# simultaneous equations

• May 20th 2008, 10:21 AM
djstar
simultaneous equations
hi, i am doing a electronic engineering course and have come up with the simultaneous equations thati need to answer. any help on it? many thanks. liam

10x - 10 + 20y - 20 = 0

+20-20y+40(y-x)= 0
• May 20th 2008, 10:31 AM
Moo
Hello,

Quote:

Originally Posted by djstar
hi, i am doing a electronic engineering course and have come up with the simultaneous equations thati need to answer. any help on it? many thanks. liam

10x - 10 + 20y - 20 = 0

+20-20y+40(y-x)= 0

In the first equation, divide by 10, then group :

$10x-10+20y-20=0 \Longleftrightarrow x-1+2y-2=0 \Longleftrightarrow x-3+2y=0$

In the second equation, divide by 20, then develop :

$20-20y+40(y-x)=0 \Longleftrightarrow 1-y+2(y-x)=0 \Longleftrightarrow 1+y-2x=0 \implies y=2x-1$

Therefore, by substituting in the first equation :

$x-3+2(2x-1)=0 \implies x-3+4x-2=0 \implies 5x-5=0 \implies \boxed{x=1}$

$y=2x-1=2(1)-1=\boxed{1}$
• May 20th 2008, 10:32 AM
mathceleb
Quote:

Originally Posted by djstar
hi, i am doing a electronic engineering course and have come up with the simultaneous equations thati need to answer. any help on it? many thanks. liam

10x - 10 + 20y - 20 = 0

+20-20y+40(y-x)= 0

Simplifying your 2 equations, you get:

$10x + 20y = 30$

$-40x + 20y = -20$

From here, I know 3 ways to solve. For the elimination method, go here:

2 Unknowns by Elimination Calculator

Enter those 2 equations and press the solve button.

I get (x,y) = (1,1) Plugging those back into the original equations, those work.

You could also use the substitution method here:

2 Unknowns by Substitution Calculator

or Cramer's method here:

2 Unknown Cramer's Method Calculator

Let me know if you have questions.