# Thread: [Alt. Algebra] Please help with Systems of Linear Equations strategy.

1. ## [Alt. Algebra] Please help with Systems of Linear Equations strategy.

Hello all,

I just spent (embarrassingly) 20 minutes working out this problem and I think the answer is $\displaystyle (-5,-\frac{10}{3})$
$\displaystyle x=\frac{3}{2}y$
$\displaystyle x=3y+5$

I've been practicing them but they all take me so long because I feel like I have to hunt and peck to find something that works. I sort stumbled around and finally came to the answer by solving $\displaystyle x=3y+5$ for y and converting the fraction to $\displaystyle x=3y$. There must be a step-by-step (like PEMDAS or BODMAS) strategy to solve these systems quickly and I have to master by next Monday.

I appreciate any help on this, thank you.

2. ## Re: [Alt. Algebra] Please help with Systems of Linear Equations strategy.

Originally Posted by alexcordero
$\displaystyle x=\frac{3}{2}y$ (1)
$\displaystyle x=3y+5$ (2)
Now do Equation (2) - Equation (1). I do this because the coefficient of x for both is 1, and hence cancel
$\displaystyle 0=\frac{3}{2}y+5$
$\displaystyle y = \frac{-10}{3}$

With this value for y, I can plug back into equation (1) to get
$\displaystyle x = -5$

You just want to subtract equations to get an equation with one variable only.
There are other techniques, such as converting the system of linear equations to a matrix and finding the reduced row echelon form. (RREF)
Sometimes you can multiply both equations by numbers so that the leading coefficient of one variable in both equations are equal, to cancel.

3. ## Re: [Alt. Algebra] Please help with Systems of Linear Equations strategy.

Originally Posted by alexcordero
$\displaystyle x=\frac{3}{2}y$ , $\displaystyle x=3y+5$
\begin{align*}\dfrac{3}{2}y&=3y+5 \\3y&=6y+10\\-3y&=10\\y&=-\dfrac{10}{3}\\x&=-5 \end{align*}

4. ## Re: [Alt. Algebra] Please help with Systems of Linear Equations strategy.

I tried to this and it didn't work. Looking at your solution I can see why and now I'm really embarrassed. When I cleared the fraction, I didn't carry the two out to the other numbers like you did. But the reason I didn't is because I cleared the fraction by itself "$\displaystyle \frac{3}{2}y = 3y$" and then tried to plug it in to the x variable in the other equation. Kinda' hard to explain what I did. Yours makes sense and it's beautifully simple. Thank you.

5. ## Re: [Alt. Algebra] Please help with Systems of Linear Equations strategy.

aNUTter look:

Originally Posted by alexcordero
$\displaystyle x=\frac{3}{2}y$ [1]
$\displaystyle x=3y+5$ [2]
[1]: 3y = 2x

[2]: x = 2x + 5 : x = -5