# Why is this system of equations unsolvable?

The system is: \begin{cases} 3a + 5b = 2\\ 15a + 15b = ? \end{cases}

Can't we just do: \begin{gather} 3a = 2 - 5b\\[2ex] a = \frac{2 - 5b}{3} \end{gather}

Then we plug in $a$ in terms of $b$ into the second equation, which gives: \begin{gather} \frac{15 (2 - 5b)}{3} + 15b = ?\\ 10 - 25b + 15b = ?\\ -10b = -10\\ b = 1 \end{gather}

Then we plug in $b$ into the first equation, to get: \begin{gather} 3a + 5 = 2\\ 3a = -3\\ a = -1 \end{gather}

We've solved for both $a$ and $b$ and we see that $15a + 15b$ is $0$, therefore $? = 0$.

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Let $a=0$, $b=\frac{2}{5}$. Then $3a+5b=2$ and $15a+15b=6$. Let $a=\frac{2}{3}$ and $b=0$. Then $3a+5b=2$ and $15a+15b=10$. Thus $15a+15b$ is not determined. You could use still other $a$ and $b$, like $a=-1$, $b=1$ to get other values of $15a+15b$. – André Nicolas Apr 10 '14 at 5:23

## 5 Answers

Going from line 9 to line 10 you assumed that $?=0$, so it should be no surprise that you ended up with $?=0$.

I'm really not sure what the question means but perhaps it means "if $3a+5b=2$, can you find the value of $15a+15b$?"

If this is the question then the answer is "no you can't" because $?$ could actually be any real number. For example, if $a=1$ and $b=-\frac{1}{5}$, then $?=12$.

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+1 for “I'm really not sure what the question means” ;) – Carsten S Apr 10 '14 at 6:20

You have three unknowns and two equations. It should be immediately obvious that this isn't solvable.

As David accurately pointed out, you accidently set ? = 0 in your work.

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Unless you've studied linear algebra, there's no reason why that should be immediately obvious. – Jack M Apr 12 '14 at 0:05
@Jack M I meant no disrespect to the OP, I just thought he may not have realized that "?" Is just any old unknown. I completely disagree with you though. There is a reason that it is immediately obvious. You need no linear algebra to understand that, for example, y = 2x is a function, that y and x do not have single numberic values you can solve for, and that simply giving a name 'z' to the sum of x and y, so that we have x + y = z too, does not change anything, it does not somehow allow us to solve for numeric values because we gave a name to something, we did not conjure more information. – Jonathan Hebert Apr 12 '14 at 0:29

Looking at it from a geometric pespective, we have (a,b) lies in the line $3x+5y=2$. According to the question we have to find the the value of $15a+15b$. Hence we have to complete the equation of the line $15x+15y=c$. This implies that we have to find the point of intersection of the two lines which is not possible.

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Up until here it's correct: $$10 - 25b + 15b = ?$$ The problem is the next step: $$-10b = -10$$ You replaced the $?$ with $0$, automatically assuming that. If you automatically assume that, of course you can show it (kinda redundant). If you continued to use $?$ you should get: $$-10b = -10+?$$

The $?$ is an unknown that's not given to you, just like $a$ and $b$, so you treat it like another variable, such as $c$. Now that you have $3$ variables in a system of $2$ linear equations you can easily see why there is no solution.

In summary, $? = c$, $?\ne 0$.

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You've find a solution imposing $?=0$ but you could decide that $?=10$ and find $a=\frac{2}{3}, b=0$, the system cannot be solved because the value of $a$ and $b$ depends on that of $?$

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