Question

How do you solve the following linear system: # x=1-y , x-y=5 #?

Answer 1

`color(blue)(x=3)`
A large amount of detail given for the method on purpose!
Using my methods you should be able to find the value of y`-> color(brown)("Practice makes it much faster")`

Explanation:

`color(brown)("You will get used to this and be able to do simple ones in your head")`

Given:

`x=1-y...........................................(1)`
`x-y=5..........................................(2)`

In each case make y the dependant variable (answer) then equate to each other through y to find x

`color(blue)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")`
`color(red)("For (1)")`

Add y to both sides giving:

`x+y=1-y+ycolor(white)(xxx) "Note that" -y +y=0 " so disappears!"`
`x+y=1`

Subtract `x` from both sides
`x-x+y=1-x`
`y=1-x..................................(1_a)`

`color(blue)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")`
`color(red)("For (2)")`

Add y to both sides giving:
`x-y+y=5+y`
`x=5+y`

Subtract 5 from bot sides giving:
`x-5 = 5-5+y`
`x-5=y`

Reverse so that y is on the left
`y=x-5.......................................(2_a)`

`color(blue)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")`
`"Equation "(1_a) = color(blue)(y) = "Equation "(2_a)`

`color(brown)(1-x)=color(blue)(y)=color(brown)(x-5)`
`color(brown)(1-x)=color(brown)(x-5)`

Add `x` to both sides

`1-x+x=x+x-5`
`1=2x-5`

Add 5 to both sides

`1+5=2x-5+5`

`2x=6`

`color(brown)(underline("DIVIDE"))` BOTH SIDES BY 2

`2/2x=6/2`

`color(blue)(x=3)`

`color(blue)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")`

Using the methods I have shown you first substitute `x=3` in equations #(1_a) "or " (2_a) then manipulate to find y