1. System of Linear Equation

To solve a system of 3 linear equations, a student splits it into 2 subsystems as follows:

SYSTEM 1
(1/2)x-5y=-20
x-(4/5)y=6

SYSTEM 2
(3/5)x-2y=-4
(1/2)x-5y=-20

Solve both sub systems. and state what your solution (if one exists) tells you about the original system of 3 linear equations.

I've tried solving the 1st one and the 2nd by using the elimination method..

S1=
(1/1/2)x-5.8=-14
=(1/1/2)x=-14+5.8
=(1/1/2)x=-8.2
im stuck from this point..how do i get rid of the 1/1/2 so it ll just be x.

S2=
(1/2)x-5y=-20
(3/5)x-2y=-4
===========
1/1/10=7y=-24
-7y=-25.1
y=-25.1/7
y=-3.5857

is this correct for system 2?

what will the solution tell me about the original system of 3 linear equations.?

2. Re: System of Linear Equation

I don't understand what you've done with S1... You should just multiply the equation by 2 and then subtract the second equation from the first...

\displaystyle \begin{align*} x - 10y &= -40 \\ x - \frac{4}{5}y &= 6 \\ \\ -\frac{46}{5}y &= -46 \\ y &= 5 \\ \\ x - 10(5) &= -40 \\ x - 50 &= -40 \\ x &= 10 \end{align*}

3. Re: System of Linear Equation

Originally Posted by Prove It
I don't understand what you've done with S1... You should just multiply the equation by 2 and then subtract the second equation from the first...

\displaystyle \begin{align*} x - 10y &= -40 \\ x - \frac{4}{5}y &= 6 \\ \\ -\frac{46}{5}y &= -46 \\ y &= 5 \\ \\ x - 10(5) &= -40 \\ x - 50 &= -40 \\ x &= 10 \end{align*}
I was using the elimination method where i would add the two equations and solve until i get x and then substitute the x value..guess i messed up that part