We have:
![A=\left[\begin{array}{cc}3&-5\\4&1\end{array}\right]\qquad X=\left[\begin{array}{c}a\\b\end{array}\right]\qquad C=\left[\begin{array}{c}2\\10\end{array}\right]](https://tex.z-dn.net/?f=%20A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-5%5C%5C4%261%5Cend%7Barray%7D%5Cright%5D%5Cqquad%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5C%5Cb%5Cend%7Barray%7D%5Cright%5D%5Cqquad%20C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C10%5Cend%7Barray%7D%5Cright%5D)
so:
![A\cdot X=C\\\\\\ \left[\begin{array}{cc}3&-5\\4&1\end{array}\right]\cdot\left[\begin{array}{c}a\\b\end{array}\right]=\left[\begin{array}{c}2\\10\end{array}\right]\\\\\\ \left[\begin{array}{c}3a-5b\\4a+1b\end{array}\right]=\left[\begin{array}{c}2\\10\end{array}\right]\\\\\\ \boxed{\begin{cases}3a-5b=2\\4a+b=10\end{cases}}](https://tex.z-dn.net/?f=A%5Ccdot%20X%3DC%5C%5C%5C%5C%5C%5C%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-5%5C%5C4%261%5Cend%7Barray%7D%5Cright%5D%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5C%5Cb%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C10%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3a-5b%5C%5C4a%2B1b%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C10%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7B%5Cbegin%7Bcases%7D3a-5b%3D2%5C%5C4a%2Bb%3D10%5Cend%7Bcases%7D%7D)
And the solution of the system of equations:
I think A) 18 is the right answer.
Bill and Joe have 60 dollars between them. Bill has half as much as Joe.
so Bill = B and Joe = J
20 (B) + 40 (J) = 60
Bill has half the amount of Joe, and so 40/2 = 20
subtract 60 from 20 (the answer for Bill) and you get 40, (the answer for Joe)
hope this helps
Answer:
<h2>5</h2>
<em>Solution,</em>
<em>For </em><em>kite </em><em>adjacent,</em>
<em>Sides </em><em>are </em><em>equal</em>
<em>
</em>
<em>hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>