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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
+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
+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)   Aug 12, 2019