MHB How can you solve this equation using an alternate method?

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To solve the equation, first multiply through the parentheses and rearrange the terms so that all $x$ terms are on one side and constant terms on the other. The equation simplifies to $-\frac{7}{9}x = -21$, leading to $x = 27$ after dividing by the coefficient of $x$. An alternate method involves multiplying the entire equation by 9, simplifying, and arriving at the same solution of $x = 27$. The discussion emphasizes the importance of handling fractions correctly when combining like terms. Ultimately, both methods confirm that the solution to the equation is $x = 27$.
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First you have to multiply through the parentheses on the left. Then you have to bring all the terms that are related to $x$ to the left side of the equation and all the constant terms to the right side. Then you can solve for $x$.
 
mathmari said:
First you have to multiply through the parentheses on the left. Then you have to bring all the terms that are related to $x$ to the left side of the equation and all the constant terms to the right side. Then you can solve for $x$.

Can you show me with the numbers?
 
xSilentShuriken said:
Can you show me with the numbers?

Sure! Multiplying through the parentheses:
\begin{align*}-\frac{1}{9}(x-27)+\frac{1}{3}(x+3)=x-17 & \Rightarrow -\frac{1}{9}\left (x+(-27)\right )+\frac{1}{3}(x+3)=x-17 \\ & \Rightarrow \left (-\frac{1}{9}\right )x+\left (-\frac{1}{9}\right )\cdot (-27)+\frac{1}{3}x+\frac{1}{3}\cdot 3=x-17 \\ & \Rightarrow -\frac{1}{9}x+3+\frac{1}{3}x+1=x-17 \end{align*}

Bringing all the terms that are related to $x$ to the left side of the equation and all the constant terms to the right side:
\begin{align*}-\frac{1}{9}x+3+\frac{1}{3}x+1=x-17 & \Rightarrow -\frac{1}{9}x+\frac{1}{3}x-x=-17-1-3 \\ & \Rightarrow \left (-\frac{1}{9}+\frac{1}{3}-1\right )x=-21 \\ & \Rightarrow -\frac{7}{9}x=-21\end{align*}

Solving for $x$ by dividing the equation by the coefficient of $x$:
\begin{equation*}-\frac{7}{9}x=-21 \Rightarrow x=\frac{-21}{-\frac{7}{9}} \Rightarrow x=\frac{21}{\frac{7}{9}} \Rightarrow x=21\cdot \frac{9}{7} \Rightarrow x=27\end{equation*}
 
mathmari said:
Sure! Multiplying through the parentheses:
\begin{align*}-\frac{1}{9}(x-27)+\frac{1}{3}(x+3)=x-17 & \Rightarrow -\frac{1}{9}\left (x+(-27)\right )+\frac{1}{3}(x+3)=x-17 \\ & \Rightarrow \left (-\frac{1}{9}\right )x+\left (-\frac{1}{9}\right )\cdot (-27)+\frac{1}{3}x+\frac{1}{3}\cdot 3=x-17 \\ & \Rightarrow -\frac{1}{9}x+3+\frac{1}{3}x+1=x-17 \end{align*}

Bringing all the terms that are related to $x$ to the left side of the equation and all the constant terms to the right side:
\begin{align*}-\frac{1}{9}x+3+\frac{1}{3}x+1=x-17 & \Rightarrow -\frac{1}{9}x+\frac{1}{3}x-x=-17-1-3 \\ & \Rightarrow \left (-\frac{1}{9}+\frac{1}{3}-1\right )x=-21 \\ & \Rightarrow -\frac{7}{9}x=-21\end{align*}

Solving for $x$ by dividing the equation by the coefficient of $x$:
\begin{equation*}-\frac{7}{9}x=-21 \Rightarrow x=\frac{-21}{-\frac{7}{9}} \Rightarrow x=\frac{21}{\frac{7}{9}} \Rightarrow x=21\cdot \frac{9}{7} \Rightarrow x=27\end{equation*}
How does the negative 1/9 and the 1/3 equal to negative 7/9
 
xSilentShuriken said:
How does the negative 1/9 and the 1/3 equal to negative 7/9

So that we cann add/subtract fractions the denominator must be the same.

We have that the coefficient of $x$ is the following: $$-\frac{1}{9}+\frac{1}{3}-1 = -\frac{1}{9}+\frac{1}{3}\cdot \frac{3}{3}-1\cdot \frac{9}{9} =-\frac{1}{9}+\frac{3}{9}-\frac{9}{9} =\frac{-1+3-9}{9}=\frac{-7}{9}$$
 
As an alternate method, I would begin by multiplying though by 9 to obtain:

$$-(x-27)+3(x+3)=9x-153$$

Distribute:

$$-x+27+3x+9=9x-153$$

Combine like terms:

$$189=7x$$

Divide through by 7:

$$x=27$$ :D
 

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