Help with another Math problem

[cowtowney]

x + y = 65
x = 3y + 5

solve

[OrioleFan]

3x+5+x=65
4x=65-5
4x=60
x=15

x+y=65
15+y=65
y=65-15
y=50

[filospinato]

x + y = 65

3*x + 5 = y

Two unknowns, two equations, substitute and solve.

[IYAS9YAS]

15 and 50
X + Y = 65
Y = 3X + 5
Substitute for Y = X + 3X + 5 = 65
4X + 5 = 65
4X = 60
X=15
15 + Y = 65
Y = 50
50 / 15 = 3 with remainder of 5

[Tarpon]

crap, B/5 should be B/4

[Mad Dawgg]

"Where do you get the “4”?"

"4" is right after "3" and just before "5"...

[US Navy Vet]

Here is a GREAT Algebra web site(It helped a rock like me through it when I was 43 YO): purplemath.com

[SAJ]

sum of two numbers is 65 And if the larger number is divided by the smaller number then the quotient is 3 and the remainder is 5. Find the numbers. Let the 2 numbers be 'a' and 'b'.

By specification in the problem: a + b = 65

Also by specification: b = 3 * a + 5 , where b > a

Then, by substitution in the first equation: a + 3 * a + 5 = 65

Or, a + 3 * a = 60 (substracting 5 fr/both sides)

Or, 4a = 60, and thus a = 15

Replacing the value of a into either equation, we see that b = 50.

FReegards!

[netmilsmom]

I adore all of you VERY MUCH!!!!!
Oh thank you, thank you, thank you.

(I gave up FR for Lent but I felt this was an emergency. See you all on Easter!)

[E. Pluribus Unum]

X = 50
Y = 15

[the_devils_advocate_666]

I miss math, that was fun! Put me on your ping list for future Algebra problems.

[brityank]

Thanks; knew someone would have the correct procedure. :^)

[IndianChief]

50 and 15

x = larger no
y = lesser no

data : x+y=65
3y+5=x

So 3y+5+y=65
4y=65-5=60

so y = 60/4 = 15

x = 3 x 15 + 5 =45 + 5
=50

I hope I haven’t flunked it

[houeto]

"X" always equals one... unless it doesn't... then it equals something else...

Man, that sounds like something straight out of Yogi Berra's playbook!

[r9etb]

A couple of folks have laid out the basic problem. The key to explaining it is in making clear the second part: A/B = 3r5.

How do you express that remainder as a number (a fraction, actually)?

You express the remainder as a fraction this way: 3r5 is the same as saying 3 + 5/B. You can show this using long division.

Thus:

A/B = 3r5 = 3 + 5/B.

A/B = 3 + 5/B

Multiply both sides by B:

A = 3B + 5

and the rest is set.

[ZGuy]

My math was ruined because no teacher ever said this one thing to me: "Arithmetic is about getting the right answers and mathematics is about learning the methods" or something like that. Had I understood that, I would have applied myself to the correct goal. As it was I still thought the goal was to come up with the right answers, so all through High School I gave them the right answers and they always said "Show your work" and I'd say "I did it in my head" but I couldn't write it out the way they wanted (and I thought they were stupid for not giving me credit when the answer was correct.)

One sentence would have changed everything, and no one ever told me.

[justlurking]

You have two simultaneous equations:

The sum of two numbers is 65

X + Y = 65

And if the larger number is divided by the smaller number then the quotient is 3 and the remainder is 5

(3 * X) + 5 = Y

Now, substitute second equation into first:

X + (3 * X) + 5 = 65

Simplify it:

(4 * X) + 5 = 65

Subtract 5 from both sides:

(4 * X) = 60

And divide both sides by X:

X = 15

Now substitute the 15 back into the first equation:

15 + Y = 65

Subtract it 15 from both sides:

Y = 50

[NTHockey]

>>I believe I saw this on an IRS worksheet.<<

Except that it was written: “If line 1 is greater than zero, enter number. Otherwise, enter number, skip Part II, go to Part III. Enter the smaller of line 38, Form 1040 A, or Line 16, Form 1040EZ or line 18. Take one-half, multiply by 85% (.85), subtract from line 19.......”

[justlurking]

Oops, a typo:

And divide both sides by X

should be:

And divide both sides by 4:

[JRios1968]

50 and 15