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suppose that x and y are both integers. find the solution of 10x^2+11xy+3y^2=7

Sagot :

konrad509
[tex]10x^2+11xy+3y^2=7\\ 10x^2+5xy+6xy+3y^2=7\\ 5x(2x+y)+3y(2x+y)=7\\ (5x+3y)(2x+y)=7\\\\ \underline{1.\ 5x+3y=1 \wedge 2x+y=7}\\ 5x+3y=1\\2x+y=7\\\\ 5x+3y=1\\ y=7-2x\\\\ 5x+3(7-2x)=1\\ 5x+21-6x=1\\ -x=-20\\ x=20\\\\ y=7-2\cdot20=7-40=-33[/tex]

[tex]\underline{2.\ 5x+3y=7 \wedge 2x+y=1}\\ 5x+3y=7\\2x+y=1\\\\ 5x+3y=7\\ y=1-2x\\\\ 5x+3(1-2x)=7\\ 5x+3-6x=7\\ -x=4\\ x=-4\\\\ y=1-2\cdot(-4)=1+8=9\\\\ [/tex]

[tex]\underline{3.\ 5x+3y=-1 \wedge 2x+y=-7}\\ 5x+3y=-1\\ 2x+y=-7\\\\ 5x+3y=-1\\ y=-7-2x\\\\ 5x+3(-7-2x)=-1\\ 5x-21-6x=-1\\ -x=20\\ x=-20\\\\ y=-7-2\cdot(-20)=-7+40=33[/tex]

[tex]\underline{4.\ 5x+3y=-7 \wedge 2x+y=-1}\\ 5x+3y=-7\\ 2x+y=-1\\\\ 5x+3y=-7\\ y=-1-2x\\\\ 5x+3(-1-2x)=-7\\ 5x-3-6x=-7\\ -x=-4\\ x=4\\\\ y=-1-2\cdot4=-1-8=-9 [/tex]


[tex]x=20 \wedge y=-33\\x=-4 \wedge y=9\\x=-20 \wedge y=33\\x=4 \wedge y=-9[/tex]
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