QUESTION IMAGE
Question
next, subtract 4h from both sides of the equation to get rid of the 4h on the right, while keeping the equation balanced.
\frac{3}{2}(5 + 2h) = 2 + 2h
2 \cdot \frac{3}{2}(5 + 2h) = 2 \cdot (2 + 2h)
3(5 + 2h) = 4 + 4h
15 + 6h = 4 + 4h
- 4h \quad \quad \quad = - 4h
\square \quad \quad \quad = \square
Step1: Identify left/right side terms
Left side: $15+6h$; Right side: $4+4h$
Step2: Subtract $4h$ from both sides
Left: $15+6h-4h = 15+2h$
Right: $4+4h-4h = 4$
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Left box: $15+2h$
Right box: $4$