1/7(35+4z)=5+35-z

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Solution for 1/7(35+4z)=5+35-z equation: 


z in (-oo:+oo)

(1/7)*(4*z+35) = 5-z+35 // - 5-z+35

(1/7)*(4*z+35)+z-35-5 = 0

1/7*(4*z+35)+z-35-5 = 0

1/7*(4*z+35)+z-35-5 = 0

11/7*z-35-5+5 = 0

11/7*z-30-5 = 0

11/7*z-35 = 0

11/7*z-35 = 0

11/7*z-35 = 0 // + 35

11/7*z = 35 // : 11/7

z = 35/11/7

z = 245/11

z = 245/11

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