Solve the simultaneous equations 2x+y=8.......................................(1) 4x^2+3y^2=52.......................(2) ?

2 Answers
Dec 20, 2017

x= 3.5 and y =1
OR
x = 2.5 and y = 3

Explanation:

2x+y=8.......................................(1)

4x^2+3y^2=52.......................(2)

(1) => y=8-2x

(2) => 4x^2+3(8-2x)^2=52

=> 4x^2 +3(64 - 32x +4x^2) =52

=> 4x^2 + 192 - 96x + 12x^2 = 52

=> 16x^2 -96x + 140 = 0

=> 4(4x^2 - 24x +35) = 0

=> 4x^2 -24x +35 = 0
Solving this quadratic equation, we get:

=> (x-3.5)(x-2.5) = 0

=> x= 3.5 or x = 2.5

Substitute this value of x in equation (1):

Case 1: Taking x= 3.5

=> 2x+y=8

=> 2(3.5) + y = 8

=> y = 8-7 =1

OR

Case 2 : Taking x= 2.5

2(2.5) + y = 8

=> y = 8- 5 = 3

therefore x= 3.5 and y =1

OR

x = 2.5 and y = 3

Dec 20, 2017

y=3 and x=5/2 or y=1 and x=7/2

Explanation:

2x+y=8
4x^2+3y^2=52

2x=8-y=>4x^2=(8-y)^2=color(blue)(64-16y+y^2)

color(blue)(64-16y+y^2)+3y^2=52
4y^2-16y+12=0

dividing everything by 4(for symplifiying calculations):

y^2-4y+3=0

y=(4+-sqrt(4^2-4*1*3))/2

y=(4+-sqrt(4))/2

y=(4+-2)/2

y=3 and x=5/2 or x=1 and x=7/2