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Question
- $2 + \frac{1}{4}(x + 12) - \frac{1}{8}x = 6 + \frac{1}{2}(x - 10) - \frac{3}{8}x$
Step1: Simplify left side
First, simplify \(2\cdot\frac{1}{4}(x + 12)-\frac{1}{8}x\).
\(2\cdot\frac{1}{4}(x + 12)=\frac{1}{2}(x + 12)=\frac{1}{2}x+6\).
So left side becomes \(\frac{1}{2}x + 6-\frac{1}{8}x\).
Combine like terms: \(\frac{1}{2}x-\frac{1}{8}x=\frac{4}{8}x-\frac{1}{8}x=\frac{3}{8}x\), so left side is \(\frac{3}{8}x + 6\).
Step2: Simplify right side
Simplify \(6+\frac{1}{2}(x - 10)-\frac{3}{8}x\).
\(\frac{1}{2}(x - 10)=\frac{1}{2}x-5\).
So right side becomes \(6+\frac{1}{2}x-5-\frac{3}{8}x\).
Simplify constants: \(6 - 5 = 1\).
Combine like terms: \(\frac{1}{2}x-\frac{3}{8}x=\frac{4}{8}x-\frac{3}{8}x=\frac{1}{8}x\), so right side is \(\frac{1}{8}x + 1\).
Step3: Solve for \(x\)
Now equation is \(\frac{3}{8}x + 6=\frac{1}{8}x + 1\).
Subtract \(\frac{1}{8}x\) from both sides: \(\frac{3}{8}x-\frac{1}{8}x + 6=1\)
\(\frac{2}{8}x + 6=1\) → \(\frac{1}{4}x + 6=1\).
Subtract 6 from both sides: \(\frac{1}{4}x=1 - 6=-5\).
Multiply both sides by 4: \(x=-5\times4=-20\)? Wait, wait, let's check again. Wait, maybe I made a mistake. Wait, let's re - do step 1 and 2.
Wait, original equation: \(2\cdot\frac{1}{4}(x + 12)-\frac{1}{8}x=6+\frac{1}{2}(x - 10)-\frac{3}{8}x\)
Step1: Left side: \(2\times\frac{1}{4}(x + 12)=\frac{1}{2}(x + 12)=\frac{1}{2}x+6\). Then subtract \(\frac{1}{8}x\): \(\frac{1}{2}x+6-\frac{1}{8}x=\frac{4}{8}x-\frac{1}{8}x + 6=\frac{3}{8}x + 6\). Correct.
Right side: \(6+\frac{1}{2}(x - 10)-\frac{3}{8}x=6+\frac{1}{2}x-5-\frac{3}{8}x=(6 - 5)+(\frac{1}{2}x-\frac{3}{8}x)=1+\frac{4}{8}x-\frac{3}{8}x=1+\frac{1}{8}x\). Correct.
So equation: \(\frac{3}{8}x + 6=\frac{1}{8}x + 1\)
Subtract \(\frac{1}{8}x\) from both sides: \(\frac{3}{8}x-\frac{1}{8}x+6 = 1\) → \(\frac{2}{8}x+6 = 1\) → \(\frac{1}{4}x=1 - 6=-5\) → \(x=-5\times4=-20\). Wait, but let's check by plugging back.
Wait, maybe I misread the original equation. Wait, the original equation is \(2\cdot\frac{1}{4}(x + 12)-\frac{1}{8}x=6+\frac{1}{2}(x - 10)-\frac{3}{8}x\)? Wait, maybe the first term is \(2+\frac{1}{4}(x + 12)\) instead of \(2\cdot\frac{1}{4}(x + 12)\)? Oh! That's the mistake. Let's re - do with \(2+\frac{1}{4}(x + 12)-\frac{1}{8}x\).
Step1 (corrected): Simplify left side (correcting the first term)
Left side: \(2+\frac{1}{4}(x + 12)-\frac{1}{8}x\).
\(\frac{1}{4}(x + 12)=\frac{1}{4}x + 3\).
So left side: \(2+\frac{1}{4}x + 3-\frac{1}{8}x\).
Simplify constants: \(2 + 3=5\).
Combine like terms: \(\frac{1}{4}x-\frac{1}{8}x=\frac{2}{8}x-\frac{1}{8}x=\frac{1}{8}x\). So left side: \(\frac{1}{8}x + 5\).
Step2 (corrected): Simplify right side (same as before)
Right side: \(6+\frac{1}{2}(x - 10)-\frac{3}{8}x=1+\frac{1}{8}x\) (as before).
Step3 (corrected): Solve for \(x\)
Now equation: \(\frac{1}{8}x + 5=\frac{1}{8}x + 1\).
Subtract \(\frac{1}{8}x\) from both sides: \(5 = 1\). Which is a contradiction. So no solution, \(\varnothing\).
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\(\varnothing\) (white)