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I have two variables x and y. They are governed by the two equations -

$$x+y = 1$$ and

$$2x + 3y = 5$$

Upon solving the two equations, I get $x = -2$ and $y = 3$.

Then, the value of $5x + 4y$ would be $2$. Another way to get this value would be to use the equation

$$k(x+y) + j(2x + 3y) = x + y$$

Solving for k and j, I get $k = 7$ and $j = -1$ and thus, the value of $5x + 4y$ would be $7(1) - 5(-1)$ which is again equal to 2. While this is simple enough, I found a problem with three variables - x, y and z. The two equations governing these variables are

$$3x + 4y + 10 z = 252$$ and

$$x - 2y +z = 0$$

and I ned to find the value of $x + y +z$.

Using the method illustrated above, that is,

$$k(3x + 4y + 10z) + j (x - 2y + z) = x + y +z$$

and I get three equations in $k$ and $j$ which are

$$3k + j = 1$$ $$4k - 2j = 1$$ $$10k + j = 1$$

Solving the first two equations, I get $k = \frac{3}{10}$ and $j = \frac{1}{10}$.

Solving the second and third equations, I get $k = \frac{1}{8}$ and $j = \frac{-1}{4}$.

Solving the first and third equations, I get $k = 0$ and $j = 1$.

The values of $x + y + z$ in each of the above solutions would be $75.6$, $31.5$ and $0$, none of which is the right answer.

Obviously, the method from the previous illustration fails here as I get three distinct values of k & j, none of which is correct. Is it because I have made a silly arithmetic error here that I am not able to detect, or is it that I have 3 equations in 2 variables, leading to multiple solutions?

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    $\begingroup$ When you chose to solve a subset of two equations : the first two, the first and third, and the last and third, then you get values of $k,j$ accordingly. However, none of them are correct, since they don't satisfy all three solutions. What this means, of course, is that $x+y+z$ cannot be written linearly as a combination of $x-2y+z$ and $3x+4y+10z$. Then $x+y+z$ can take any value as desired. $\endgroup$ Dec 5, 2017 at 4:07
  • $\begingroup$ does this mean that I cannot uniquely solve for x, y and z from the above two equations? $\endgroup$ Dec 5, 2017 at 4:12
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    $\begingroup$ No, I've solved it and got $y = \frac 7{24}(x+36), z = -5x\frac{5x}{12}$. Thus, $x+y+z$ is some function of $x$, so varying $x$ will allow us to vary $x+y+z$ freely. $\endgroup$ Dec 5, 2017 at 4:15

1 Answer 1

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Notice that

$$\begin{bmatrix} 3 & 4 & 10 \\1 & -2 & 1 \\1 & 1 & 1\end{bmatrix}$$ is non-singular.

Hence we can set $x+y+z$ to be anything and we can recover a unique solution of $x,y,z$.

The first two equations doesn't uniquely determined the value of $x+y+z$. It can be set to any value.

Your trick works if the third row lies in the span of the first two rows which is not the case here.

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