The sum of two numbers exceeds thrice the smaller by 2.if the difference between them is 19,find the numbers
8 years ago
8 years ago
agree
1 year ago
Let's assume the two numbers as x and y, where x is the smaller number.
According to the problem, the sum of the two numbers exceeds thrice the smaller by 2:
x + y = 3x + 2
We also know that the difference between them is 19:
y - x = 19
From the second equation, we can express y in terms of x:
y = x + 19
Now we can substitute the value of y in the first equation:
x + (x + 19) = 3x + 2
Simplifying the equation:
2x + 19 = 3x + 2
Bringing the variables to one side:
2x - 3x = 2 - 19
-x = -17
Multiplying both sides by -1 to make x positive:
x = 17
Now substituting this value of x back into y = x + 19:
y = 17 + 19
y = 36
Therefore, the two numbers are 17 and 36.
11 months ago
To solve this problem, we will set up a system of equations. Let's call the two numbers x and y.
From the given information, we can write two equations:
Equation 1: The sum of two numbers exceeds thrice the smaller by 2.
x + y = 3x + 2
Equation 2: The difference between them is 19.
x - y = 19
Now, we can solve this system of equations using the substitution method or elimination method.
Let's solve using the elimination method.
Multiply Equation 2 by -1 to simplify it:
-x + y = -19
Now, add Equation 1 and -Equation 2:
x + y + (-x + y) = (3x + 2) + (-19)
Simplifying:
2y = 3x - 17
Now, we have a new equation:
2y = 3x - 17
We can substitute this equation back into Equation 1:
x + y = 3x + 2
Substituting 2y for 3x - 17:
x + 2y = 3x + 2
Now, solve for x in terms of y:
x = 2y - 2
Substitute this back into Equation 2:
x - y = 19
Substituting 2y - 2 for x:
2y - 2 - y = 19
Simplifying:
y - 2 = 19
Now, solve for y:
y = 19 + 2
y = 21
Substitute this value of y back into Equation 1 to find x:
x + y = 3x + 2
Substituting 21 for y:
x + 21 = 3x + 2
Now, solve for x:
x - 3x = -19
-2x = -19
x = -19 / -2
x = 9.5
Therefore, the two numbers are 9.5 and 21.