Question

can someone help me with this problem?

Answer 1

Solve the following systems of a circle and a line by finding the intersection point(s), if any,
algebraically.

2x + y = 15
(x - 2) + (y - 1) = 25

Answer 2

do I substitute the value for y into the circle equation and solve
for x?

Answer 3

get y by itself

Answer 4

nvm

Answer 5

I assume you mean
\[ 2x + y = 15\\ (x - 2)^2 + (y - 1)^2 = 25
\]
it is a bit painful. but you can first solve for y in the first equation
\[ y = 15-2x\]
and replace y in the 2nd equation with this expression:
\[ (x - 2)^2 + (15-2x - 1)^2 = 25 \]
now solve for x (you may be either 2 ,1 or 0 real solutions for x)
then find y using y = 15-2x

Answer 6

thank you

Answer 7

do you want me to let you know what I get when I'm done working it out?

Answer 8

you can post the answer and I'll check it.

Answer 9

okay

Answer 10

do I turn (x-2)^2 into x^2 - 4x +4?

Answer 11

I simplified (15-2x-1)^2 and got 4x^2 - 56x + 196

Answer 12

so far, so good.
x^2 - 4x +4 + 4x^2 - 56x + 196 = 25

Answer 13

I combine like terms and got

5x^2 - 60x + 200 = 25

Answer 14

do I sub 25 from both sides now?

Answer 15

yes, add -25 on both sides. then divide both sides (and all terms) by 5 to simplify it a bit

Answer 16

okay

Answer 17

5x^2 - 60x + 175 = 0

x^2 - 12x + 175 = 0

Answer 18

you should divide 175 by 5 also
you are doing
\[ \frac{5x^2 - 60x + 175 }{5} = \frac{0}{5} \\
\frac{5x^2 }{5} + \frac{-60x}{5} + \frac{175}{5} = 0
\]

Answer 19

oh okay

Answer 20

x^2 - 12x + 35 = 0

Answer 21

yes, and that factors

Answer 22

am I done?

Answer 23

what did you get ?

Answer 24

x = 7,5

Answer 25

or x = 5,7

Answer 26

yes. but they want the intersection points, so you want (x,y) pairs
you find y using y= 15-2x

Answer 27

fyi, here is a graph of the solution

Answer 28

okay thanks