Question

Solve the system of equations. x+y=4 y=x^2 - 8x + 16 a) {(-3,7).(-4, 8)} b) [(4,0)} c) {(3,1),(4,0) d) {(3,7). (4.0)} e) none

Answer 1

Answer:  The required solution of the given system is

(x, y) = (3, 1)  and  (4, 0).

Step-by-step explanation:  We are given to solve the following system of equations :

$x+y=4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\y=x^2-8x+16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

From equation (i), we have

$x+y=4\\\\\Rightarrow y=4-x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of y from equation (iii) in equation (ii), we get

$y=x^2-8x+16\\\\\Rightarrow 4-x=x^2-8x+16\\\\\Rightarrow x^2-8x+16-4+x=0\\\\\Rightarrow x^2-7x+12=0\\\\\Rightarrow x^2-4x-3x+12=0\\\\\Rightarrow x(x-4)-3(x-4)=0\\\\\Rightarrow (x-3)(x-4)=0\\\\\Rightarrow x-3=0,~~~~~~~x-4=0\\\\\Rightarrow x=3,~4.$

When, x = 3, then from (iii), we get

$y=4-3=1.$

And, when x = 4, then from (iii), we get

$y=4-4=0.$

Thus, the required solution of the given system is

(x, y) = (3, 1)  and  (4, 0).