anonymous
  • anonymous
a+c = -1 b+ac+d = 1 bc+ad = -1 bd = 1
Algebra
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
bahrom7893
  • bahrom7893
Ok here goes... hopefully I won't make any arithmetic errors. From the first one: \[a+c= -1; \space c = -1-a\]From the last one: \[b*d = 1; \space d = \frac{1}{b}\]2nd equation gives: \[b+a*c+d = 1\]\[b+a*(-1-a)+\frac{1}{b}= 1\]\[b-a-a^2 + \frac{1}{b}=1\]3rd equation:\[b*c+a*d=-1\]\[b*(-1-a)+\frac{a}{b}=-1\]Let's see if we can solve 2nd and 3rd equations
bahrom7893
  • bahrom7893
\[b^2-ab-a^2b+1=1\]\[b^2-ab-a^2b=0\]\[b*(b-a-a^2)=0\]
bahrom7893
  • bahrom7893
3rd equation:\[-b^2-ab^2+a=-1\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

bahrom7893
  • bahrom7893
Ok, now I'm stuck. @amistre64 @satellite73 , any ideas?
anonymous
  • anonymous
try using the last equation to get d in terms of b, and the first equation to get c in terms of a. this should let you plug into the second and third equations to have a system of 2 variables and 2 functions. solve that and then use those value to solve the first and last equation for d and b.
anonymous
  • anonymous
From a+c = -1, we get a = -1 - c ---(1) From bd = 1, we get b= 1/d ---(2) Sub (1) and (2) into b+ac+d = 1 and bc+ad = -1 \[\frac{1}{d} + (-1 - c ) c + d =1\]\[\frac{1}{d} c + (-1-c) d = (-1)\] How are you going to handle these equations?
anonymous
  • anonymous
You can pick one equation, solve for either c or d and substitute into the other. From there you will be able to solve for a variable. Then you plug into the first equation you picked to get the second variable. Then you can solve both (1) and (2) easy.
anonymous
  • anonymous
It is easier said than done.
anonymous
  • anonymous
From \(\frac{1}{d} + (-1 - c ) c + d =1\) \[\frac{1}{d} + (-1 - c ) c + d =1\]\[\frac{1}{d} + d =1+ (1 + c ) c\]
anonymous
  • anonymous
\[\frac{1}{d} + d =1+ (1 + c ) c\]\[\frac{1}{d} + d =(c+\frac{1}{2})^2+\frac{3}{4}\]\[\frac{1}{d} + d -\frac{3}{4}=(c+\frac{1}{2})^2\]\[(c+\frac{1}{2}) =\pm \sqrt{\frac{1}{d} + d -\frac{3}{4}}\]\[c =-\frac{1}{2}\pm \sqrt{\frac{1}{d} + d -\frac{3}{4}}\]Wow!!~
anonymous
  • anonymous
hummm... I cant find a way to make this a linear system and I don't know a good way to solve a non-linear system that I would want to explain. Wolfram alpha is coming up with some very ugly numbers for this one too so I don't think this is a easy one to solve. What class is this for?
anonymous
  • anonymous
Class..... amateur :D
anonymous
  • anonymous
@bahrom7893 The first line in your second comment here should be \[b^2−ab−a^2b+1=b\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.