Question

How do you solve the system of equations `y-3 = (x-1)^2` and `y-x=4` algebraically?

Answer 1

The Soln. Set `={(0,4),(3,7)}.`

Explanation:

`y-x=4 rArr y=x+4.`

Subst.ing, in, `y-3=(x-1)^2,` we have,

`x+4-3=(x-1)^2," i.e., "x+1=x^2-2x+1,` or,

`x^2-3x=0.`

`:. x(x-3)=0 rArr x=0, or, x=3.`

`y=x+4, &, x=0 rArr y=4, and,`

`y=x+4, &, x=3 rArr y=7.`

These roots satisfy the given eqns.

Hence, the Soln. Set `={(0,4),(3,7)}.`

Answer 2

`x=0` or `x=3` and `y=4` or `y=7`

Explanation:

You can rearrange the second equation to get just `y`
`y=x+4`
the you can rearrange the first equation to get all the unknown numbers on one side
`0=(x-1)^2-y+3`
then expand
`0=x^2-2x+4-y`

the second equation is equal to `y` therefore we can substitute this equation into the first equation everywhere where there is a `y`

`0=x^2-2x+4-(x+4)`
then solve
`0=x^2-2x+4-x-4`
`0=x^2-3x`
therefore `x=0` or `x=3`
substitute back into original equations to find y

`y=x+4`
`y=4` or `y=7`

Answer 3

`(0,4),(3,7)`

Explanation:

Simplify the first equation to `y=(x-1)^2+3=x^2-2x+4`
We can then plug this into the second equation to get
`x^2-2x+4-x=4`

Then simplify
`x^2-3x+4=4`
`x^2-3x=0`

Then use the quadratic formula to solve
`\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(0)}}{2(1)}`
`\frac{3\pm\sqrt{9-0}}{2}`
`\frac{3\pm3}{2}`

Solving we get
`\frac{3+3}{2}=6/2=3`
`\frac{3-3}{2}=0/2=0`