Wyznacz : Z Po kolei z czynnościami, Dziekuje :)
1) a(a+z)+cz=d
2) az+b=cz+d
3) 3(z+a)+bz=cz+d
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1)![a(a+z)+cz=d\\a^{2}+az+cz=d\\az+cz=d-a^{2}\\z(a+c)=d-a^{2}\\z=\frac{d-a^{2}}{a+c} a(a+z)+cz=d\\a^{2}+az+cz=d\\az+cz=d-a^{2}\\z(a+c)=d-a^{2}\\z=\frac{d-a^{2}}{a+c}](https://tex.z-dn.net/?f=a%28a%2Bz%29%2Bcz%3Dd%5C%5Ca%5E%7B2%7D%2Baz%2Bcz%3Dd%5C%5Caz%2Bcz%3Dd-a%5E%7B2%7D%5C%5Cz%28a%2Bc%29%3Dd-a%5E%7B2%7D%5C%5Cz%3D%5Cfrac%7Bd-a%5E%7B2%7D%7D%7Ba%2Bc%7D)
2)![az+b=cz+d\\az-cz=d-b\\z(a-c)=d-b\\z=\frac{d-b}{a-c} az+b=cz+d\\az-cz=d-b\\z(a-c)=d-b\\z=\frac{d-b}{a-c}](https://tex.z-dn.net/?f=az%2Bb%3Dcz%2Bd%5C%5Caz-cz%3Dd-b%5C%5Cz%28a-c%29%3Dd-b%5C%5Cz%3D%5Cfrac%7Bd-b%7D%7Ba-c%7D)
3)![3(z+a)+bz=cz+d\\3z+3a+bz-cz=d\\3z+bz-cz=d-3a\\z(3+b-c)=d-3a\\z=\frac{d-3a}{3+b-c} 3(z+a)+bz=cz+d\\3z+3a+bz-cz=d\\3z+bz-cz=d-3a\\z(3+b-c)=d-3a\\z=\frac{d-3a}{3+b-c}](https://tex.z-dn.net/?f=3%28z%2Ba%29%2Bbz%3Dcz%2Bd%5C%5C3z%2B3a%2Bbz-cz%3Dd%5C%5C3z%2Bbz-cz%3Dd-3a%5C%5Cz%283%2Bb-c%29%3Dd-3a%5C%5Cz%3D%5Cfrac%7Bd-3a%7D%7B3%2Bb-c%7D)