первом просто подставляешь значения y=3x+1
х=5 тогда у=16
х=7 тогда у=-20
х=1/3 тогда у=2
во втором решаем как обычное линейное уравнение
1)-4=3х+1
-5=3х
х=-1 и 2/3
2)11=3х+1
10=3х
х=3 и 1/3
3)8=3х+1
7=3х
х=2 и 1/3
11n-7n+1-6n+8= -2n+9= -2*16+9= -32+9= -23
(х-3)(х-7)-2х(3х-5)= х²-7х-3х+21-6х²+10х=-5х²-10х+21+10х=-5х²+21
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ОДЗ: x-5≠0⇒x≠5
![\frac{x^2-25}{x-5}\geq0\\\frac{(x-5)(x+5)}{x-5}\geq0\\x+5\geq0\\x\geq-5](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2-25%7D%7Bx-5%7D%5Cgeq0%5C%5C%5Cfrac%7B%28x-5%29%28x%2B5%29%7D%7Bx-5%7D%5Cgeq0%5C%5Cx%2B5%5Cgeq0%5C%5Cx%5Cgeq-5)
Включая ОДЗ: x∈[-5;5)U(5;+∞)
![|x-4|\leq2\\x-4\leq2\ \ \ \ \ -(x-4)\leq2\\x\leq6\ \ \ \ \ \ \ \ \ \ \ x-4\geq-2\\x\leq6\ \ \ \ \ \ \ \ \ \ \ x\geq2\\x\in[2;6]](https://tex.z-dn.net/?f=%7Cx-4%7C%5Cleq2%5C%5Cx-4%5Cleq2%5C+%5C+%5C+%5C+%5C+-%28x-4%29%5Cleq2%5C%5Cx%5Cleq6%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+x-4%5Cgeq-2%5C%5Cx%5Cleq6%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+x%5Cgeq2%5C%5Cx%5Cin%5B2%3B6%5D)
![\left \{ {{\frac{x^2-25}{x-5}\geq0} \atop {|x-4|\leq2}} \right. \Rightarrow \left \{ {{x\in[-5;5)\cup (5;+\infty)} \atop {x\in[2;6]}} \right.\Rightarrow x\in[2;5)\cup(5;6]](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B%5Cfrac%7Bx%5E2-25%7D%7Bx-5%7D%5Cgeq0%7D+%5Catop+%7B%7Cx-4%7C%5Cleq2%7D%7D+%5Cright.+%5CRightarrow+%5Cleft+%5C%7B+%7B%7Bx%5Cin%5B-5%3B5%29%5Ccup+%285%3B%2B%5Cinfty%29%7D+%5Catop+%7Bx%5Cin%5B2%3B6%5D%7D%7D+%5Cright.%5CRightarrow+x%5Cin%5B2%3B5%29%5Ccup%285%3B6%5D)