ahhhhh help

The graph of the equation 3x + 4y = 15 is a line. The graph of the equation x^2 + y^2 = 9 is a circle. The lines and circle intersect at two points.

To find the coordinates of these points, we can solve the system of equations.

We can do this by first subtracting 3x from both sides of the equation 3x + 4y = 15. This gives us:

4y = 15 - 3x

We can then divide both sides of this equation by 4. This gives us:

y = \frac{15}{4} - \frac{3}{4}x

We can then substitute this equation for y in the equation x^2 + y^2 = 9. This gives us:

x^2 + \left( \frac{15}{4} - \frac{3}{4}x \right)^2 = 9

We can then simplify this equation. This gives us:

x^2 + \frac{225}{16} - \frac{45}{2}x + \frac{9}{16} = 9

We can then combine constant terms on the left-hand side of the equation. This gives us:

x^2 - \frac{45}{2}x + \frac{162}{16} = 0

We can then factor the left-hand side of the equation. This gives us:

\left( x - \frac{27}{4} \right)^2 = 0

We can then take the square root of both sides of the equation. This gives us:

x - \frac{27}{4} = 0

We can then solve for x. This gives us:

x = \frac{27}{4}

We can then substitute this value of x into the equation y = \frac{15}{4} - \frac{3}{4}x to find the y-coordinate of the point of intersection. This gives us:

y = \frac{15}{4} - \frac{3}{4} \cdot \frac{27}{4}

This simplifies to:

y = -\frac{9}{2}

Therefore, the point of intersection of the two graphs is (27/4, -9/2).

I'm sorry... that's not correct... but thanks for trying!

Hi HumenBeing,

We have two equations in two variables.

So, from the first equation: \(3x+4y=15\), we can choose to make whichever variable the "subject" (I.e. Isolated)
So, for example, I choose x:  \(x=\dfrac{15-4y}{3}\)

Next, we substitute this in the second equation: \(x^2+y^2=9\)

Therefore, \((\dfrac{15-4y}{3})^2+y^2=9 \\ \iff \dfrac{1}{9}(15-4y)^2+y^2 = 9 \space \space \space \space \text{Focus more on this step} \\ \iff \dfrac{1}{9}(225-120y+16y^2)+y^2=9 \\ \text{It would be much easier if we got rid of fractions, so why don't we multiply both sides by 9?} \\ \iff 225-120y+16y^2+9y^2=81 \\ \iff (225-81)-120y+(16+9)y^2=0 \\ \iff 144-120y+25y^2=0 \\ \iff 25y^2-120y+144=0 \\ \text{This is a Quadratic Equation, we can solve it either by factoring or by the quadratic formula} \\ \text{But, notice this is a perfect square:} \\ (5y-12)^2=0 \\ \iff y=\dfrac{12}{5} \text{Now, we have found the only value of y} \\ \text{To find the x-coordinate, we just substitute this value of y in the very first equation where we isolated "x"} \\ \implies x=\dfrac{15-4y}{3}=\dfrac{15-4*\dfrac{12}{5}}{3}=\dfrac{9}{5} \)

Therefore, the point is: \((\dfrac{9}{5},\dfrac{12}{5})\)

To check this answer, go to desmos.com and write both equations, notice the line is tangent to the circle at (1.8,2.4) which is exactly the point that we have just found!  

I hope this helps.

HumenBeing is just trying to cheat on homework.