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

2 Answers
Mar 2, 2018

Answer:

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)#

Mar 2, 2018

Answer:

#(-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]}