Question

How do solve the following linear system?: `8x + 3y= -9 , x - 3y = 1`?

Answer 1

See the entire solution process below:

Explanation:

Step 1) Solve the second equation for `x`:

`x - 3y = 1`

`x - 3y + color(red)(3y) = 1 + color(red)(3y)`

`x - 0 = 1 + 3y`

`x = 1 + 3y`

Step 2) Substitute `1 + 3y` for `x` in the first equation and solve for `y`:

`8x + 3y = -9` becomes:

`8(1 + 3y) + 3y = -9`

`(8 * 1) + (8 * 3y) + 3y = -9`

`8 + 24y + 3y = -9`

`8 + 27y = -9`

`-color(red)(8) + 8 + 27y = -color(red)(8) - 9`

`0 + 27y = -17`

`27y = -17`

`(27y)/color(red)(27) = -17/color(red)(27)`

`(color(red)(cancel(color(black)(27)))y)/cancel(color(red)(27)) = -17/27`

`y = -17/27`

Step 3) Substitute `-17/27` for `y` in the solution to the second equation at the end of Step 1 and calculate `x`:

`x = 1 + 3y` becomes:

`x = 1 + (3 xx -17/27)`

`x = 1 - 17/9`

`x = (9/9 xx 1) - 17/9`

`x = 9/9 - 17/9`

`x = -8/9`

The solution is: `x = -8/9` and `y = -17/27` or `(-8/9, -17/27)`

Answer 2

`x = -8/9; y = -17/27`

Explanation:

`8x + 3y = -9" "` Equation 1

`x - 3y = 1" "` Equation 2

Eliminate `y` by adding the two equations. The result becomes

`9x = -8`

Hence,

`x = -8/9`

Find y by substituting the already known `x` to any of the original equation (or both just to double-check)

Using Equation 1,

`8(-8/9) + 3y = -9`

`-64 + 27y = -81`

`27y = -17`

Hence,

`y = -17/27`

To double-check, use also Equation 2 to get `y`

`-8/9 - 3y = 1`

`-8 -27y = 9`

`-17 = 27y`

Hence,

`y = -17/27`