substituting again..many thanks

[mpkoellein]

thanks to everyone...I was able to move past simple fractions in solving by substitution...now a rational equation that I'm not getting how to get rid of the 3y in the first equaltion....I came up with 11x-9y=51 for the second equation...thanks for your time....

4x-7y/3y + 1/3 = 0

5x-17= 4x/3 - 3y

 

[mmm4444bot]

4x-7y/3y + 1/3 = 0

5x-17= 4x/3 - 3y

Hi there:

You need to learn how to properly type algebraic ratios. I suspect that what you've actually typed above is not what you intend.

Here is what your first equation means:

\(\displaystyle 4x \;-\; \frac{7y}{3y} \;+\; \frac{1}{3} \;=\; 0\)

Here is what your second equation means:

\(\displaystyle 5x \;-\; 17 \;=\; \frac{4x}{3} \;-\;3y\)

It's easy to properly type algebraic ratios, to clearly show what belongs in numerators versus denominators.

Use grouping symbols. In other words, enclose numerators and denominators within parentheses.

For example:

5x - 17 = 4x/(3 - 3y)

means the following

\(\displaystyle 5x \;-\; 17 \;=\; \frac{4x}{3 \;-\;3y}\)

and

(4x - 7y)/(3y + 1/3) = 0

means the following

\(\displaystyle \frac{4x \;-\; 7y}{3y \;+\; \frac{1}{3}} \;=\; 0\)

Please retype your two equations for us, using proper grouping symbols, so that we can read the actual equations.

You can also go to THIS SITE to learn more about how to properly type mathematical expressions.

 

[soroban]

Hello, mpkoellein!

\(\displaystyle \begin{array}{cccc} \dfrac{4x-7y}{3y} + \dfrac{1}{3} \;= \; 0 & [1] \\ \\[-2mm] 5x-17 \;=\; \dfrac{4x}{3} - 3y & [2] \end{array}\)

\(\displaystyle \text{Multiply [1] by }3y\!:\quad 4x - 7y + y \:=\:0 \quad\Rightarrow\quad 4x - 6y \:=\:0 \qyad\Rightarrow\quad y \:=\:\tfrac{2}{3}x \quad[3]\)

\(\displaystyle \text{Multiply [2] by 3: }\;\;15x - 51 \:=\:4x - 9y \quad\Rightarrow\quad 11x + 9y \:=\:51 \quad [4]\)

\(\displaystyle \text{Substitute [3] into [4]: }\;11x + 9\left(\tfrac{2}{3}x\right) \:=\:51 \quad\Rightarrow\quad 17x \:=\:51 \quad\Rightarrow\quad\boxed{ x \:=\:3}\)

\(\displaystyle \text{Substitute into [3]: }\;y \:=\:\tfrac{2}{3}(3) \quad\Rightarrow\quad\boxed{ y \:=\:2}\)