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Solve the simultaneous equations, 2x+y-5=0 and x^2-y^2=3

2x+y-5=0, y=5-2x (put into second equation)

x²-y²=3, substituting in we get, x²-(5-2x)²=3, expand, x²-(25+4x²-20x)=3, simplify, x²-25-4x²+20x=3, 0=3x²-20x+28, put into brackets by seperating the two factors, 0=3x²-6x-14x+28, 0=(3x-14)(x-2), therefore x=14/3 and x=2, y=1 and y=-13/3

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Answered by Joshua S. • Maths tutor

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