-2
$\begingroup$

Which math tool am I supposed to use to solve this system?**

Given these coefficients , which is the combination of $(x , y , d)$ that satisfies these equations?

$$ \left\{ \begin{aligned} 5x+7y &=d \\ 2x+y&=d\\ \end{aligned} \right. $$

  • 1
    You can solve for $x,y$ in terms of $d\,$, but you won't be able to (univocally) determine all $3$ unknowns $x,y,d$ from just two equations, since the system has infinititely many solutions. – 2017-02-01
  • 1
    Are you looking for integer solutions? – 2017-02-01
  • 2
    You can try $x=y=d=0$. – 2017-02-01
  • 0
    **The** combination? It is a homogeneous system, hence if $(x,y,d)$ works, $(kx,ky,kd)$ works just as well. – 2017-02-01
  • 0
    You know that $d=d$, with this you can determine a relation between x and y. The problem here is that you have 3 unknowns and only 2 equations, you won't be able to determine any solution without arbitrarily picking values. – 2017-02-01
  • 0
    @DrkVenom This is not true. For any given $d$, you can determine $x$ and $y$ as functions of $d$. Doing what you suggest and setting the two equations equal is the wrong way to approach this problem, see the answer of fsblajinha below for the correct way of doing it. Of course, there is no numerical solutions for $x$ and $y$ independent of $d$ which can be derived from this system of equations. – 2017-02-01
  • 0
    @TimonG, there are typically three ways taught to solve a system of linear equations prior to matrix mathematics: Addition, subtraction and substitution. Equating d to d is the version of substitution. It just sets $5x+7y=2y+y$, you can then form the relation of x to y, or y to x. It doesn't matter what you pick for d (hence arbitrarily) you will get the same relation and hence the same thing fsblajinha stated. – 2017-02-01
  • 0
    Sure, you could set $d=d$ but then you still need to substitute the relationship between $x$ and $y$ which you get back into the original equations and solve yet another equation for $x(d)$ and $y(d)$ but this is a convoluted method imo and a backwards way to tackle this kind of problem. – 2017-02-01

1 Answers 1

0

$ \left\{ \begin{aligned} 5x+7y &=d \\ 2x+y&=d\\ \end{aligned} \right. $ $\Leftrightarrow $ $ \left\{ \begin{aligned} 5x+7y &=d \\ 14x+7y&=7d\\ \end{aligned} \right. $ $\Rightarrow 9x=6d \Rightarrow x=\frac{2}{3}d \Rightarrow y=-\frac{1}{3}d$

  • 0
    You can do the second thing when $d=0$ as well. – 2017-02-01
  • 0
    @G. Sassatelli is true! – 2017-02-01