Question 908655: One integer is 5 less than 4 times another. Their product is 21. Find the integers. Found 2 solutions by ewatrrr, algebrapro18:Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! Note: (21/n)n = 21)
Numbers 21/n and n
Question States***
(21/n) = 4n - 5
21 = 4n^2 - 5n
4n^2 - 5n - 21 = 0
(4n + 7) (n - 3) "Find the integers" n = -7/4 will Not work
n = 3 and the other number is 7
You can put this solution on YOUR website! First lets let y equal one number and x equal the other unknown number. We need to come up with equations involving x and y. The first sentence as an equation looks like y = 4x-5. The other equation would be xy=21. So now we have a system of equations that we can solve. I'm going to solve it using substitution. The system is as follows:
y= 4x-5
xy = 21
Substituting y = 5-4x into the second equation we get:
xy=21
x(4x-5)=21
Now we can solve that equation for x:
x(4x-5) Distribute
4x^2-5x=21 Subtract 21 from both sides
4x^2-5x-21 = 0 Factoring the left hand side
(4x+7)(x-3)=0 Set each factor equal to 0
4x+7 = 0 and x-3 = 0
Solving 4x+7=0:
4x+7=0 Subtract 7 from both sides
4x=-7 Divide both sides by 4
x=-7/4
Since the question specified that both numbers are integers we know that x can not be -7/4th so we eliminate this solution.
Solving x-3=0:
x-3=0 add 3 to both sides
x = 3
Now we can Plug x = 3 into the first equation(y=4x-5) and solve for y:
y=4x-5 Plug in x = 3
y=4(3)-5 Simplify
y = 12-5 Simplify
y = 7
Checking our work we see that 7 is 5 less than 4 times 3 and that 7*3 is 21.
(