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.
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
+128474 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)
