# Solving a system of equations

Could anyone help me to solve the following system of equations:

\begin{align*} x-y&=17 \\ \frac{4}{3} x+ \frac {3}{2} y &= 0 \end{align*}

How should I go about solving this, I am stuck.

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What have you tried? –  Simon Markett May 20 '12 at 18:00
The three most common methods are elimination, substitution, and graphing. I would try either of the first two, after multiplying the second equation by $6$... –  The Chaz 2.0 May 20 '12 at 18:02
I multiplied 3 and 2 to the zero –  James Pedeston May 20 '12 at 18:02
Try solving for $y$ in the first equation. Then substitute that into the second to get a single variable equation, and solve for $x$. With $x$, you can easily find $y$ because $x-y=17$. –  Argon May 20 '12 at 21:48

From the first equation by adding to both sides $y$ you get $x=17+y$. Next try to substitute obtained equation for $x$ in the second equation.

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Welcome to math stackexchange! I suggest learning how to use Tex to make those look like:

$$x-y=17$$ $$\frac{4}{3}x+\frac{3}{2}y=0$$

It makes everyone's lives easier.

I don't want to give an answer right away (to things that look like homework problems we usually don't), but what I will suggest is to try and eliminate one of the variables. Solve for a variable in one equation, and plug that into the other one. OR, add some multiple of the first equation to the second equation to make one of the variables cancel out. Can you show us what you've tried so far? I don't want to give an answer without seeing some of your work, as that defeats the whole point of doing math problems in the first place.

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Here is an example:

\begin{align} x + y &= 5 \\ 2x + 3y &= 10 \end{align}

There are 2 ways I might try to solve this. The first equation tells me that $x = 5 - y$. The second tells me that $2x + 3y = 10$, so plugging in for $x$, I get that $2(5-y) + 3y = 10 - 2y + 3y = 10 + y = 10$. And thus $y = 0$. Then we also get (from either original equation) that $x = 5$.

The other way might be to notice that $x + y = 5$ is the same as $2x + 2y = 10$. So then I might subtract $2x + 2y = 10$ from $2x + 3y = 10$ to see that $(2x + 3y) - (2x + 2y) = y = 0 = (10 - 10)$. And again, $y=0$, $x = 5$.

Can you apply these methods to this system?

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From the second equation $\frac{4}{3}x = - \frac{3}{2} y$. Multiplying both sides with $\frac{3}{4}$, we get $x= -\frac {9}{8}y$. By putting x into first equation we get $-\frac {9}{8}y-y = 17$. Which is $-\frac {17}{8}y=17$ and $y=-8$. From first equation, $x-(-8) = 17$ then $x=9$.

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Here are two comments on technique since the solution has been given.

1) One nice thing you can do if working with problems like this by hand is to clear out the fractions. In this case, multiplying the second equation by 6 gives

\begin{align} x-y&=17 \\ 8x+9y&=0 \end{align} You can then proceed by multiplying the top equation by 9 to get \begin{align} 9x-9y&=17*9 \\ 8x+9y&=0 \end{align}

I chose to multiply by 9 since now I can add the equations together to eliminate the $y$ variable. $$17x=17*9 \implies x = 9$$ and $y=-8$ follows by substituting back into $x-y=17$. You should -verify- your results by checking $x=9$ and $y=-8$ satisfy $8x+9y=0$ as well.

               The solution must satisfy both equations.
Always check this.


2) This is a nice thing to notice about the pair of equations you have, namely, one of them is equal to $0$. This makes things much easier to solve, since we can immediately find $y = -\frac{8}{9}x$ which shows us how to eliminate the $y$ from the first equation. Then, $$x+\frac{8}{9}x = 17$$ gives $$\frac{17}{9}x = 17$$ which shows $x=9$.

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