Question

How do you solve this system of equation: `x + 3y = 1 and - 3x - 3y = - 15`?

Answer 1

The solutions are `x=7` and `y=-2`

Explanation:

By substitution

`{(x+3y=1),(-3x-3y=-15):}`

`<=>`, `{(x=1-3y),(x+y=5):}`

`<=>`, `{(x=1-3y),(1-3y+y=5):}`

`<=>`, `{(x=1-3y),(1-2y=5):}`

`<=>`, `{(x=1-3y),(2y=-4):}`

`<=>`, `{(x=1-3*-2=7),(y=-2):}`

Or graphically

graph{(x+3y-1)(x+y-5)=0 [-5.704, 12.076, -4.195, 4.694]}

Answer 2

`x=7, \ \ \ y=-2`

Explanation:

Given equations

`x+3y=1\ ......(1)` &

`-3x-3y=-15`

`x+y=5\ .........(2)`

Subtracting (2) from (1), we get

`x+3y-(x+y)=1-5`

`2y=-4`

`y=-2`

setting the value of `y` in (1), we get

`x+3(-2)=1`

`x-6=1`

`x=7`

Hence the solution is `x=7, \ \ \ y=-2`