SOLUTION: the difference between two numbers is 10.When eight times the smaller number is added to 40, the result i ten less than three times the larger number. find the numbers

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Question 550023: the difference between two numbers is 10.When eight times the smaller number is added to 40, the result i ten less than three times the larger number. find the numbers
Found 2 solutions by nyc_function, lwsshak3:
Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
Let x and y be the numbers. You can use any letters.
Let x = larger number
Let y = smaller number
x - y = 10
8x + 40 = 3y - 10

You have a system if linear equations in two variables.

I will solve x - y = 10 for x and plug the answer into the other equation to find y.

x - y = 10
x = y + 10
8(y + 10) + 40 = 3y - 10
8y + 80 + 40 = 3y - 10
8y + 120 = 3y - 10
8y - 3y = -120 - 10

5y = -130

y = -130/5

y = -26

To solve for x, replace y with -26 in either of the two equations above and simplify.

I will use x - y = 10.

Let y = -26

x - (-26) = 10

x + 26 = 10
x = 10 - 26

x = -16

The smaller number is -26 and the bigger number is -16.

The number -16 is BIGGER than -26 because it is closer to zero on the real number line.




Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
the difference between two numbers is 10.When eight times the smaller number is added to 40, the result i ten less than three times the larger number. find the numbers
**
x=smaller no.
y=larger no.
x-y=10
y=x+10
..
8x+40=3y=3(x+10)-10=3x+30-10
5x=-50+30=-20
x=-4
y=x+10=6
..
ans:
smaller no.=-4
larger no.=6