Question

How do you solve the following system?: `4x + 5y = 2 , y + 5 = 3x`

Answer 1

`x = 27/19` and `y = -14/19`

Explanation:

Let's isolate `y` in the second equation

`y = 3x - 5`

Now we can solve for `x` in the other equation

`4x + 5 (3x - 5) = 2`

`4x + 15x -25 = 2`

`19x = 27`

`x = 27/19`

Now we solve for `y`

`y = 3x - 5`

`y = 3 (27/19) - 5`

`y = 81/19 - 5/1 xx 19/19`

`y = 81/19 - 95/19`

`y = -14/19`

To check our work, let's substitute `27/19` for `x` and `-14/19` for `y` in the first equation and solve.

`4(27/19) + 5(-14/19) = 2`

`108/19 - 70/19 = 38/19`

`38/19 = 2 = 2`!

We were right!

Answer 2

`color(blue)(x = 27/19), color(purple)(y = -14 / 19)`

Explanation:

`4x + 5y = 2` Eqn (1)

`y + 5 = 3x`

`y = 3x - 5` Eqn (2)

Substituting value of y term in Eqn (1) in terms of x,

`4x + 5(3x - 5) = 2`

`4x + 15x - 25 = 2`

Rearranging variables on L H S, constants on R H S,

`4x + 15x = 27`

`19x = 27` or `x = 27/19`

Substituting value of x in Eqn (2),

`y = 3x - 5 = (3 * (27 / 19)) - 5 = (81 - 95) / 19 = -14/19`