4x^2-y^2=35 and 2x+y=7 ?

To solve this system of equations, we can use substitution or elimination method.

Let's solve using substitution method:

From the second equation, we can express y in terms of x:
y = 7 - 2x

Now substitute y in the first equation:
4x^2 - (7 - 2x)^2 = 35
4x^2 - (49 - 28x + 4x^2) = 35
4x^2 - 49 + 28x - 4x^2 = 35
28x - 49 = 35
28x = 84
x = 3

Now substitute x back into the second equation to find y:
2(3) + y = 7
6 + y = 7
y = 1

Therefore, the solution to the system of equations is x = 3 and y = 1.

Given 4x^2-y^2=35 and 2x y=7
$$\Rightarrow (2x)^2-y^2=35$$ $$\Rightarrow y^2=4x^2-35$$ substituting the value of y^2 in 2xy=7 $$2x\sqrt{4x^2-35}=7$$ $$4x^2-35=\frac{49}{4x^2}$$ $$16x^4-140x^2-49=0$$ $$(4x^2-49)(4x^2+7)=0$$ $$\Rightarrow 4x^2-49=0 , 4x^2+7=0$$ $$\Rightarrow (2x-7)(2x+7)=0 , 4x^2=-7$$ $$\Rightarrow 2x=7 , 2x=-7 , 4x^2=-7$$ $$\Rightarrow x=\frac72 , x=-\frac72 , x^2=-\frac74$$ $$\Rightarrow x=\frac72 , x=-\frac72 , x=\pm \frac{\sqrt{7}i}{2}$$