Question

Решите систему уравнения.
x²-y²=8
x+y=4

Answer 1

Ответ:

х=4-у х=4-у х=4-у х=4-у х=4-1 х=3

(4-у)²-у²=8 16-8у+у²-у²=8 -8у=8-16 у=1 у=1 у=1

Answer 2

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$\displaystyle \bf \left \{ {{x^2-y^2=8} \atop {x+y=4 \ \ \ }} \right. \ \ \ \left \{ {{x^2-y^2=8} \atop {x=4-y \ \ \ }} \right. \ \ \ \left \{ {{(4-y)^2-y^2=8} \atop x=4-y \ \ \ \ \ \ \ \ \ \ }} \right. \\\\\\\left \{ {{y=1 \ \ \ \ \ } \atop {x=4-1}} \right. \ \ \ \left \{ {{y=1} \atop {x=3}} \right.$

$\displaystyle \bf \Big( 4-y\Big)^2-y^2=8\\\\16-8y+y^2-y^2=8\\\\16-8y=8\\\\-8y=8-16\\\\-8y=-8\\\\y=1$

$\boxed{ \displaystyle \bf \Big(3 \ ; 1\Big)}$