2x = 12 – 2y and 6y = 36 – 6x
The teacher offered $20 to the first student to solve these equations for x and y. Just how generous was the teachers offer? Show working.
2007-10-29 19:48:38 · 11 answers · asked by Josh M 1 in Science & Mathematics ➔ Mathematics
Step by step procedure:
2x = 12 - 2y
6y = 36 + 6x
6y = 36 + 6x
-36 -36
6y - 36 = 6x
/6 /6
y - 6 = x
2x = 12 - 2y
2(y-6) = 12 - 2y
2y - 12 = 12 - 2y
+2y - 12 +2y - 12
4y = 0
y = 0
6y = 36 - 6x
6(0) = 36-6x
0 = 36-6x
+6x +6x
6x = 36
/6 /6
x = 6
x = 6 and y = 0
2007-10-29 19:52:14 · answer #1 · answered by Jean-Francois 5 · 0⤊ 3⤋
assuming he/she meant you to find a set of solutions that satisfy both ....
the answer is... y = any real number...
why...???
6x = 3*2x, so 6y = 36 - 6x
6y = 36 - 3( 12 -2y)
6y = 36 - 36 + 6y, or 6y = 6y
and then.. y = y
thus any value of y will work... the two equation are really the same [ that is what the result y = y indicates ]... similarly for solving for the x value, you will get... x = x ... [ or by subtracting, 0 = 0 ]
This is because the two eqns are the SAME LINE,and not two parallel lines, since one equation is only a multiple of the other one, [ with some extra re-writing ] , and with such a system there are an INFINITE number
of solutions.... for ex. at x = 6, y = 0, and x = 2, y = 4 , and so on ... so you could never list them all ... thus it would take an infinite amount of time to write out all the solutions.... you'll never get your $20 !!!! .... :-(
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don't forget to choose someone for Best Answer, whoever you decide, since we do like the points...
2007-10-30 02:54:32 · answer #2 · answered by Mathguy 5 · 0⤊ 0⤋
both r same equations ...
6y = 36 - 6x
=> 6x = 36 -6y
=> 3(2x = 12 - 2y)
btw ... 2x = 12 - 2y => x+y = 6
about the teacher being generous, one can argue ...
if he wants a unique solution, there is none ... => not generous
if he accepts the infinite number of solution that exist ... => generous (?)
2007-10-30 02:55:52 · answer #3 · answered by Yash 2 · 0⤊ 0⤋
2x = 12 - 2y
add 2y to both sides
2x + 2y = 12
2(x + y) = 12
divide 2 to both sides
x + y = 6
x = 6 - y
now use substitution
6y = 36 - 6x
6y = 36 - 6(6-y)
6y = 36 - 36 + 6y
subtract +6y to both sides
0 = 0
infinite
2007-10-30 03:02:41 · answer #4 · answered by ru4real130 2 · 0⤊ 0⤋
the two equations are of parallel lines,
they would never intersect..
2x=12-2y
or 2x+2y=12 ------------1
and the other equation is 6x+6y=36------------2
if you multiply equation 1 with 3 you get equation 2..
hence there is no solution to this question..
The teacher was not generous but smart enough.
2007-10-30 02:52:51 · answer #5 · answered by crashbird 2 · 0⤊ 0⤋
6y=36-6x so y=6-x
so if y=6-x
2x=12-2(6-x) where y=6-x
2x =12-12-2x
2x =-2x
x=-x
Therefore this equation cannot be solved as x cannot equal an opposite number and the same for Y.
2007-10-30 02:56:05 · answer #6 · answered by Ruby D 1 · 0⤊ 0⤋
Simplifying eqn 1 to have y on LHS: y = 6 - x
Simplifying eqn 2 to have y on LHS: y = 6 - x
They are the same equation. You cannot exactly solve one equation in two unknowns. The only solution is the equation itself, which is the line 'y = 6 - x'.
That is, a straight line with slope -1 passing though the y-axis at y=6.
2007-10-30 02:54:37 · answer #7 · answered by Anonymous · 0⤊ 0⤋
can i have 5 dollars if you happen to win?loL...
here is the solution to your simple algebra problem..
1st:
(12-2y)/2=> 6-y=x
2nd:
(36-6y)/6=> 6-y=x
substitution:
to get the value of x, substitute the resulting values of x,that is:
x=6-y-(6+y)
x=o
for y,use any of the two original equations:
6x=36-6y
6(0)=36-6y
6y=36
y=6
2007-10-30 03:27:19 · answer #8 · answered by rHai 2 · 0⤊ 0⤋
Just from inspection, both equations reduce to x + y = 6, which makes it an indefinite set of equations with no unique solutions. This is a non-problem, and your teacher's $20 is safe with him. Yes, I bet the teacher is a man, and he is probably snickering. Waste of time.
2007-10-30 03:05:53 · answer #9 · answered by Scythian1950 7 · 0⤊ 0⤋
simple...
2x = 12-2y
divide both sides by 2
then
x =6-y
For y,
6y=36-6x
divide both sides by 6
y=6-x
2007-10-30 02:52:55 · answer #10 · answered by Anonymous · 0⤊ 0⤋