Question

How do you solve the following linear system: `3y+8x=7 , 8x=y+1`?

Answer 1

`x=(5/16)` and `y=(3/2)`

Explanation:

Firstly, re-arrange the equations so that the `x` and `y` and their respective coefficients are on the lhs of the equal sign, and the constants on the rhs. Number the equations accordingly:

`8x+3y=7` `(1)`
`8x-y=1` `(2)`

subtract `(2)` from `(1)`:

`8x-8x=0` immediately allows us to solve for `y`:

`3y-(-y)=4y` and `7-1=6`

`4y=6`
`y=(6/4)=(3/2)`
Next, substitute `y=(3/2)` back into equation `(2)`, giving:
`8x-(3/2)=1`
`8x=1+(3/2)=(5/2)`
`x=(5/2)/8`
`x=(5/16)`
The solution is: `x=(5/16)` and `y=(3/2)`

CHECK SOLUTION (IMPORTANT): Substitute these values back into equations `(1)` and `(2)` to make sure you arrive at `7` for equation `(1)` and `1` for equation `(2)`.