prove that where a,b,c,d,e are real numbers with (a) not equal to zero. if this linear...

prove that where a,b,c,d,e are real numbers with (a) not equal to zero. if this linear equation ax+by=c has the same solution set as this one : ax+dy=e, then they are the same equation.

Expert Solution

solutions

Let the first solution set to be x=(c-by)/a, let y=0 gives the solution set (c/a,0)

Substitute that solution set to the second equations and you will get a(c/a) + d*0 = e so c = e. Let y =1 gives the solution ((c-b)/a,1) and substitute the solution to the second equation and get d = b.

You have to make the assumption that the 2 equations does intersect at y= 1 and y = 2. I guess this is a reasonable assumption since we know that they are linear equations and that a does not equal zero, so the both equations span the entire number line. I feel that there should be a more general approach to this problem. The most important thing for me is that to identify the my flaw in understanding the problem. If ax + by = c and ax + dy = e both share the same solution set, then it means that the solution set is located at where the equations intersect (whether they intersect once or more times). The solution set they both share is ((c-e)/(b-d)), (b/a)((c-e)/(b-d)).This is the general expression for where the 2 equations intersect. The question tells us to prove that c = e, d = b.
hence proved

milcah answered 1 year ago

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