Question

How do you solve the system y= -2x + 2 and 6x + 2y = 3?

Answer 1

`x=-1/2`, `y=3`

Explanation:

Substitute in the second equation the expression for `y` given by the first, i.e. `y=-2x+2`, obtaining

`6x+2(-2x+2)=3`

Expanding, you have

`6x-4x+4=3 \iff 2x=-1 \iff x=-1/2`.

Once `x` is known, you can obtain `y` from the foretold relation

`y=-2x+2 \iff y=-2(-1/2)+2 = 1+2=3`

Answer 2

`color(blue)("Solution: "y=1 ,color(white)(xxx) x=-1/2)`

Explanation:

Given: `y=-2x+2 .............................(1)`
`color(white)(xxxx) 6x+2y=3..................................(2)`

I am going change equation (2) into the format of `y=`something

Then I can equate both equations to each other through `y`
removing that variable. We will then only have `x`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Consider equation (2) "color(green)("Method shown in detail")`

Subtract `color(blue)(6x)` from both sides isolating the y-term

`color(brown)((6x+2y)color(blue)(-6x) =(3)color(blue)(-6x))`

`color(green)("The brackets are there only to show what is being altered or")`
`color(green)("grouped to make things easier to see what is happening")`

`(6x-6x)+2y = 3 -6x`

But `6x-6x =0` giving

`0+2y=3-6x`

`2y=3-6x`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Divide both sides by 2 so that we have y on its own")`
Not that `divide 2 -> times 1/2`

`color(brown)((2y) color(blue)( times 1/2) = (3-6x)color(blue)(times 1/2)`

`2/2 times y = 3/2 -6/2x`

`y=-3x+3/2.........................................(2_a)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Combining equations the remove the variable "y)`

`Equation (1) = y = Equation(2_a)`

`-2x+2 = y= -3x+3/2`

`-2x+2 = -3x+3/2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Collecting like terms")color(green)(" Less detail now")`

`3x-2x=3/2-2`

`x=-1/2.........................................(3)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Substitute (3) into (1) to find y")`

`y=-2x+2 -> y=(-2)(1/2)+2`

`y=1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solution: "y=1 , x=-1/2)`