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?
4 answers
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.
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.
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
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