Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $? = 0$.

share|cite|improve this question
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
up vote 7 down vote accepted

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 $?=12$.

share|cite|improve this answer
+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.

share|cite|improve this answer
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.

share|cite|improve this answer

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

share|cite|improve this answer

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 $?$

share|cite|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.