Question

\(\frac{3x+1}{18}\) + \(\frac{2y}{12}\)=\(\frac{2}{9}\) và x-y =-1

Answer 1

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cho 3 k lun

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Answer 2

Ta có :

   \(\frac{3x+1}{18}+\frac{2y}{12}=\frac{2}{9}\)

   \(\frac{3x+1}{18}+\frac{y}{6}=\frac{2}{9}\)

   \(\frac{3x+1}{18}+\frac{3y}{18}=\frac{2}{9}\)

   \(\frac{3x+1+3y}{18}=\frac{2}{9}\)

\(\frac{3\left(x+y\right)+1}{18}=\frac{2}{9}\)

\(\left[3\left(x+y\right)+1\right]\times9=18\times2\)

\(3\left(x+y\right)+1=18\div9\times2\)

\(3\left(x+y\right)+1=4\)

\(3\left(x+y\right)=3\)

\(x+y=1\)(1)

Mà x - y = -1 (2)

   Từ (1) và (2) suy ra \(x=\left(1-1\right)\div2\)

                                    \(x=0\)

   Lại có : x - y = -1

  hay 0 - y = -1

              y = 1

  Vậy x = 0 , y = 1