What is another division problem that has a quotient of 3 and a remainder of 28???
1 answer:
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Translate the info given into equations:
<span>A positive integer is 15 less than another -----> x = y - 15
</span><span>If 4 times the reciprocal of the smaller integer is subtracted from the reciprocal of the larger integer, then the result is − 3/4
</span>------> (1/y) - (4/x) = -3/4
Use substitution, replace 'x' with 'y-15' in 2nd equation
We know that both x and y must be positive integers, so y =-1 is not a solution.
If y=20, then x = 20-15 = 5
Final answer:
The pair of integers is (5,20)
<span>A positive integer is 15 less than another -----> x = y - 15
</span><span>If 4 times the reciprocal of the smaller integer is subtracted from the reciprocal of the larger integer, then the result is − 3/4
</span>------> (1/y) - (4/x) = -3/4
Use substitution, replace 'x' with 'y-15' in 2nd equation
We know that both x and y must be positive integers, so y =-1 is not a solution.
If y=20, then x = 20-15 = 5
Final answer:
The pair of integers is (5,20)
Two numbers that gives twice the sum of a number and 3 times a second number is 4. The difference of ten times the second number and five times the first is 90 are -10 and 4
Given :
Twice the sum of a number and 3 times a second number is 4. The difference of ten times the second number and five times the first is 90.
Let a and b be the two unknown numbers
Lets frame equation using the given statements
Twice the sum of a number and 3 times a second number is 4.
the difference of ten times the second number and five times the first is 90
Now use these two equations to solve a and b
Add both the equations
Now find out 'a'
The two numbers are -10 and 4
Learn more : brainly.com/question/13856304