Question

solve the simultaneous equations.
x+y=2
4y^2-x^2=11

Answer 1

The solution set = { (1/3, 5/3), (5, -3) }

Step-by-step explanation:

Given :

$x + y = 2 \: \: ...(i) \\ 4 {y}^{2} - {x}^{2} = 11 \: ...(ii)$

From (i) substitute x = 2 - y in equation (ii) we get,

$4 {y}^{2} - {(2 - y)}^{2} = 11 \\ \implies \: 4 {y}^{2} - (4 + {y}^{2} - 4y) = 11 \\ \implies3 {y}^{2} + 4y - 15 = 0 \\ \implies \: 3 {y}^{2} + 9y - 5y - 15 = 0 \\ \implies3y(y + 3) - 5(y + 3) = 0 \\ \implies \: (3y - 5)(y + 3) = 0 \\ \implies \: 3y - 5 = 0 \: or \: y + 3 = 0 \\ \implies \: y = \frac{5}{3} \: or \: y = - 3$

When y = 5/3, x = 2 - 5/3 = 1/3

when y = -3, x = 2 - (-3) = 5

Hence, the solution set = { (1/3, 5/3), (5, -3) }

Answer 2

Answer:

x=5,y=-3  and x=1/3,y=5/3

Step-by-step explanation:

x+y=2

x=2-y===>1

4y²-x²=11==>2

Substitute equation 1 in equation 2

4y²-(2-y)²=11

4y²-(2²-2(2)(y)+y²)=11

4y²-(4-4y+y²)=11

4y²-4+4y-y²=11

3y²+4y-4=11

3y²+4y-4-11=0

3y²+4y-15=0

a=3

b=4

c=-15

Product=a×c=3×(-15)=-45

Sum=b=4

9×(-5)=-45

9-5=4

3y²+9y-5y-15=0

3y(y+3)-5(y+3)=0

(y+3)(3y-5)=0

y+3=0

y=-3

3y-5=0

3y=5

y=5/3

Substitute Y in equation 1

x=2-y

x=2-(-3)  --->(y=-3)

x=5,y=-3

x=2-y

y=5/3

x=2-5/3

x=(6-5)/3

x=1/3,y=5/3