The sum of two number is 40.Two times the smaller number exceeds the larger number by 38.Find the number
let the smaller be x
then the larger is 40-x
"Two times the smaller number exceeds the larger number by 38"
---> 2x > (40-x) by 38
2x = 40-x + 38
3x = 78
x = 26
the smaller is 26
the larger is 14
contradiction!
Your wording should have been
"Two times the larger number exceedss the smaller number by 38
then
2(40-x) - x = 38
80 - 2x-x = 38
-3x = -42
x = 14
the smaller is 14, the larger is 26
check:
2(26) = 52
does 52 exceed 14 by 38 ? YES
Let's assume the smaller number to be "x" and the larger number to be "y".
According to the information given,
1. The sum of the two numbers is 40, so we can write the equation: x + y = 40.
2. Two times the smaller number exceeds the larger number by 38, which can be written as: 2x = y + 38.
To find the values of x and y, we can solve this system of equations. Here's one way to do it:
1. Rearrange the first equation to express y in terms of x: y = 40 - x.
2. Substitute this expression for y in the second equation: 2x = (40 - x) + 38.
3. Simplify and solve the equation for x: 2x = 78 - x.
Adding x to both sides: 3x = 78.
Dividing by 3: x = 26.
4. Substitute the value of x back into the first equation to find y: x + y = 40.
26 + y = 40.
Subtracting 26 from both sides: y = 14.
Therefore, the smaller number is 26 and the larger number is 14.