Question

How do you solve the system of equations `3x - 10y = - 20` and `6x - 8y = 20`?

Answer 1

`{x=10,y=5]`

Explanation:

`3x-10y=-20" "(1)`

`6x-8y=20" "(2)`

`"you can multiply the both side of equation (1) by 2"`

`color(red)2(3x-10y)=color(red)(2)(-20)`

`"rearrange "`

`6x-20y=-40" "(3)`

`"now subtract (3) from (2)"`

`6x-8y-(6x-20y)=20-(-40)`

`"rearrange the equation"`

`cancel(6x)-8y-cancel(6x)+20y=20+40`

`12y=60`

`"divide both side by 12"`

`(cancel(12)y)/cancel(12)=60/12" ; "y=60/12`

`color(green)(y=5)`

`"now use (1) or (2)"`

`3x-10*color(green)(5)=-20" "(1)`

`3x-50=-20`

`3x=-20+50`

`3x=30`

`"divide both side of equation by 3"`

`(cancel(3)x)/cancel(3)=30/3`

`x=30/3`

`x=10`

Answer 2

Different approach

`x=10"; "x=5` as solutions to this system of equations

Explanation:

Given:

`3x-10y=-20...........................Equation(1)`
`6x-8y=20..................................Equation(2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method: use one equation to express what y is 'worth' (dependant variable). Then substitute for y in the other equation so that you can solve for `x`.

Consider Equation(1)

`10y=3x+20`

`y=3/10x+2 ...........................Equation(1_a)`

Substitute for y in `Equation(2)` using `Equation(1_a)`

`6x-8y=20" "->" "6x-8(3/10x+2)=20`

`" "6x-24/10x-16=20`

`" "36/10x=36`

`" "1/10x=1`

`" "color(purple)(x=10)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I chose equation(1) as the number would be easier to work out.

Substitute `x=10` into equation(1) giving

`3x-10y=-20" "->" "3(10)-10y=-20`

`" "-10y=-50`

`" "color(purple)(y=+5)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~