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Solve the following linear inequality: 

3(x + 4) ≥ 5x – 123

Sagot :

konrad509
[tex]3(x + 4) \geq 5x - 123 \\ 3x+12\geq5x-123\\ 2x\leq135\\ x\leq\frac{135}{2} [/tex]
[tex]3(x+4)\geq5x-123\\3\cdot x+3\cdot4\geq5x-123\\3x+12\geq5x-123\ \ \ \ /-12\\3x\geq5x-135\ \ \ \ /-5x\\-2x\geq-135\ \ \ /:(-2) < 0\ then\ "\geq"\ change\ "\leq"\\x\leq67.5\\\\x\in(-\infty;\ 67.5 >[/tex]
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