SOLUTION: In cramers rule and elimination and substitution Y=-1 Y=-5/2x+4

Algebra ->  Matrices-and-determiminant -> SOLUTION: In cramers rule and elimination and substitution Y=-1 Y=-5/2x+4       Log On


   

Question 999769: In cramers rule and elimination and substitution
Y=-1
Y=-5/2x+4
Found 2 solutions by solver91311, MathLover1:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Put the equations in standard form:

Eq 1:


Eq 2:

Cramer's Rule











Elimination

Multiply Eq 1 by -1

Eq 1 X -1:

Add the two equations:

Eq 3:

Solve Eq 3 for :

The value of is discerned by inspection of Eq 1.

Substitution

The value of is discerned by inspection of Eq 1.

Substitute the value of into equation 2:



And solve for

John

My calculator said it, I believe it, that settles it

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
1.
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 2 variables



system%280%2Ax%2B1%2Ay=-1%2C2.5%2Ax%2B1%2Ay=4%29



First let A=%28matrix%282%2C2%2C0%2C1%2C2.5%2C1%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 4 which are highlighted here:
system%280%2Ax%2B1%2Ay=highlight%28-1%29%2C2.5%2Ax%2B1%2Ay=highlight%284%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=%280%29%281%29-%281%29%282.5%29=-2.5. 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.



---------------------------------------------------------



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%2C1%2Chighlight%284%29%2C1%29%29



Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=%28-1%29%281%29-%281%29%284%29=-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=%28-5%29%2F%28-2.5%29=2



So the first solution is x=2




---------------------------------------------------------


We'll follow the same basic idea to find the other solution. Let's reset by letting A=%28matrix%282%2C2%2C0%2C1%2C2.5%2C1%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%2C0%2Chighlight%28-1%29%2C2.5%2Chighlight%284%29%29%29



Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=%280%29%284%29-%28-1%29%282.5%29=2.5.



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=%282.5%29%2F%28-2.5%29=-1



So the second solution is y=-1




====================================================================================

Final Answer:




So the solutions are x=2 and y=-1 giving the ordered pair (2, -1)




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




solutions:
x=2
y=-1+
2.
elimination
y=-1+.....eq.1
%285%2F2%29x%2By=4....eq.2
-----------------------------------subtract eq.1 from eq.2
%285%2F2%29x%2By-y=4-%28-1%29
%285%2F2%29x=4%2B1
5x%2F2=5
5x=5%2A2
x=%28cross%285%29%2A2%29%2Fcross%285%29
x=2
solutions:
x=2
y=-1+



3.substitution
y=-1+.....eq.1
%285%2F2%29x%2By=4....eq.2
-----------------------------------substitute -1 for y in eq.2 annd solve for x
%285%2F2%29x%2B%28-1%29=4....eq.2
%285%2F2%29x-1=4
5x%2F2=4%2B1
5x=5%2A2
x=%28cross%285%29%2A2%29%2Fcross%285%29
x=2
solutions:
x=2
y=-1+