Answer:
16 and 35
Solution:
Suppose the integers are: x, y. and y is greater than x.
x + y = 51 ................(eq. 1)
y= 2 x + 3 ................(eq. 2)
substitute eq. 2 in eq 1:
x + 2 x + 3 = 51
3 x + 3 = 51
add -3 to both sides in last equation:
3 x = 51 - 3
3 x = 48
divide both sides by 3 in last line:
3 x/ 3 = 48 / 3
x = 16
substitute the value of x in eq. 2:
y = 2 x + 3
y= 2 * 16 + 3
y = 35
9 and 10 9 + 2(10) = 29
The two integers are 7 and 14. 7 x 14 = 98.
let x be the first integer. therefore (x-3)=(36-x)*2. (36-x)*2 expanded=(72-2x) (x-3)=(72-2x). 3x-3=72. 3x=75 x=25. 36-25=11. Answer: the smaller number is 11.
They are 14, 16 and 18.
Let L equal the larger of the two integers and S the smaller. Then L=2S. Also L-1/S=1/S. Starting with this second equation and solving for L results in L=2/S. Equating the two values of L produces 2/S=2S and 1/S=S. The solutions to this last equation are S=1 or S=-1. If L is to be the larger of the two integers, then S must equal 1 and L must equal 2.
no one wants to know the answer. its freaking math
The integers are 14 and 7.
9 and 10 9 + 2(10) = 29
The two integers are 7 and 14. 7 x 14 = 98.
let x be the first integer. therefore (x-3)=(36-x)*2. (36-x)*2 expanded=(72-2x) (x-3)=(72-2x). 3x-3=72. 3x=75 x=25. 36-25=11. Answer: the smaller number is 11.
They are 14, 16 and 18.
The numbers are 14, 16 and 18.
ok this is what it looks like to me...I definitely could be wrong. one integer is three more than twice another integer if one integer is three more than another number (x+3=y) Then you just multiply "the other number" by 2. (x+3=2*y) Sum of two integers is 36....so add them (x+y=36) Solve that equation for one number so...(y=36-x) Then you would plug that equation into the first equation where there is a y: (x + 3 = 2*(36-x))
Let L equal the larger of the two integers and S the smaller. Then L=2S. Also L-1/S=1/S. Starting with this second equation and solving for L results in L=2/S. Equating the two values of L produces 2/S=2S and 1/S=S. The solutions to this last equation are S=1 or S=-1. If L is to be the larger of the two integers, then S must equal 1 and L must equal 2.
a = b + 7; 2a = b2 - 1. Substituting: 2b + 14 = b2 - 1 or b2 - 2b - 15 = 0 Factors are (x - 5)(x + 3) b = 5, a = 12. Also works with b = -3 and a = 4
-10 and -11.
Nice little problem !Call the consecutive integers ' x ' and ' x+1 '.The sum is [ x + 2(x+1) ] = Ax + 2x + 2 = A3x + 2 = ASubtract 2 from each side:3x = A - 2Divide each side by 3:x = (A-2)/3 and that's the smaller of the two integers.The larger one is (x+1) = (A+1)/3