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Himpunan penyelesaian dari sistem persamaan eksponen adalah...

Himpunan penyelesaian dari sistem persamaan eksponen open curly brackets table attributes columnalign left end attributes row cell 7 to the power of x plus 2 end exponent equals 49 to the power of y minus 1 end exponent end cell row cell 10 to the power of x squared plus 5 end exponent equals 1000 to the power of y end cell end table close adalah...

  1. open curly brackets open parentheses negative 1 half comma 7 over 4 close parentheses comma open parentheses negative 2 comma negative 3 close parentheses close curly brackets

  2. open curly brackets open parentheses negative 1 half comma negative 7 over 4 close parentheses comma open parentheses 2 comma 3 close parentheses close curly brackets

  3. open curly brackets open parentheses negative 1 half comma 7 over 4 close parentheses comma open parentheses 2 comma 3 close parentheses close curly brackets

  4. open curly brackets open parentheses negative 1 half comma 7 over 4 close parentheses comma open parentheses negative 2 comma 3 close parentheses close curly brackets

  5. open curly brackets open parentheses negative 1 half comma 7 over 4 close parentheses comma open parentheses 2 comma negative 3 close parentheses close curly brackets

Jawaban:

7 to the power of x plus 2 end exponent equals 49 to the power of y minus 1 end exponent  7 to the power of x plus 2 end exponent equals 7 to the power of 2 y minus 2 end exponent  x plus 2 space equals 2 y minus 2  x minus 2 y equals negative 4 space..... left parenthesis 1 right parenthesis  10 to the power of x squared plus 5 end exponent equals 1000 to the power of y  x squared plus 5 equals 3 y  x squared minus 3 y equals negative 5 space.... left parenthesis 2 right parenthesis

Sebelumnya kita perlu mengeliminasi variabel y dari persamaan (1) dan (2).

x minus 2 y space equals space minus 4 space vertical line space cross times 3 space rightwards arrow space 3 x minus 6 y equals negative 13  x squared minus 3 y equals negative 5 space vertical line space cross times 2 space rightwards arrow space fraction numerator 2 x squared minus 6 y equals negative 10 over denominator blank end fraction minus  3 x minus 2 x squared equals negative 2  2 x squared minus 3 x minus 2 equals 0  left parenthesis 2 x plus 1 right parenthesis left parenthesis x minus 2 right parenthesis equals 0  x equals negative 1 half space logical or space x equals 2

Untuk x=negative 1 half maka

x-2y = -4
negative 1 half - 2y = -4
2y= 7 over 2 rightwards arrowy= 7 over 4

Untuk x= 2 maka

x-2y= -4
2- 2y = -4
2y = 6 rightwards arrow y = 3

Jadi himpunan penyelesaian kedua persamaan tersebut adalah open curly brackets open parentheses negative 1 half comma 7 over 4 close parentheses comma open parentheses 2 comma 3 close parentheses close curly brackets

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