Question

How do solve the following linear system?: `4y-16=x , 2x+7=-5y`?

Answer 1

`x = -8.31`
`y = 25/13`

Explanation:

The first equation already has `x` as the subject. Therefore you substitute that in place of `x` in the second equation.

`2(4y - 16) + 7 = -5y -> 8y - 32 + 7 = -5y`

Rearrange the equation to have all unknowns on one side.

`13y = 25`

`y = 25/13`

Now substitute the `y` value into the first equation to get `x`.

`4(25/13) - 16 = x`

`x = -8.31`

Answer 2

`(x,y)=color(red)(""(-108/13,25/13))`

Explanation:

Given
[1]`color(white)("XXX")4y-16=x`
[2]`color(white)("XXX")2x+7=-5y`

[1]`rarr`
[3]`color(white)("XXX")x=4y-16`

Using [3] we can replace `x` in [2] with `4y-16`
[4]`color(white)("XXX")2(4y-16)+7=-5y`

Expanding and simplifying the left side of [4]
[5]`color(white)("XXX")8y-25=-5y`

Adding `(5y+25)` to both sides of [5]
[6]`color(white)("XXX")13y=25`

Dividing both sides of [6] by `13`
[7]`color(white)("XXX")y=25/13`

Using [7] we can replace `y` in [1] with `25/13`
[8]`color(white)("XXX")4 * 25/13-16=x`

Simplifying [8]
[9]`color(white)("XXX")x=-108/13`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The ugliest part is probably verifying this result by putting the derived values for `x` and `y` back into the original equations to check that the results are correct.

...with values like these, I would suggest you go to the effort to verify the results (despite the ugliness). Personally I used a spread sheet for this but you might want to use some other method.

Answer 3

`color(red)(y=25/13` or `color(red)(1 12/13`,`color(red)(x=-108/13` or`color(red)(-8 4/13`

Explanation:

`4y-16=x`----`color(red)((1))`

`2x+7=-5y`----`color(red)((2)`

`4y-x=16`----`color(red)((3)`

`-5y-2x=7`----`(color(red)(4))`

`color(red)((3) xx 5`

`20y-5x=80`----`color(red)((5)`

`color(red)((4) xx 4`

`-20y-8x=28`----`color(red)((6)`

`color(red)((5)+(6)`

`-13x=108`

`color(red)(x=-108/13` or`color(red)(-8 4/13`

Substitute `color(red)(x=-108/13` in (2)

`2(color(red)(-108/13))+7=-5y`

`-216/13+7=-5y`

Multiply both sides by 13

`-65y=-216+91`

`-65y=-125`

`y=cancel125^25/cancel65^13`

`color(red)(y=25/13` or `color(red)(1 12/13`

Check:

Substitute `color(red)(y=25/13` and `color(red)(x=-108/13` in (1)

`4(color(red)(25/13))-16=color(red)(-108/13`

`100/13-16=-108/13`

`(100-208=-108)/13`

Multiply both sides by 65

`100-208=-108`

`color(red)(-108=-108`