the sum of two integers is 10. three times one integer is 3 less than 8 times the other integer. find the integers. (hint: if one number is x, then the other number is 10-x.) show work please
To solve the problem, let's assign variables to the two integers.
Let's assume the first integer is 'x' and the second integer is '10 - x'.
According to the problem, the sum of the two integers is 10. So we can write the equation:
x + (10 - x) = 10
Now let's simplify the equation:
x + 10 - x = 10
The 'x' term cancels out, leaving us with:
10 = 10
This equation is true, indicating that any value of 'x' and '10 - x' will satisfy the condition that the sum of the two integers is 10. Therefore, the values of 'x' and '10 - x' can be any pair of integers that add up to 10.
Now let's move on to the second part of the problem, which states that three times one integer is 3 less than 8 times the other integer. We can use the values of 'x' and '10 - x' to form another equation.
Three times one integer is 3 less than 8 times the other integer can be written as:
3x = 8(10 - x) - 3
Let's simplify this equation:
3x = 80 - 8x - 3
Combining like terms:
3x + 8x = 80 - 3
11x = 77
Now, let's solve for 'x' by dividing both sides of the equation by 11:
x = 77 / 11
Simplifying the division:
x = 7
Now, substitute 'x' back into the equation to find the other integer:
10 - x = 10 - 7 = 3
Therefore, the two integers that satisfy the given conditions are 7 and 3.
translate
"three times one integer is 3 less than 8 times the other integer" into
3x + 3 = 8(10-x)
I am sure you can solve this