What is y^2=5λ,2xy=3λ,3x+4y=11 ?

The solution to y^2=5λ,2xy=3λ,3x+4y=11 is

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y^2=5λ,2xy=3λ,3x+4y=11

y^{2} = 5 λ, 2 x y = 3 λ, 3 x + 4 y = 11

(x = \frac{11}{3}, y = 0, λ = 0)

Solution steps

y^{2} = 5 λ 2 x y = 3 λ 3 x + 4 y = 11

Isolate

λ

for

y^{2} = 5 λ : λ = \frac{y ^{2}}{5}

Plug

λ = \frac{y ^{2}}{5}

into

2 x y = 3 λ

2 x y = 3 \cdot \frac{y ^{2}}{5}

Simplify

2 x y = \frac{3 y ^{2}}{5}

Solve the system of equations:

2 x y = \frac{3 y ^{2}}{5}, 3 x + 4 y = 11 : (x = \frac{11}{3}, x = \frac{33}{49}, y = 0 y = \frac{110}{49})

Verify solutions by plugging them into the original equations

Therefore, the final solution for

y^{2} = 5 λ, 2 x y = 3 λ, 3 x + 4 y = 11

(x = \frac{11}{3}, y = 0, λ = 0)

m^{2} - 8 m + 4 = 0

0 = 6 t^{2} - 18 t + 12

12 x^{2} + 11 x = 5

8 - 9 h = - h^{2}

x^{2} - 208 = 0