Question
Mathematics
Hallan: $\frac{\left(b+d\right)c}{a}$((b+d)c)/(a)
$\frac{48{x}^{7}{y}^{10}{z}^{12}}{12{x}^{3}{y}^{5}{z}^{6}}=a{x}^{6}{y}^{6}{z}^{2}$(48x^(7)y^(10)z^(12))/(12x^(3)y^(5)z^(6))=ax^(6)y^(6)z^(2)
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Answer: $\frac{4}{y}$(4)/(y)
Solution: We can simplify the expression on the left by canceling out common factors of the numerator and denominator.
$\begin{array}{rl}\frac{48{x}^{7}{y}^{10}{z}^{12}}{12{x}^{3}{y}^{5}{z}^{6}}& =\frac{4\cdot 12{x}^{7}{y}^{10}{z}^{12}}{12\cdot {x}^{3}{y}^{5}{z}^{6}}\\ \\ & =\frac{4{x}^{4}{y}^{5}{z}^{6}\cdot 3{x}^{3}{y}^{5}{z}^{6}}{{x}^{3}{y}^{5}{z}^{6}\cdot 4}\\ \\ & =3{x}^{4}\\ \end{array}${:[(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:
$\begin{array}{rl}3{x}^{4}& =a{x}^{6}{y}^{6}{z}^{2}\\ \\ a& =\frac{3{x}^{4}}{{x}^{6}{y}^{6}{z}^{2}}\\ \\ a& =\frac{3}{{x}^{2}{y}^{6}{z}^{2}}\\ \end{array}${:[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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