linear equation

[Ericca]

Solve the following system of linear equations. Verrify your solution.

4x + 3y = 13

5x + y - 8 = 0

[Inspector Gadget]

4x + 3y = 13
5x + y = 8

Multiply the 2nd equation by -3...this is done in preparation for the next step, you multiply by a number to get one of the variables to cancel each other out...
4x + 3y = 13
-15x + -3y = -24

Now, you can add the two rows together...
-11x + 0y = -11
x = 1

Now, put it in either of the two original equations...I'll use the 2nd...
5(1) + y - 8 = 0
5 + y - 8 = 0
y - 3 = 0
y = 3

And, you can check both equations...
4(1) + 3(3) = 4 + 9 = 13
5(1) + 3 - 8 = 5 + 3 - 8 = 0

[Inspector Gadget]

Or, you can do it this way...

4x + 3y = 13
5x + y - 8 = 0, which is the same as: y = -5x + 8

Then, substitute y= - 5x + 8 into the first equation...

4x + 3(-5x + 8) = 13
4x - 15x + 24 = 13
-11x = -11
x = 1

Then, put it back into the original equation...
y = -5x + 8
y = -5(1) + 8
y = -5 + 8
y = 3

And the verification is the same as the one in the first post I made since we got the same solutions.

[Chrono]

I prefer do it by elimination. Man, that stuff was fun!

[honestrosewater]

Chrono,
If you liked that, you might enjoy using the factoring method to solve quadratic equations. For quadratic equations in standard form
a*x^2+b*x+c=0
where a does not equal 0, the factoring method finds the roots of the above equation by finding the m, n, p, and q : (m*x+n)*(p*x+q)=a*x^2+b*x+c=0.
Since (m*x+n)*(p*x+q)=0, (m*x+n)=0 and/or (p*x+q)=0. And so once you find m, n, p, and q, you can solve (m*x+n)=0 and (p*x+q)=0 for x, and those two x are the roots of your quadratic equation.
I used to like choosing nonzero rational a, b, and c at random and then factoring a*x^2+b*x+c=0. This is the same as finding the n, m, p, and q that satisfy
a=m*p
b=(m*q)+(n*p)
c=n*q
I found it fun and relaxing to solve these by trial and error. Maybe you'll like it too.
Happy thoughts
Rachel
P.S. You might want to let a=1 the first few times, and just try to find
n+q=b
n*q=c
You can also try factoring polynomials of higher degree. Have fun :)

[NSX]

Chrono,
If you liked that, you might enjoy finding roots of quadratic equations. Just randomly pick three nonzero real numbers, a, b, c, and try to find real roots of ax^2+bx+c=0, i.e., the m, n, p, and q : (mx+n)*(px+q)=ax^2+bx+c. I used to have fun doing this the "hard" way- it was quite relaxing, but then again I'm quite odd ;)
Happy thoughts
Rachel

what if you couldn't factor a, b, & c to be in that form?

[honestrosewater]

NSX,
I am having a bad night. I will edit my previous post and answer your question.
Rachel

[recon]

For simple simultaneous equations like these

4x + 3y = 13

5x + y - 8 = 0

I find it easier to solve by trial and error. I mean, x can't be more than 3 because 4x + 3y = 13. It only takes me 5 seconds to find that x = 1 and y = 3 by this method.

[honestrosewater]

NSX,
I would just pick different a, b, and c :) it is for fun, after all.

But if you want to know if you can always solve by factoring, I don't know :redface:
One should be able to figure it out, but I am not in the right mind to do it now. Perhaps later :) You could start by noting that
a*x^2+b*x+c=0, a[not=]0
always has solutions if you let a, b, c, and x be complex.
Happy thoughts
Rachel
EDIT- I cheated and found that several sites say that not all quadratic equations can be solved by factoring. Of course, I'm still not sure why this is.

[Muzza]

I find it easier to solve by trial and error. I mean, x can't be more than 3 because 4x + 3y = 13.

Oh? That's only true if you're looking for positive, integer solutions (and I see no mention of these restrictions). Constructing /a/ solution to that equation with x > 3 is simple, take x = 10^6 and y = -1333329 for example.