Math Question

[Bucky623]

My son asked for my help with his math. I was able to help him with some of the problems but this one has me stumped.

"The difference of two numbers is 6. The quotient of the two numbers is 3 if you divide the smaller into the larger. What are the two numbers?"

 

[Mr Evilwrench]

9 and 3. 9-3=6, 9/3=3.

 

[Que]

7.5 and 1.5

I forgot the stupid minus sign!

 

[dcr465]

x-y=6 and x/y=3 so x=3y then 3y - y =6 ... 2y=6 ... y=3 thus x = 3(3) = 9

 

[Expat]

nerd thread...

 

[Kurr]

3 and 9

3+6=9
9/3=3

 

[radar8756]

x - y = 6
x/y = 3 ... x = 3y
3y - y = 6 ... 2y = 6 ... y = 3
x - 3 = 6 ... x = 9

9 - 3 = 6
9/3 = 3

 

[TB1999]

69

 

[pudly]

x-y=6 and x/y=3 so x=3y then 3y - y =6 ... 2y=6 ... y=3 thus x = 3(3) = 9

for showing your work

 

[sj kahr k40]

Common core answer: yellow and purple

 

[Bucky623]

Thank you everyone.

It’s keeps getting harder for me to help my 11 year old with his math.

 

[Rookie]

69

Nope, 77. You get 8 more.

 

[jsharmon7]

Pi....mmmmmm pie....

 

[CathyInBlue]

I'm fluent in Algebra. I shall translate.

A is a number, which is larger than B.

eqn. 1) A - B = 6

eqn. 2) A/B = 3

Therefore, 3) A = 3B, by solving eqn. (2) for A.

Therefore, 4) 3B - B = 6, by substituting eqn. (3) into eqn. (1); 5) 2B = 6, by combining like terms in eqn. (4); and finally 6) B = 3, by dividing both sides of eqn. (5) by the common factor of 2.

Therefore, 7) A = 3(3), by substituting eqn. (6) into eqn. (3); and finally 8) A = 9, by multiplying.

Therefore, A is 9 and B is 3, Q.E.D.

 

[findingZzero]

Al-jebr meaning "reunion of broken parts") is useful. I can still do it. So can you. My brain feels better after that.
I see you've calmed down. Still can't find the post that started the broo-ha-ha. I feel your pain.

 

[patience0830]

I'm fluent in Algebra. I shall translate.

A is a number, which is larger than B.

eqn. 1) A - B = 6

eqn. 2) A/B = 3

Therefore, 3) A = 3B, by solving eqn. (2) for A.

Therefore, 4) 3B - B = 6, by substituting eqn. (3) into eqn. (1); 5) 2B = 6, by combining like terms in eqn. (4); and finally 6) B = 3, by dividing both sides of eqn. (5) by the common factor of 2.

Therefore, 7) A = 3(3), by substituting eqn. (6) into eqn. (3); and finally 8) A = 9, by multiplying.

Therefore, A is 9 and B is 3, Q.E.D.

QFT

 

[pudly]

Congratulations to all the INGOers who came up with the correct answer. I now pronounce you "Smarter than a fifth grader".

 

[churchmouse]

I'm fluent in Algebra. I shall translate.

A is a number, which is larger than B.

eqn. 1) A - B = 6

eqn. 2) A/B = 3

Therefore, 3) A = 3B, by solving eqn. (2) for A.

Therefore, 4) 3B - B = 6, by substituting eqn. (3) into eqn. (1); 5) 2B = 6, by combining like terms in eqn. (4); and finally 6) B = 3, by dividing both sides of eqn. (5) by the common factor of 2.

Therefore, 7) A = 3(3), by substituting eqn. (6) into eqn. (3); and finally 8) A = 9, by multiplying.

Therefore, A is 9 and B is 3, Q.E.D.

This is math for an 11 year old........

 

[Mr Evilwrench]

Um, yeah? I was doing calculus at 16, just because I had to go through six (!) semesters of what they called algebra (yet included differential equations etc) and two of geometry. Otherwise I would have been doing it even earlier, like 13. On one of my bus routes I had a middle school kid I'd drop off from HS where he was taking calculus. I was sympatico. Yeah, some of us are better at it than others. I didn't splain mine so well, but it was so intuitive it just fell out of my brain, as enfeebled as it is.

 

[JeremiahJohnson]

As Cat said, this is Algebra, not math.

If he's getting this question in a math class, that's pretty tough. Simple algebra, but difficult math. If I didn't know algebra I'd start trying numbers and eventually get 9 and 3 just because the answer is simple with high probability of success, but those kinds of math questions get pretty complicated quickly if the answer is not an easy one (and algebra is not an option).