SOLUTION: Solve the system by the method of substitution. {x^2+y^2=25 {3x-4y=0 a. no real solution b.(3,4),(3,-4) c.(4,3),(-4,-3) d.(4,3) e.(4,3)(-4,3),(-4,-3),(4,-3)

Question 470832: Solve the system by the method of substitution.
{x^2+y^2=25
{3x-4y=0
a. no real solution
b.(3,4),(3,-4)
c.(4,3),(-4,-3)
d.(4,3)
e.(4,3)(-4,3),(-4,-3),(4,-3)
Answer by hinaahsan(5) (Show Source): You can put this solution on YOUR website!
x^2 + y^2 =25 --> (1)
whereas, 3x-4y=0 -->(2)

using equation 2
3x=4y
x=4y/3 -->(3)
using value of x from equation (3) and subsituting in equation (1)
(4y/3)^2 + y^2 =25
16y^2/9 +y^2 =25
(16y^2/9 +9y^2/9) = 25
25y^2/9 =25
y^2/9 =25/25
y^2/9 =1
y^2 = 9
(y)^2 = (3)^2
y = 3 -->(4)
using the value of y from equation (4) in equation (3) to get the value of x
x= 4y/3
x= 4(3)/3
x=4
Ans: (x,y)=(4,3)
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