Question

How do you solve the system `y= 1/2x + 1` and `y= -2x + 6`?

Answer 1

`x=3`,`y=0`
`Sol^n= (3,0)`

Explanation:

`y= 1/2x+1`

`2y=x+2`----------(i)

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

Substitute eq ii in eq i

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

`-4x+12=-2x+6`

`2x=6`

`ul(x=3)`

Put `x=3` in eq ii

`y=-6+6`

`ul(y=0)`

`Sol^n= (3,0)`

Answer 2

`2,2)`

Explanation:

`color(red)(y)=1/2x+1to(1)`

`color(red)(y)=-2x+6to(2)`

Since both equations are expressed with `color(red)(y)` as the subject, we can equate the right sides.

`rArr1/2x+1=-2x+6`

To eliminate the fraction, multiply ALL terms on both sides by 2, the denominator of the fraction.

`(cancel(2)^1xx x/cancel(2)^1)+(2xx1)=(2xx-2x)+(2xx6)`

`rArrx+2=-4x+12`

add 4x to both sides.

`x+4x+2=cancel(-4x)cancel(+4x)+12`

`rArr5x+2=12`

subtract 2 from both sides.

`5xcancel(+2)cancel(-2)=12-2`

`rArr5x=10`

divide both sides by 5

`(cancel(5) x)/cancel(5)=10/5`

`rArrx=2`

To find the corresponding value of y, substitute x = 2 into either (1) or (2). I've chosen equation (2)

`x=2toy=(-2xx2)+6=-4+6=2`

`"solution is "(2,2)`