Question

How do you solve the simultaneous equations `y = x^2- 2` and `y = 3x + 8`?

Answer 1

x = 5 , y= 23 and x = -2 , y = 2.

Explanation:

Here, we have a quadratic and a linear equation which can be solved by the substitution method.

From the second equation, know `y = 3x+8`
put this in place of `y` in the first equation. We get,
`3x + 8 = x^2-2`
Rearrange
`x^2 -3x -10 = 0`
It can be factorized :
`x^2 - 5x +2x - 10 =0`
`x(x-5)+ 2(x-5) = 0`
`(x-5)(x+2) = 0`

Therefore, `x =5` and `x = -2` are the roots of equation.
Plugging it into the second equation we get `y` as:
`y = 3(5) +8 = 23`
`y = 3(-2) + 8 = 2`

So we get the solution as `(5,23) (-2,2)`

Answer 2

`(-2,2),(5,23)`

Explanation:

`"Since both equations give y in terms of x we can equate"`
`"them"`

`rArrx^2-2=3x+8`

`"rearrange into standard form"`

y`rArrx^2-3x-10=0larrcolor(blue)"in standard form"`

`"the factors of - 10 which sum to - 3 are - 5 and + 2"`

`rArr(x-5)(x+2)=0`

`"equate each factor to zero and solve for x"`

`x-5=0rArrx=5`

`x+2=0rArrx=-2`

`"substitute these values into "y=3x+x`

`x=5rArry=(3xx5)+8=23`

`x=-2rArry=(3xx-2)+8=2`

`"the solutions are "(5,23)" and "(-2,2)`
graph{(y-3x-8)(y-x^2+2)((x+2)^2+(y-2)^2-0.04)((x-5)^2+(y-23)^2-0.04)=0 [-10, 10, -5, 5]}