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5(x+7)=5x+35
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If you're asking if that's correct. It is.
5(x+7)=5x+35 5x+35=5x+35
why is it different than this problem 5(x-7)=3(x-2)+2x 5x35=3x-6+2x 5x=5x-6+35 5x-5x=35-6 x cancels out so no solution
You're correct in thinking that no solutions exist.
For the first problem: \[5(x-7)=5x-35\;\; \Longrightarrow \;\;5x-35=5x-35\]So you get \(0=0\). This is always true, so \(x\) has infinite solutions. In the second problem, you can simplify to get \(0=29\). This is obviously not true. Therefore there are no solutions.
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KingGeorege is right. I agree with King
thanks guys
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