How do you solve for the intersection points?

x^2+y^2=25
4y=3x

I know the points are (4,3) and (-4,-3) but how do you get this algebraically?

Here, no quadratics required, luckily.

solving the second equation for y in terms of x
y = (3/4) * x

substituting into the first equation
x^2 + y^2 = 25
yields
x^2 + ( (3/4) * x)^2 = 25
x^2 + (9/16)*(x^2)=25
combining terms
(25/16) * (x^2)= 25
x^2 = 25 * (16/25)
x^2 = 16
x = +-4
substitute back into the original
x^2 is always 16
so
16 + y^2 = 25
y^2 = 9
y = +-3

Use the second equation to express one variable in terms of the other variable. Substitute that into the first expression. Expand and combine the terms. Solve by inspection of the quadratic forumula.

Oops! I meant to say:

Solve by inspection OR the quadratic forumla.

Yea... I've been trying that but I get fraction numbers and I think I'm doing it wrong.

To solve for the intersection points algebraically, you need to find the values of x and y that satisfy both equations simultaneously. Here's how you can do it step by step:

1. Start by substituting the value of y from the second equation into the first equation:
4y = 3x
y = (3/4)x

2. Substitute this expression for y in the first equation:
x^2 + (3/4)x^2 = 25
Simplify the equation:
(16/16)x^2 + (12/16)x^2 = 25
(28/16)x^2 = 25
Multiply both sides by 16:
28x^2 = 400
Divide both sides by 28:
x^2 = 400 / 28
x^2 = 14.2857
Take the square root of both sides:
x = ±√(14.2857)

3. Now, substitute the values of x into the second equation to find the corresponding y-values:
For x = √(14.2857):
4y = 3(√(14.2857))
y = 3√(14.2857) / 4

For x = -√(14.2857):
4y = 3(-√(14.2857))
y = -3√(14.2857) / 4

So, the two sets of intersection points between the two equations are approximately (√(14.2857), 3√(14.2857) / 4) and (-√(14.2857), -3√(14.2857) / 4).

To get the specific points you mentioned, (4,3) and (-4,-3), it seems there might be a calculation or rounding error.