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Hallan: ( b + d ) c a ((b+d)c)/(a)
48 x 7 y 10 z 12 12 x 3 y 5 z 6 = a x 6 y 6 z 2 (48x^(7)y^(10)z^(12))/(12x^(3)y^(5)z^(6))=ax^(6)y^(6)z^(2)
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Answer: 4 y (4)/(y)
Solution: We can simplify the expression on the left by canceling out common factors of the numerator and denominator.
48 x 7 y 10 z 12 12 x 3 y 5 z 6 = 4 12 x 7 y 10 z 12 12 x 3 y 5 z 6 = 4 x 4 y 5 z 6 3 x 3 y 5 z 6 x 3 y 5 z 6 4 = 3 x 4 {:[(48x^(7)y^(10)z^(12))/(12x^(3)y^(5)z^(6))=(4*12x^(7)y^(10)z^(12))/(12*x^(3)y^(5)z^(6))],[],[=(4x^(4)y^(5)z^(6)*3x^(3)y^(5)z^(6))/(x^(3)y^(5)z^(6)*4)],[],[=3x^(4)],[]:}
Therefore, we have 3 x 4 = a x 6 y 6 z 2 3x^(4)=ax^(6)y^(6)z^(2). Solving for a a, we get:
3 x 4 = a x 6 y 6 z 2 a = 3 x 4 x 6 y 6 z 2 a = 3 x 2 y 6 z 2 {:[3x^(4)=ax^(6)y^(6)z^(2)],[],[a=(3x^(4))/(x^(6)y^(6)z^(2))],[],[a=(3)/(x^(2)y^(6)z^(2))],[]:}
Finally, substituting a a into the original expression and simplifying, we get:
\begin{align*}
\frac{(b + d)c}{a} &= \frac{b + d}{a} \cdot c \\
&= \frac{b + d}{\frac{3}{x{2}y{6}z^{2}}} \cdot c \\
&= (b + d) \cdot \frac{x{2}y{6}z^{2}}{3c} \\
&= \frac{bx{2}y{6}z^{2}}{3c} + \frac{dx{2}y{6}z^{2}}{3c} \
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