8 + y
8 + 6y
5
y = 3x+8 x + y = 72 x = 16 y = 56
Let this numbers be x and y, then x + y = 52 x - y = 8 x = 8 + y x + y = 52 substitute 8 + y for x into the first equation; 8 + y + y = 52 8 + 2y = 52 subtract 8 to both sides; 2y = 44 divide by 2 to both sides; y = 22 x = 8 + y substitute 22 for y into the equation; x = 8 + 22 x = 30 Thus this numbers are 22 and 30. 22 & 30
8 + y
8 + 6y
5
Y = 4 (x+8)
The word sum is the clue. It means the problem will be an addition problem.y + 3(8) = ?y + 24 = ?
8y+6y is 14y
y = x +8 where x is the number
x+y=18x^2+y^2=164 x=10 and y=8 or vice versa
It is an algebraic equation.
This question equates to, find x and y such that x + y = 8 x - y = 8 Solving the second equation for x: x = 8 + y Substituting in to the first equation: 8 + y + y = 8 2y = 0 y = 0 Solving for x: x - 0 = 8 x = 8. The two numbers are 8 and 0.
Twice the sum of 'x' and 'y' . . . 2(x+y) The sum of twice 'x' and 'y' . . . (2x+y)
x+y=12 x-y =4 x=12-4 =8 8+y=12 y=12-8 =4 so x=8 y=4 8+4=12 8-4=4