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

[AGlas9837]

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]

Re: Points of intersection

Hello, AGlas9837!

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

. . \(\displaystyle x^2+y^2 \:=\:25\,\text{ and }\,2x+y\:=\:10\)

\(\displaystyle \text{You made a }really\text{ silly mistake . . .}\)

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

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


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

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

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

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

\(\displaystyle \text{Got it?}\)

 

[AGlas9837]

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