How can you solve this equation using an alternate method?

[xSilentShuriken]

View attachment 7002

 

[mathmari]

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 $x$.

 

[xSilentShuriken]

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?

 

[mathmari]

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*}

 

[xSilentShuriken]

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

 

[mathmari]

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: $$-\frac{1}{9}+\frac{1}{3}-1 = -\frac{1}{9}+\frac{1}{3}\cdot \frac{3}{3}-1\cdot \frac{9}{9} =-\frac{1}{9}+\frac{3}{9}-\frac{9}{9} =\frac{-1+3-9}{9}=\frac{-7}{9}$$

 

[MarkFL]

As an alternate method, I would begin by multiplying though by 9 to obtain:

\(\displaystyle -(x-27)+3(x+3)=9x-153\)

Distribute:

\(\displaystyle -x+27+3x+9=9x-153\)

Combine like terms:

\(\displaystyle 189=7x\)

Divide through by 7:

\(\displaystyle x=27\) :D