Reviewing Algebra II (Conclusion) Solving Simultaneous Equations

 In Part 5 we conclude this series on Review of Algebra II by examining three methods for solving linear simultaneous equations - which form an important part of  any proper Algebra II course.   Three primary modes of solution include: i) By substitution (often the simplest), ii) graphically and iii) using determinants.  Each of which will be shown in the sample problem.

  The last is especially important given determinants actually have their origin in connection with the problem of solving simultaneous linear equations.

A pair of values, that satisfies each of two equations in two unknowns (say one for x, the other for y) is a simultaneous solution of the two equations.  

Take the following simultaneous equation, with unknowns in x, y:

3x + y  =  12 and x - 3y = 14

To solve by substitution, we take the 1st eqn. and solve for y:

y =   12 - 3x

Then subst. this y into the 2nd equation.

x-  3( 12 - 3x) = 14

And solve for x:

x - 36 + 9x = 14

10x  =  14 + 36 =  50

x = 5

Now subst. x = 5 back into 2nd eqn.

x - 3y = 14

Þ

5  -  3y  = 14  Or:  3y = 5 - 14 = -9

Then:  y =  -9 / 3  

y =  -3

The pair can also be solved graphically yielding:

The graphs are seen to intersect at x = 5 and y = -3.

These simultaneous equations can also be solved using determinants and indeed determinants were developed originally to solve such equations. For a generic pair of equations:  

a₁x +  b₁y =    c₁

a₂x +  b₂y =   c₂

We have the main determinant:

Þ

D = a₁ b₂ -  a₂ b₁

However, for the solution x we need the ratio of the determinants, i.e.

Where for the numerator:  D2 = c₁ b₂ -  c₂ b₁

For the solution to y we have:

The pair of our original eqns.:  3x + y  =  12 and x - 3y = 14

Yields:  a₁ = 3,   b₁ = 1  and   c₁  = 12

a₂ = 1,   b₂ = -3  and   c₂  = 14

Form main determinant from  a₁ , b₁, a₂, b₂ :

Þ

=   (3)  (-3) – (1)(1) = - 9 -1 = -10

Solve for x:

Given top determinant: D2 =  c₁ b₂ -  c₂ b₁

= (12) (-3) – (14)(1) =  -36 -14 = -50

Then:  x = -50/-10 = 5 

Solve for y:

Given D2 = a₁ c₂ -  a₂ c₁

= (3) (14) – (1) (12) =  42 – 12 = 30

Then: y = D2/ D =30/ -10  = -3 

A shorter (but less instructive, i.e. faithful to the origin) method of finding the x,  y solutions entails simply replacing the respective x and y-columns in the main determinant with the constants (i.e. answers, 12 and 14).

Then for the single x- determinant (D_x), after replacing the x-column with 12 and 14:

Þ   (12)(-3) -  14 (1) = -36 -14 = -50

And for the single y- determinant (D_y) ,  after replacing the y-column with 12 and 14:

Þ   (14)(3) -  12 (1) = 42 -12 =  30 

Then the respective solutions are:

Practice Problems:

1) Solve the system below by substitution:

3x - y = 4

5x + 2y = 3

2)Solve the linear system below graphically and by determinants:

2x - y = 10

5x + 2y = 7

3) Solve the system below by substitution and by determinants:

4x - y = 5

3x + 2y = 12

4) Solve the pair of equations below graphically, by substitution and using determinants:

2x – 3y = 8,  

3x + y = 1