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.

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

Either I forgot or never did learn to do it well. I need to solve the following system:

$$9a+3b+c=0$$ $$25a-5b+c=0$$ $$a-b+c=12$$

Google shows me this page with some instructions:, I decided to follow them.

The first one, is "add the first equation with the third one, this will eliminate an x-term". So I assume that, in my context, this will eliminate at least one term when I try it.

Adding the first with the third one: $$9a+3b+c=0$$ $$a-b+c=12$$ I get: $$10a-2b+2c$$ Aw... No term was removed. So something's not well.

Either my system is wrong or I am not following the instructions well. If you want to know where my system comes from, it is from the following question:

Determine the quadratic function such that $f(3) = 0$, $f(-5) = 0$ and $f(-1)=12$.

If I'm not mistaken, this involves solving the system I got above.

Can you tell me what did I do wrong following those instructions? I'm not really looking for the solution - instead, I'd prefer to understand how to do this.

share|cite|improve this question

I will answer your question

How to solve systems of three [linear] equations?

Don't necessarily take such instructions literally: what you refer to (adding the first equation to the third) was probably correct for the example used in the particular problem demonstrated.

Essentially, solving a system of linear equations (aka Gaussian Elimination) is just like using elementary row operations in matrices, except you have the variables. But how you proceed depends on the coefficients:

$$9a+3b+c=0\tag{1}$$ $$25a-5b+c=0\tag{2}$$ $$a-b+c=12\tag{3}$$

To get you started:

Add $-9(3)$ to $(1)$: that eliminates $a$...

$\;\;-9a + 9 b - 9c = -108$
$+\; 9a + 3b + c = 0$
$= 0\; + 12b - 8c = -108$
$= 3b - 2c = -26\quad\quad\quad\quad (R_1)$

From here you can solve for for $b$ in terms of $c$, then back-substitute, etc.

Or, you can eliminate, say, $c$ altogether to solve for $b$:

For example, you can use $-25(\text{equation }\;3) + (\text{equation}\;2)$ to get a second equation without the $a$ variable...

$-25a + 25b - 25c = 300$
$+\; 25a - 5b + \;\;c = \;\;0$
$= 0\;\;20b - 24c = 300$
$= 5b - 6c = 75\quad\quad\quad\quad (R_2)$

Now continue the process using $3(R_1)+(R_2)$ to get rid of the $c$-term...

Then back substitute to solve for $b$, and then using $b, c$ solve for $a$.

share|cite|improve this answer
RHS should be -108, I believe. Need to multiply both sides of $(3)$ by $-9$. – icurays1 Dec 7 '12 at 2:59

I will answer your question

Determine the quadratic function such that $f(3) = 0$, $f(-5) = 0$ and $f(-1) = 12$.

Any quadratic polynomial will have only two roots. From the question, you know that the only two roots are $3$ and $-5$. Hence, we have $$f(x) = a(x-3)(x+5)$$ In addition, we are given thaat $f(-1) = 12$. This implies $$a(-4)(4) = 12 \implies a = -\dfrac34$$ Hence, $$f(x) = -\dfrac34 (x-3)(x+5) = \dfrac34(3-x)(x+5)$$ To solve by your method, subtract equation $3$ from $1$ and $3$ from $2$ to eliminate $c$. If we do this, we then get that $$8a+4b = -12$$ $$24a-4b = -12$$ Add the above two equations to get $$32a = - 24 \implies a = - \dfrac34$$ We have $8a+4b = -12 \implies 4b = - 12 +6 = -6 \implies b = - \dfrac32$. Now that we have values for $a$ and $b$ plug it in any of the three original equations you had to get that $c = \dfrac{45}4$.

share|cite|improve this answer

@Marvis' answer is the best way to go about getting that quadratic, but to answer "how do I solve this system of linear equations", the answer is: you can do other operations besides adding two equations to each other. In particular, you can add a multiple of one equation to another. So, if you wanted to eliminate $a$ from the first equation, you could multiply the third equation by $-9$ then add it to the first: $$ \begin{align*} -9(a-b+c)&=-9\cdot 12\\ + (9a+3b+c)&=0\\ \Rightarrow -9a+9a+9b+3b-9c+c&=-9\cdot 12 \end{align*} $$

The first equation would then read:

$$ 12b-8c=-108 $$

From here you could put $b$ in terms of $c$, and substitute, etc. This process is called Gaussian elimination. There are lots of good YouTube videos on the subject as well.

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.