one number is 5 more than another number.3 times the greater plus twice the lesser is 30
x = y+5
3x + 2(y+5) = 30
3x + 2x = 30
5x = 30
x = 6
y+5 = 6
y = 1
Let's assume the lesser number as x and the greater number as y.
According to the given information, we can form two equations:
1) "One number is 5 more than another number":
y = x + 5
2) "3 times the greater plus twice the lesser is 30":
3y + 2x = 30
We can solve this system of equations by substitution or elimination method.
Let's solve it using the substitution method:
- Replace y in equation 2 with the value from equation 1:
3(x + 5) + 2x = 30
Expanding the equation:
3x + 15 + 2x = 30
Combine like terms:
5x + 15 = 30
Subtract 15 from both sides:
5x = 15
Divide both sides by 5:
x = 3
Now, substitute this value of x back into equation 1 to find the value of y:
y = x + 5
y = 3 + 5
y = 8
Therefore, the lesser number (x) is 3 and the greater number (y) is 8.
To solve this problem, let's assign variables to the two numbers. Let's call the greater number "x" and the lesser number "y".
According to the problem statement, we can create two equations:
1) "One number is 5 more than another number" can be translated to: x = y + 5
2) "3 times the greater plus twice the lesser is 30" can be translated to: 3x + 2y = 30
We now have a system of two equations with two variables. We can use the substitution or elimination method to solve for both x and y.
Let's use the substitution method:
From equation 1, we can express x in terms of y: x = y + 5
Substitute this expression for x in equation 2:
3(y + 5) + 2y = 30
Distribute the 3: 3y + 15 + 2y = 30
Combine like terms: 5y + 15 = 30
Subtract 15 from both sides: 5y = 15
Divide both sides by 5: y = 3
Now that we have found the value of y, we can substitute it back into equation 1 to find x:
x = y + 5
x = 3 + 5
x = 8
Therefore, the two numbers are x = 8 and y = 3.