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.

Is this equation solvable? It seems like you should be able to get a right number! If this is solvable can you tell me step by step on how you solved it. $$\begin{align} {2a + 5b} & = {20} \end{align}$$

My thinking process: $$\begin{align} {2a + 5b} & = {20} & {2a + 5b} & = {20} \\ {0a + 5b} & = {20} & {a + 0b} & = {20} \\ {0a + b} & = {4} & {a + 0b} & = {10} \\ {0a + b} & = {4/2} & {a + 0b} & = {10/2} \\ {0a + b} & = {2} & {a + 0b} & = {5} \\ \end{align}$$

The problem comes out to equal: $$\begin{align} {2(5) + 5(2)} & = {20} \\ {10 + 10} & = {20} \\ {20} & = {20} \end{align}$$

since the there are two different variables could it not be solved with the right answer , but only "a answer?" What do you guys think?

share|improve this question
Are $a$ and $b$ supposed to be integers or real numbers? I don't really know a lot about number theory, but I know that if they're real numbers then there are infinite solutions. –  Javier Badia Nov 5 '12 at 2:05
There are infinite solutions either way (for this one anyways). See here. –  EuYu Nov 5 '12 at 2:07
Anyway, I don't understand the thinking process. How do you get from $2a+5b =20$ to $0a+5b=20$? That's certainly not a valid step. –  Javier Badia Nov 5 '12 at 2:09
I have no vaild reason why I put 2a to 0a. My thinking process was that if i had just magicly put it to 0 that i could get "a answer." but that I look at it makes me feel dumb! –  Hobbs Nov 5 '12 at 13:57

5 Answers 5

up vote 2 down vote accepted

You have what is known as a linear diophantine equation. An equation of the form $$ax + by = c$$ is solvable in $x$ and $y$ if and only if $\gcd(a,\ b)\mid c$. In your particular case the equation is solvable.

You've generated one solution already, the pair $(x,\ y)=(5,\ 2)$. All the other solutions are then given by $$(x,\ y)=\left(5 + 5k,\ 2-2k\right)$$ for $k\in \mathbb{Z}$.

share|improve this answer

Also, this equation has solutions in the integers in the greatest common divisor of 2 and 5 is 1, which divides 20.

To get an explicit solution, use the Euclidean algorithm. (BTW these are called Diophantine equations if you want to do further reading on them)

share|improve this answer

Note that $2a$ must have as its unit digit $0, 2, 4, 6,$ or $8$ (because it's even!).

Similarly, note that $5b$ must have as its unit digit $0$ or $5$.

With a bit of thinking, you can see that for $2a + 5b$ to be $20$, we need to ensure that $2a$ has a unit digit of $0$ (hence $a$ is a multiple of $5$).

So let $a$ be a multiple of $5$. That is, let $a = 5k$, for some integer $k$.

Then $2a + 5b = 20$ becomes $10k + 5b = 20$, so that $5b = 20 - 10k$.

Dividing both sides of our last equation by $5$, we have $b = 4 - 2k$.

This gives you all the possible answers for $(a, b)$, namely, $(5k, 4-2k)$.

For example, when $k = 1$ you get $(5 \cdot 1, 4 - 2 \cdot 1) = (5, 2)$, which is the answer you came to.

share|improve this answer

Generally one can use the Extended Euclidean algorithm, but that's overkill here. First note that since $\rm\,2a+5b = 20\:$ we see $\rm\,b\,$ is even, say $\rm\:b = 2n,\:$ hence dividing by $\,2\,$ yields $\rm\:a = 10-5n.$

Remark $\ $ The solution $\rm\:(a,b) = (10-5n,2n) = (10,0) + (-5,2)\,n\:$ is the (obvious) particular solution $(10,0)\,$ summed with the general solution $\rm\,(-5,2)\,n\,$ of the associated homogeneous equation $\rm\,2a+5b = 0,\:$ i.e. the general form of a solution of a nonhomogeneous linear equation.

share|improve this answer

There are infinite solutions to this problem. Why? Fix a value of a. say that you wanted a to be 1. Then you get the equation 2+5b=20. 5b=18 so b=17/5. lets do it more generally. 2a+5b=20 so then 5b=20-2a and b=4-2a/5. If b=4-2a/5. Simply plug it into the original equation to get 2a+5(2-a/5)=20 which then gets to 20=20. so then for any a: (a,4-2a/5) will be an answer, you need both a and b to be integers. just take an a that is a multiple of 5 and you will have two integer solutions.

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.