Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm not sure what this is called, but I'll write the problem below. If anyone can help me, as well as tell me where I can get a review of this topic I'd appreciate it.

$${\left\{{\begin{align}&2x = 13 - 5y\\&6x + 15y - 39 = 0\end{align}}\right.}$$

share|improve this question
You have a system of linear equations. Unfortunately, this time your equations are degenerate, so there is not a single solution. The Wikipedia link will get you started. –  Ross Millikan Mar 27 '14 at 18:44
Begin by learning Gauss elimination. That allows you to solve all systems of linear equations. It is an algorithm with just two or three steps that repeat a few times and you get the solution. –  Mlazhinka Shung Gronzalez LeWy Mar 27 '14 at 18:51

3 Answers 3

$$(\mathrm S_e):\left\{\begin{align}&2x = 13 - 5y\\&6x + 15y - 39 = 0\end{align}\right.$$ This is a system of equations, more specifically, this is a system of linear equations of two variables $x$ and $y$. Linear since they describe lines, we will discuss this in some seconds. So with those systems, our main goal is to find a pair of numbers $(x,y)$ that satisfy both equations. You can think of the first equation in the system as representing a line in a Cartesian coordinate system, and the other representing another line in that same coordinate system. Now, if those two lines intersect at a point $\mathcal P=(\alpha,\beta)$ then that means that the pair $(\alpha,\beta)$ satisfy the two equations describing those lines which means that they solve the system of equations $(\Pi)$. If however, those two lines are parallel, then this means that $(\Pi)$ have no solutions since they don't share a point in common. If those two lines are really just the same, then this means that $(\Pi)$ have an infinite number of solutions which can be determined by solving one of the equations in $(\Pi)$.

$\,\,\,$enter image description here

We can solve those systems using multiple ways, like Gaussian substitution, or by writing $x$ in terms of $y$ and replacing it in the other equation. We can even solve it using matrix! But for this particular example, we wont use any of them. Why? Since it is easy to see that: $$\left\{\begin{matrix}2x = 13 - 5y\\6x + 15y - 39 = 0\end{matrix}\right.\overset{\times3}\implies \left\{\begin{matrix}6x = 39 - 15y\\6x + 15y - 39 = 0\end{matrix}\right.\implies\left\{\begin{matrix}6x + 15y - 39 = 0\\6x + 15y - 39 = 0\end{matrix}\right.$$ So we have a system of two equations, which are really the same, so we can reduce that system to only one equation which of course is: $$6x+15y-39=0.$$ To solve this, we either have to write $x$ with respect to $y$ or in the opposite way. Anyway, I'll choose the second. $$6x + 15y - 39 = 0\iff 15y=39-6x\iff y=\frac{3(13-2x)}{15}=\frac{13-2x}5$$ So the set of solutions to $(\mathrm S_e)$ is: $$\text{Set of Solutions to $(\mathrm S_e)$}= \left\{\left(x,\frac{13-2x}5\right)\,\left|\,\right.x\in\mathbb R\right\}$$

To learn more about system of linear equations, systems of inequalities, with two variables, with 3 variables... check Khan|Academy's section on the topic. Have fun!

I hope this helps.
Best wishes, $\mathcal H$akim.

share|improve this answer

This is a system of equations. You could probably spend a good ten years reading Linear Algebra texts, which would review this and much, much more... :)

But since you presumably don't have ten years, here's one solution:

Multiplying the first equation by three and rearranging, $$6x+15y=39$$ So the two equations are actually the same. The solution set is all the points on the line.

share|improve this answer
@ABC Um and then there's vectors, and matrices, and bases, and dimensions, and spans, and decompositions, and spectral theorems, and multilinear and abstract and on and on... Two weeks? That would be one extraordinary students! –  user138335 Mar 27 '14 at 18:45
@ABC For what course would you cover all that in two weeks?! If this is a freshman-level Linear Algebra, what do you do for the remainder of the semester? –  user138335 Mar 27 '14 at 18:50
It is not a linear algebra course. It was a 'Mathematics' course. That is why in two weeks you teach linear algebra. And it wasn't even for university, it was high-school. –  Mlazhinka Shung Gronzalez LeWy Mar 27 '14 at 18:55
Thanks! Pretty helpful. No, not two weeks... NONE of my college mates remember ONE simple course. My professor says she's been studying for 20 years and she doesn't know it all. She knows a lot well, but not one thing completely. Maybe for students who have a 600 average. Most students will take years and years to know even the basics really well. –  Mike Wentworth Mar 27 '14 at 18:57
@ABC WHAT??!! In which country do "regular students" in high school learn about QR decompositions?? (Do say - I'm moving there) –  user138335 Mar 27 '14 at 18:58

This is a system of linear equations.

But actually, the first equation has the same solutions as the second. We can see this by multiplying both sides by 3 and rearranging the terms. So it is really all about one of the equations. All solutions are on the line:

$$y = \frac{-2}{5} + \frac{13}{5}$$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.