Answer:
1 solution; (1,4)
Step-by-step explanation:
Whenever two lines intersect, their solution is the point that the two share in common. Here, it's (1,4), so that is the answer.
Answer:
(a)
(b)
Step-by-step explanation:
(a) For using Cramer's rule you need to find matrix 
and the matrix 
for each variable. The matrix is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get more easily.
![\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Ctherefore%20A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%265%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D)
To get 
, replace in the matrix A the 1st column with the results of the equations:
![B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]](https://tex.z-dn.net/?f=B_1%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%265%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D)
To get 
, replace in the matrix A the 2nd column with the results of the equations:
![B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]](https://tex.z-dn.net/?f=B_2%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C1%262%5Cend%7Barray%7D%5Cright%5D)
Apply the rule to solve 
:
In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable in one of the equations and solve for 
:
(b) In this system, follow the same steps,ust remember 
is formed by replacing the 3rd column of A with the results of the equations:
![\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Ctherefore%20A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%5D)
![B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]](https://tex.z-dn.net/?f=B_1%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%5D)
![B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]](https://tex.z-dn.net/?f=B_2%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%260%261%5C%5C0%260%262%5Cend%7Barray%7D%5Cright%5D)
![B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]](https://tex.z-dn.net/?f=B_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%261%5C%5C1%262%260%5C%5C0%261%260%5Cend%7Barray%7D%5Cright%5D)
Answer:
if we take the number of quarters Barrett have to be x we know Penelope has twice that ie 2x. so in total they have 2x+x = 60, 3x = 60; so x= 60/3 = 20.
Barrett has 20 quarters.
