2x+5y-13=03x+5y=12

1.	2x+5y-13=03x+5y=12
Answer» X=-1,y=3 Get the variable with the smaller coefficient by itself:2x+5y−13=02x+5y−13=0⟹2x=−5y+13⟹2x=−5y+13⟹2x=−4y+12+1−y⟹2x=−4y+12+1−y⟹2x2=−4y\xa02+122+1−y2⟹2x2=−4y\xa02+122+1−y2⟹x=−2y+6+1−y2.⟹x=−2y+6+1−y2.Next lettingt=1−y2t=1−y2⟹2t=1−y⟹2t=1−y⟹y=1−2t.⟹y=1−2t.Let\xa0tt\xa0be an integer, which means that\xa0yy\xa0is an integer. Thenx=−2y+6+1−y2x=−2y+6+1−y2=−2(1−2t)+6+t=−2(1−2t)+6+t=−2+4t+6+t=−2+4t+6+t=4+5t.=4+5t.Check these two answers:2x+5y−13=02x+5y−13=0⟹2(4+5t)+5(1−2t)−13⟹2(4+5t)+5(1−2t)−13=8+10t+5−10t−13=8+10t+5−10t−13=13+10t−10t−13=0✓=13+10t−10t−13=0✓∴∴\xa0integer solutions to the given equation have the formx=4+5t,\xa0y=1−2tx=4+5t,\xa0y=1−2twhere\xa0tt\xa0is an integer.So there is a countable infinitude of integer solutions to the given equation and more than\xa04.4.Choose integer values of\xa0tt\xa0to obtain specific answers. For example,\xa0t=10t=10\xa0yields the answerx=4+5(10)=4+50=54,\xa0y=1−2(10)=1−20=−19.x=4+5(10)=4+50=54,\xa0y=1−2(10)=1−20=−19.Select four different values of\xa0tt\xa0to get four distinct solutions.To obtain positive answers for\xa0xx\xa0and\xa0y,y,\xa0must havex=4+5t>0\xa0and\xa0y=1−2t>0x=4+5t>0\xa0and\xa0y=1−2t>0⟹5t>−4\xa0and\xa01>2t⟹5t>−4\xa0and\xa01>2t⟹t>−45\xa0and\xa0t<12⟹t>−45\xa0and\xa0t<12⟹−45

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