2x + 3y = 13, x^2 + y^2 = 78

Question

solve the simultaneous equation, giving the answer in their simplest surd form..

Answer 1

Hi Rodela,

2x+3y=13......Eq-1

x^2+y^2=78......Eq-2

from Eq-1, y=13-2x/3

substituting value of y in Eq-2,

we get, x^2+(13-2x/3)^2=78

x^2+169/9-52x/9+4x^2/9=78

9x^2+169-52x+4x^2=78*9

9x^2-52x+169=702

13x^2-52x-533=0

13(x^2-4x-41)=0

x^2-4x-41=0

x=-b+sq root of (b^2-4ac)/2a  or -b-sq root of (b^2-4ac)/2a........(Quadratic eq formula for a^2+bx+c=0))

substituting the value of a=1,b=-4,c=-41

x=-(-4)+sq root of (-4^2-4*1*-41)/2*1=4+sq root of (16+164)/2=4+sq root of 180/2=4+13.42/2=8.71

                    or

x=4-13.42/2=-9.42/2=-4.71.

   y=13-2x/3, substituting the valueof x in eq,

if x=8.71, y=13-17.42/3=-1.47

or if x=-4.71, y=13+9.42/3=7.47

 

so x=8.71, y=-1.47

or x=-4.71, y=7.47.Check answers.

 

Meena from Strongsville, OH

Answer 2

2x + 3y = 13
x^2 + y^2 = 78

(2x)^2 = (-3y + 13)^2

4x^2 = 9y^2 - 2*3*13y + 13*13
+ 4y^2 = + 4y^2

4(x^2 + y^2) = 4(78) = 4*6*13 = 13y^2 - 6*13y + 13*13

24 = y^2 - 6y + 13

0 = y^2 - 6y - 11

y = (-b ± √(b^2 - 4ac))/(2a)

= (- -6 ± √(2*3*2*3 - 2*2(1)(-11)))/2

Dividing by 2 is same as dividing by √(4):

y = 3 ± √(3*3 + 11) = 3 ± √(4*5)

y = 3 ± 2 √(5)

x = (13 - 3y)/2 = (13 - 3(3 ± 2 √5)))/2

x = (13 - 9 ± (-6) √(5)))/2

x = 2 ± (-3) √(5)

Solutions are:
(2 - 3 √(5),3 + 2 √(5)), and
(2 + 3 √(5),3 - 2 √(5))

 

Checked with GeoGebra.