# Thread: Help with System of Equations question!

1. ## Help with System of Equations question!

edit: my bad it was 20 not 23!

It starts getting into fractions and I'm just having a really tough time I have no idea what I'm doing wrong... I can usually do the ones without fractions but with just makes it so much harder. I've been doing this question for like 2 hours (srs). One time I got everything to work out right except the 2nd equation because of negatives (answers didn't have fractions like they do now) so I had to re-do it and haven't come close since.

a + b + c = 20

2a + b -2c = -53

3a - 4b + c = 59

can anyone help me out and show me how to do it? I always do substitution but if elimination is easier im open to that to thanks

2. ## Re: Help with System of Equations question!

Hi Auto,

If you are skeptical about dealing with fractions and want one stop solutions you can go ahead with the Cramer's ruleCramer's rule - Wikipedia, the free encyclopedia.
Applying to your problem at hand you will see that
$\Delta =$
| 1 1 1 |
| 2 1 -2 |
| 3 -4 1 |
$\Delta_a =$
| 20 1 1 |
| -53 1 -2 |
| 59 -4 1 |
$\Delta_b =$
| 1 20 1 |
| 2 -53 -2 |
| 3 59 1 |

$\Delta_c =$
| 1 1 20 |
| 2 1 -53 |
| 3 -4 59 |

a = $\Delta/\Delta_a$
b = $\Delta/\Delta_b$
c = $\Delta/\Delta_c$

could not get a proper symbol for determinant so bare with the parallel lines.

3. ## Re: Help with System of Equations question!

Here I suggest a method of substitution that involves no fractions.
Adding eq 1, 2,3 we have
6a - 2b = 26 ==> 3a - b = 13
Subtracting eq 1 from eq 3 we have
2a - 5b = 39
from these two equation we have a = 2, b = -7 and c = 25

Kalyan

4. ## Re: Help with System of Equations question!

Hello, auto!

I didn't see that many fractions . . .

$\begin{array}{cccc}a + b + c &=& 20 & [1] \\2a + b -2c &=& \text{-}53 & [2] \\ 3a - 4b + c &=& 59 & [3] \end{array}$

$\text{From [1]: }\:c \:=\:20 - a - b$

$\begin{array}{cccccccccc}\text{Substitute into [2]:} & 2a + b - 2(20 - a - b) &=& \text{-}53 & \Rightarrow & 4a + 3b &=& \text{-}13 & [4] \\ \text{Substitute into [3]:} & 3a - 4b + (20 - a - b) &=& 59 & \Rightarrow & 2a - 5b &=& 39 & [5] \end{array}$

$\text{From [4]: }4a + 3b \:=\:\text{-}13 \quad\Rightarrow\quad b \:=\:\frac{\text{-}13-4a}{3}\;\;[6]$

$\text{Substitute into [5]: }\:2a - 5\left(\tfrac{\text{-}13 - 4a}{3}\right) \:=\:39$

. . $2a + \tfrac{65}{3} + \tfrac{20}{3}a \:=\:39 \quad\Rightarrow\quad \tfrac{26}{3}a \:=\:\tfrac{52}{3} \quad\Rightarrow\quad \boxed{a \:=\:2}$

$\text{Substitute into [6]: }\:b \:=\:\frac{\text{-}13 - 4(2)}{3} \quad\Rightarrow\quad \boxed{b \:=\:\text{-}7}$

$\text{Substitute into [1]: }\:2 - 7 + c \:=\:20 \qud\Rightarrow\quad \boxed{c \:=\:25}$

5. ## Re: Help with System of Equations question!

ok thanks a lot guys ill learn from your examples. and haha when I did it I got a lot of fractions even going up into like 683 over 26