2x + 2y = 7, x^2 - 4y^2 = 8

Question

solve by simultaneous equation....

Answer 1

Hi Rodela,

2x+2y=7

y=7-2x/2...eq-1

Substituting the value of y in eq-2,

x^2-4(7-2x/2)^2=8

x^2-4(49-28x+4x^2/4)=8

x^2-49+28x-4x^2=8

-3x^2+28x-57=0

3x^2-28x+57=0

3x^2-19x-9x+57=0

x(3x-19)-3(3x-19)=0

(3x-19)(x-3)=0

3x-19=0  or x-3=0

x=19/3   or x=3

y=7-2x/2

y=7-2*19/3/2  or y=7-6/2

y=-17 /6           or y=1/2

x=19/3, y=-17/6......1

x=3,y= 1/2...............2

Check:for 1,LHS 0f eq-1=2x+2y=2*19/3+2*-17/6=38/3-17/3=21/3=7=RHS

for 2, LHS=2*3+2*1/2=7=RHS

Ans: (19/3,-17/6), (3,1/2)

Meena from Strongsville

Answer 2

2x + 2y = 7
x^2 - 4y^2 = 8

2x = -2y + 7
x = -y + 7/2

(-y + 7/2)^2 - 4y^2 = 8

y^2 - 7y + 49/4 - 4y^2 = 8

0 = 3y^2 + 7y + 8 - 49/4

0 = 12y^2 + 28y + 32 - 49

0 = 12y^2 + 28y - 17

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-28 ± √(28^2 - 4(12)(-17)))/(2(12))

Dividing by 2 is like dividing by √(4):

y = (-14 ± √(7*28 + (12)(17)))/(2*6)

y = (-7 ± √(7*7 + (3)(17)))/6

y = (-7 ± 10)/6 = 1/2 or -17/6

x = -y + 7/2 = 3 or 19/3

Solutions are (3,1/2) and (19/3,-17/6)

Checked with GeoGebra: http://www.wyzant.com/resources/files/266808/intersecting_line_and_hyperbola