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[tex]-7x+1\geq22\qquad\text{subtract 1 from both sides}\\\\-7x\geq21\qquad\text{change the signs}\\\\7x\leq-21\qquad\text{divide both sides by 7}\\\\x\leq-3\\-----------\\-10x+41\geq81\qquad\text{subtract 41 from both sides}\\\\-10x\geq40\qquad\text{change the signs}\\\\10x\leq-40\qquad\text{divide both sides by 10}\\\\x\leq-4\\------------\\\\x\leq-3\ or\ x\leq-4\ therefore\ x\leq-3[/tex]

Answer:

[tex]x\leq -3[/tex]

Step-by-step explanation:

We have been given a compound inequality. We are supposed to solve our given inequality.

[tex]-7x+1\geq 22\text{ or }-10x+41\geq 81[/tex]

First of all, we will solve both of our given inequalities separately, then we will combine the solution of both inequalities.

[tex]-7x+1-1\geq 22-1[/tex]

[tex]-7x\geq 21[/tex]

we know dividing or multiplying inequality by a negative number flips inequality.

[tex]\frac{-7x}{-7}\leq \frac{21}{-7}[/tex]

[tex]x\leq -3[/tex]

[tex]-10x+41\geq 81[/tex]

[tex]-10x+41-41\geq 81-41[/tex]

[tex]-10x\geq 40[/tex]

[tex]\frac{-10x}{-10}\leq \frac{40}{-10}[/tex]

[tex]x\leq -4[/tex]

Upon merging overlapping intervals, the solution for our given inequality would be [tex]x\leq -3[/tex].

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