Solve the simultaneous equations
4x^{2}+y^{2}=10, 2x-y=4 

Answer
:

2x-y=4  \\  \therefore y=2x-4

Substituting the value of y in 4x^{2}+y^{2}=10, we get

4x^{2}+\left( 2x-4\right) ^{2}=10  \\  4x^{2}+\left( 2x\right) ^{2}+2\times \left( 2x\right) \times \left( -4\right) +\left( -4\right) ^{2}=10  \\  4x^{2}+4x^{2}-16x+16=10  \\  8x^{2}-16x+16-10=0  \\  8x^{2}-16x+6=0

Dividing both sides by 2, we get

4x^{2}-8x+3=0 which can be factorised as follows;

4x^{2}-6x-2x+3=0  \\  2x\left( 2x-3\right) -1\times \left( 2x-3\right) =0  \\  \left( 2x-3\right) \left( 2x-1\right) =0

\therefore 2x-3=0 or 2x-1=0

or, 2x=3 or 2x=1

thus, x=\dfrac {3}{2} or x=\dfrac {1}{2}

Going back to the simultaneous equations, y=2x-4

Thus for x=\dfrac {3}{2}, \space y=2\times \dfrac {3}{2}-4=3-4=-1

For x=\dfrac {1}{2}, \space y=2\times \dfrac {1}{2}-4=1-4=-3

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