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Jika begin mathsize 14px style open curly brackets table attributes columnalign left end attributes row cell begin inline style fraction numerator 2 over denominator straight x plus straight y end fraction minus fraction numerator 1 over denominator straight x minus straight y end fraction equals 3 over 4 end style end cell row cell begin inline style fraction numerator 1 over denominator straight x plus straight y end fraction plus fraction numerator 2 over denominator straight x minus straight y end fraction equals 1 end style end cell end table close end style maka x + y = ....

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A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

begin mathsize 14px style fraction numerator 1 over denominator straight x plus straight y end fraction equals straight a space lalu space fraction numerator 1 over denominator straight x minus straight y end fraction equals straight b    Maka space soal space dapat space ditulis  table row cell 2 straight a minus straight b equals 3 over 4 end cell row cell straight a plus 2 straight b equals 1 end cell end table open parentheses table row cell cross times 2 end cell row cell cross times 1 end cell end table close parentheses  4 straight a minus 2 straight b equals begin inline style 3 over 2 end style  bottom enclose space space straight a plus 2 straight b equals 1 space space plus end enclose  space space space space space space space 5 straight a equals 5 over 2  space space space space space space space space space straight a equals 1 half  fraction numerator 1 over denominator straight x plus straight y end fraction equals 1 half rightwards double arrow straight x plus straight y equals 2 end style

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