Intersection of 2 graphs

Hello,as for this question,

given that y = - 3x + 9(1st), y = x^(2) + 4x + 1(2nd)

by subsituting (1st) into (2nd),

• 3x + 9 = x^(2) + 4x + 1

x^(2) + 4x + 1 = -3x + 9

x^(2) + 7x - 8 = 0

(x-1)(x+8) = 0

x = 1 or x = - 8

when x = 1,

y = - 3(1) + 9 = -3 + 9 = 6,

when x = -8,

y = -3(-8) + 9 = 24 + 9 = 33,

therefore those 2 points are (1,6) , (-8,33),

let's make (1,6) = (a,b) , (-8,33) = (c,d),

a + b + c + d = 1 + 6 + (-8) + 33 = 32,

thanks...

By solving simultaneous equations, we get: $x^2+4x+1=-3x+9$

Now solve for $x$ :

$x^2 +7x-8=0$ $(x+8)(x-1) = 0$ $x=-8 \text{ OR } x = 1$

Substituting $x$ into the original functions, the points of intersection of the two graphs are: $(-8,33)$ and $(1,6)$ . Therefore $a=-8; b=33; c=1; d=6$ and $a+b+c+d = \boxed{32}$

Note that $b=-3a+9$ and $d=-3c+9$ . Thus we want to find the value of $-2(a+c)+18$ .

Letting $x^2+4x+1=-3x+9$ , we see that $x^2+7x-8=0$ . Thus, $a+c=-7$ , and our answer is $-2\cdot -7+18=\boxed{32}$ .

x² + 4x + 1 = −3x + 9 $$ => x² + 7x − 8 = 0 $$ => x = 1, −8 $$ $$ => y = − 3x + 9 = 6 [when x = 1] $$ and y = − 3x + 9 = 33 [when x = −8] $$ $$ => Intersecting points are (1, 6), (−8, 33) $$ => a + b + c + d = 1 + 6 + (−8) + 33 = 32