The difference between two numbers is 21. If twice the smaller number is added to one-half of the larger number, the result is 38. What are the numbers?
smaller ---- x
larger ----- x+21
2x + (1/2)(x+21) = 38
each term times 2
4x + (x+21) = 76
should be easy to solve for x
To find the two numbers, let's first assign variables to the unknowns. Let's represent the smaller number as 'x' and the larger number as 'y'.
We are given that the difference between the two numbers is 21, which can be written as the equation:
y - x = 21
Next, we are told that twice the smaller number, which is 2x, added to one-half of the larger number, which is (1/2)y, equals 38. This can be expressed as the equation:
2x + (1/2)y = 38
Now we have a system of two equations with two variables. We can solve this system using either substitution or elimination method.
Let's use the substitution method:
1. Rearrange the first equation to solve for y:
y = x + 21
2. Substitute this expression for y in the second equation:
2x + (1/2)(x + 21) = 38
3. Simplify the equation:
2x + (1/2)x + (1/2)(21) = 38
2x + (1/2)x + 10.5 = 38
4x + x + 21 = 76 (multiplying both sides by 2 to clear the fraction)
5x = 55
x = 11
4. Substitute the value of x back into the first equation to solve for y:
y = x + 21
y = 11 + 21
y = 32
Therefore, the smaller number is 11 and the larger number is 32.