Solve for \( x \) e. \( x \times 1=x \) f. \( 1 \times 54=x \) \( 3 \cdot a, x+15=15+10 \) b. \( x \times 7 \times 2=5 \times 2 \times 7 \) c. \( 14 \times 11 \times 2=28 \times x \)
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First, let's break down the equations one by one. For \( e. \) \( x \times 1 = x \): This is an identity! The value of \( x \) can be any real number, and the equation holds true for all of them. For \( f. \) \( 1 \times 54 = x \): Here, simply multiply to find \( x \). So, \( x = 54 \). Next, for \( 3 \cdot a, x + 15 = 15 + 10 \): Rearranging gives \( x + 15 = 25 \); thus, \( x = 10 \). For \( b. \) \( x \times 7 \times 2 = 5 \times 2 \times 7 \): Simplifying, we see both sides equal \( 70 \), leading to \( x = 5 \). Lastly, for \( c. \) \( 14 \times 11 \times 2 = 28 \times x \): Calculate the left side to get \( 308 = 28x \), so \( x = \frac{308}{28} = 11 \). In summary, the solutions are: - \( e: x \) can be any real number - \( f: x = 54 \) - \( x = 10 \) - \( b: x = 5 \) - \( c: x = 11 \)