Question

solve for x and y (solutions of equations involving quadratic equations)
\[4x^{2}+y ^{2}=17\]
\[x ^{2}+y=5\]

Answer 1

y^2 = 17 - 4x^2 , substitute that

Answer 2

err

Answer 3

x^2 = (17 - y^2)/4

Answer 4

so (17 - y^2)/4 + y = 5

Answer 5

beat everyone else ;)

Answer 6

so you have -y^2/4 + y - 5 + 17/4 = 0

Answer 7

y=1, y=3

Answer 8

when y = 1 , x = 2 , and when y = 3 , x = sqrt 2

Answer 9

4x^2 = 17 - y^2 - equation 1

x^2 + y = 5
x^2 = 5 -y - equation 2

substitute equation to into equation 1, you will get
4(5-y) = 17 - y^2

expand everything and you will get
20 - 4y = 17 - y^2

bring everything to one side and you will get
y^2-4y+3 = 0

solve this you wil get
(y-3)(y-1)=0

so y = 3 and y =1

substitute y = 3 and y =1 into equation 2

if y = 3

x^2 = 5-3
*solve this

if y = 1

x^2 = 5-1
x = 2

Answer 10

thank you so much :)

Answer 11

no medals?