5 consecutive even integers: 2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers: 2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 504n + 2 = 6n -3234 = 2nn = 17 The smallest integer in our sequence is 2n = 2*17 = 34 Check:2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 5034 + 36 = 38 + 40 + 42 - 5070 = 120 - 5070 = 70
X+y+5=0 y=-x-5 If a solution exists y=y so we can say x^2-9x+10=-x-5 add x+5 to both sides x^2-8x+15=0 now factor x^2-3x-5x+15=0 x(x-3)-5(x-3) (x-5)(x-3) so x=3 and 5, using y=-x-5 y(3)=-8 and y(5)=-10 So the two solutions are: (3,-8) and (5,-10)
