what is x/3 + 1 = 9
\[\frac{ x }{ 3 }+ 1 =9\] \[3(\frac{ x }{ 3 }) + 1 = 3\times9\] \[x + 1 = 27\] \[x + 1 -1 = 27 -1\] x= 26
Yes, first you have to get the x alone, and do to that you must first subtract \(1\) on both sides of the equation, since you are adding \(1\) in the equation \(\rm \frac{x}{3}~ + ~1 ~= ~9\) \(\rm \frac{x}{3}~ + ~1~-~1 ~= ~9~-~1\) \(\rm \frac{x}{3}= ~9~-~1\) So what is \(9~-~1\)?
You do not do what is connected with the variable first, you always do what isn't paired with the variable when it comes to two-step equations
\(9~-~1~=~8\) So now we plug in \(8\) where \(9~-~1\) was in the equation. \(\rm \frac{x}{3}= ~8\ Now next to make the x alone we must multiply \(3\) on both sides of the equation, since we are dividing in the original equation \(\rm \frac{x}{3}\times 3= ~8\times 3\) \(\rm x~= ~8\times 3\)
Oops LaTeX fail
But all that you need to do is multiply \(8\times 3\)
Sorry my answer was wrong at the top. The person above is not 100% correct. There is no rule that states that you can't start with variable first. It does make it easier at times, but if you have a fraction it's easier to get rid of the fraction first. \[\frac{ x }{ 3 }+ 1 = 9\] \[3(\frac{ x }{ 3 }+ 1) = 3\times9\] \[x + 3 = 27\] \[x + 3 -3 = 27 -3\] \[x = 24\]
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