SOLUTION: A zoo keeper has a certain number of cages and a certain number of tigers. If she puts one tiger in each cage, she has one tiger too many. If she puts two tigers in each cage she h

Algebra ->  Human-and-algebraic-language -> SOLUTION: A zoo keeper has a certain number of cages and a certain number of tigers. If she puts one tiger in each cage, she has one tiger too many. If she puts two tigers in each cage she h      Log On


   

Question 301697: A zoo keeper has a certain number of cages and a certain number of tigers. If she puts one tiger in each cage, she has one tiger too many. If she puts two tigers in each cage she has one cage too many. How many tigers and how many cages does she have?
Answer by Deina(147) About Me  (Show Source):
You can put this solution on YOUR website!
Ok,
What do we know?
We have 1 more tiger than we do cages, so
+c+=+t+-+1+

We have 1 more cage than 1/2 of the number of tigers, so
+c+=+%281%2F2%29t+%2B+1+

Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 2 variables



system%281%2Ax%2B-1%2Ay=-1%2C1%2Ax%2B-.5%2Ay=1%29



First let A=%28matrix%282%2C2%2C1%2C-1%2C1%2C-.5%29%29. This is the matrix formed by the coefficients of the given system of equations.


Take note that the right hand values of the system are -1 and 1 which are highlighted here:
system%281%2Ax%2B-1%2Ay=highlight%28-1%29%2C1%2Ax%2B-.5%2Ay=highlight%281%29%29



These values are important as they will be used to replace the columns of the matrix A.




Now let's calculate the the determinant of the matrix A to get abs%28A%29=%281%29%28-.5%29-%28-1%29%281%29=0.5. Remember that the determinant of the 2x2 matrix A=%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29 is abs%28A%29=ad-bc. If you need help with calculating the determinant of any two by two matrices, then check out this solver.



Notation note: abs%28A%29 denotes the determinant of the matrix A.



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Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bx%5D (since we're replacing the 'x' column so to speak).


A%5Bx%5D=%28matrix%282%2C2%2Chighlight%28-1%29%2C-1%2Chighlight%281%29%2C-.5%29%29



Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=%28-1%29%28-.5%29-%28-1%29%281%29=1.5. Once again, remember that the determinant of the 2x2 matrix A=%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29 is abs%28A%29=ad-bc



To find the first solution, simply divide the determinant of A%5Bx%5D by the determinant of A to get: x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%281.5%29%2F%280.5%29=3



So the first solution is x=3




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We'll follow the same basic idea to find the other solution. Let's reset by letting A=%28matrix%282%2C2%2C1%2C-1%2C1%2C-.5%29%29 again (this is the coefficient matrix).




Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5By%5D (since we're replacing the 'y' column in a way).


A%5Bx%5D=%28matrix%282%2C2%2C1%2Chighlight%28-1%29%2C1%2Chighlight%281%29%29%29



Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=%281%29%281%29-%28-1%29%281%29=2.



To find the second solution, divide the determinant of A%5By%5D by the determinant of A to get: y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%282%29%2F%280.5%29=4



So the second solution is y=4




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Final Answer:




So the solutions are x=3 and y=4 giving the ordered pair (3, 4)




Once again, Cramer's Rule is dependent on determinants. Take a look at this 2x2 Determinant Solver if you need more practice with determinants.



& bob's your uncle!