and 2x+y=10

#### AGlas9837

##### Junior Member
I'm to find the points of intersection, if any, of the graphs of the equations:

x^2 + y^2 = 25 and 2x+y=10

I solved for y for both equations and got y=5-x and y=10-2x. Then I solved 5-x=10-2x and got x =5, giving y=0.

The answer given in my book also states that the point (3,4) is a point of intersection. How do I get this second point?

#### soroban

##### Elite Member
Re: Points of intersection

Hello, AGlas9837!

Find the points of intersection, if any, of the graphs of the equations:

. . $$\displaystyle x^2+y^2 \:=\:25\,\text{ and }\,2x+y\:=\:10$$
$$\displaystyle \text{You made a }really\text{ silly mistake . . .}$$

$$\displaystyle x^2+y^2 \:=\:25 \quad\Rightarrow\quad y^2 \:=\:25 - x^2 \quad\Rightarrow\quad y \:=\:\om\sqrt{25-x^2}$$

. . $$\displaystyle \text{Then you took the square root ??}$$

$$\displaystyle \text{"Substitution" is the method to use.}$$

$$\displaystyle \text{The second equation gives us: }\:y \:=\:10-2x$$

$$\displaystyle \text{Substitute into the first: }\;x^2 + (10-2x)^2 \:=\:25$$

. . $$\displaystyle \text{which simplifies to: }\:x^2-8x+15 \:=\:0$$

$$\displaystyle \text{Got it?}$$

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#### AGlas9837

##### Junior Member
Unfortunately, no. It's becoming more and more apparent I need to drop this class. Thanks for trying.