There are two equations.
U = age of uncle
A = age of aunt
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U x A = 3190
A + 3 = U
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Can you solve those two equations simultaneously? Or have you had that yet?
Your uncle is three years older than your aunt. The product of their ages is 3190. What are their ages? I know the answer is 55 and 58. But I can't come up with the correct formula or equation to get the answer. I've tried 3(A+3)=3190. It doesn't come up with the right answer and neither do any of the variations on that formula. What am I doing wrong? Please explain what equation/formula I should use.
6 answers
That gives me 55 for A and 58 for U.
Yes, I've had it already. I'm actually a 32 year old lawyer trying to help my teenaged sister. I forgot all of this after my sophomore year in college twelve years ago.
Is this how you do it--
(A+3)A=3190
Then it becomes A(squared)+3A=3190
Then, 4A(cubed)=3190
divide by four on both sides
Then A(cubed)=3190.
If so, on a scientific calculator, what function do I need to do to solve for A(cubed).
Or am I totally off base. Have I forgotten something from many years ago?
Is this how you do it--
(A+3)A=3190
Then it becomes A(squared)+3A=3190
Then, 4A(cubed)=3190
divide by four on both sides
Then A(cubed)=3190.
If so, on a scientific calculator, what function do I need to do to solve for A(cubed).
Or am I totally off base. Have I forgotten something from many years ago?
is this how you do it--
(A+3)A=3190
Then it becomes A(squared)+3A=3190 You are OK to here. Then
A^2 + 3A -3190 = 0 and solve the quadratic OR factor it. I solved using the quadratic formula because I have that formula put into my calculator. But it can be factored, too, (it helps to know the answer for big numbers like this) but the factors are like this.
(A -55)(A+58) = 0 (check that out to make sure it gives A^2 + 3A -3190). Then set each factor equal to zero like so
A - 55 = 0 and A = +55. The other factor is A + 58 = 0 and A = -58 which doesn't make sense so you discard that because there are no negative ages (don't we wish). So A is the aunt's age at 55 and we add 3 to that to make the uncle's age 55 + 3 = 58.
Let me know if this isn't ok. It just HAPPENS that the answer we threw away was the negative of the other age. that won't always be true.
Then, 4A(cubed)=3190
divide by four on both sides
Then A(cubed)=3190.
If so, on a scientific calculator, what function do I need to do to solve for A(cubed).
(A+3)A=3190
Then it becomes A(squared)+3A=3190 You are OK to here. Then
A^2 + 3A -3190 = 0 and solve the quadratic OR factor it. I solved using the quadratic formula because I have that formula put into my calculator. But it can be factored, too, (it helps to know the answer for big numbers like this) but the factors are like this.
(A -55)(A+58) = 0 (check that out to make sure it gives A^2 + 3A -3190). Then set each factor equal to zero like so
A - 55 = 0 and A = +55. The other factor is A + 58 = 0 and A = -58 which doesn't make sense so you discard that because there are no negative ages (don't we wish). So A is the aunt's age at 55 and we add 3 to that to make the uncle's age 55 + 3 = 58.
Let me know if this isn't ok. It just HAPPENS that the answer we threw away was the negative of the other age. that won't always be true.
Then, 4A(cubed)=3190
divide by four on both sides
Then A(cubed)=3190.
If so, on a scientific calculator, what function do I need to do to solve for A(cubed).
Yes, this helped. It took me all the way back to high school. I can't believe they're doing the quadratic formula in the eighth grade. Thanks for your help.
You're welcome. I can't believe it either. I didn't learn the quadratic in the 8th grade for sure.
1 answer