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At what point does the graph of 3x+4y=15 intersect the graph of x^2+y^2=9? Express any non-integer coordinate as a common fraction.

 Aug 12, 2019
 #1
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+1

We solve this like a system of equations.

Solve for x or y,

Once you do that you will get,

\(\begin{align*} x=\frac{9}{5} \end{align*}\)

Then you can substute this into both equations and you will get for \(y \)

You can then use this to get the final pair of cordinates,

\(\begin{align*} (\frac{9}{5},\frac{12}{5}) \end{align*}\)

 

here is the link to the graph

https://www.desmos.com/calculator 

copy and paste.

HOPE THIS HELPS

 Aug 12, 2019
 #2
avatar+106515 
+1

3x + 4y  = 15   ⇒  4y  = 15-3x    ⇒  y  =  [ 15 -3x ] / 4       (1) 

x^2 + y^2  = 9        (2)

 

Sub (1) into 2 for y  and we have

 

x^2  +  ( [15 - 3x ] / 4 )^2  = 9            simplify

 

x^2  + (1/16) ( 9x^2  - 90x + 225)  = 9          multiply through by 16

 

16x^2  + 9x^2 - 90x + 225  =  144

 

25x^2  - 90x  + 81  = 0         factor

 

(5x - 9) (5x - 9)  = 0

 

(5x- 9)^2  = 0         take the square root

 

5x  - 9  = 0      add 9 to both sides

 

5x  = 9      divide both sides by 5

 

x  = 9/5

 

And using (1)

 

y  =   [ 15 - 3(9/5) ] / 4   =  [ 15 - 27/5 ] / 4   =  [ 75 - 27 ]/ 20  =  48 / 20   =   12 / 5

 

So  the intersection point is  ( 9/5, 12/5)

 

 

cool cool cool

 Aug 12, 2019

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