Y1.............4x + 8y=16
Y2.............- x - 2y = - 4
When i multiply Ysub2 by 4 or i multiply Ysub1 by 1/4 both X and Y cancel out which doesnt help me...Thank you
2007-05-08 04:33:23 · 5 answers · asked by Kevin M 1 in Science & Mathematics ➔ Mathematics
If you multiply Y2 by 4 and add it to the first, you get 0 = 0. This means there are an infinite amount of solutions. If you were to graph these two equations as lines, you'd find that they're the same exact line. The best you can do in this case is solve one of the equations for y, say Y2: y = -(x - 4)/2. So the solutions are all in the form of
( x, -(x-4)/2) ).
If you had gotten something like 0 = 3, which is never true, then you'd have NO solutions.
2007-05-08 04:38:12 · answer #1 · answered by Anonymous · 0⤊ 0⤋
There is an error in the question. If 2nd equation is multiplied by 4, will give the first equation.
ie Y1 and Y2 are one and the same equation.
2007-05-08 11:45:35 · answer #2 · answered by Como 7 · 0⤊ 0⤋
4 timesY2 Plus Y1 Eliminates Y2 therefore solution is all real # that satisfy
2007-05-08 11:42:15 · answer #3 · answered by dwinbaycity 5 · 0⤊ 0⤋
y2 times -4 gives you 4x+8y=16 which is the same as y1
so the answer is always true for x & y real numbers.
2007-05-08 11:45:30 · answer #4 · answered by fii 3 · 0⤊ 0⤋
It shows that the equations are equal so Y1 = Y2
2007-05-08 11:39:03 · answer #5 · answered by ignoramus 7 · 0⤊ 0⤋