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

##### 2 Answers

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

put this in place of

Rearrange

It can be factorized :

Therefore,

Plugging it into the second equation we get

So we get the solution as

#### 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]}