Give two systems of equations that would be easier to solve by substitution than by elimination. then give two systems that would be easier to solve with elimination. finally, explain how you decide whether to use elimination or substitution to solve a system.

These techniques for elimination are preferred for 3rd order systems and higher.  They use "Row-Reduction" techniques/pivoting and many subtle math tricks to reduce a matrix to either a solvable form or perhaps provide an inverse of a matrix (A-1)of linear equation AX=b.  Solving systems of linear equations (n>2) by elimination is a topic unto itself and is the preferred method.  As the system of equations increases, the "condition" of a matrix becomes extremely important.  Some of this may sound completely alien to you.  Don't worry about these topics until Linear Algebra when systems of linear equations (Rank 'n')  become larger than 2.

Step-by-step explanation:

Just to add a bit more information, "Elimination" Can have a variety of other interpretations.  Elimination techniques typically refer to 'row reduction' to achieve 'row echelon form.'  Do not worry if you have not heard of these terms.  They are used in Linear Algebra when referring to "Elimination techniques"

Gaussian Elimination

Gauss-Jordan Elimination

LU-Decomposition

QR-Decomposition

These techniques for elimination are preferred for 3rd order systems and higher.  They use "Row-Reduction" techniques/pivoting and many subtle math tricks to reduce a matrix to either a solvable form or perhaps provide an inverse of a matrix (A-1)of linear equation AX=b.  Solving systems of linear equations (n>2) by elimination is a topic unto itself and is the preferred method.  As the system of equations increases, the "condition" of a matrix becomes extremely important.  Some of this may sound completely alien to you.  Don't worry about these topics until Linear Algebra when systems of linear equations (Rank 'n')  become larger than 2.

Substitution is the preferred method for 2 equations in 2 unknowns.  The constants are unimportant other than having a non-zero determinant.  It is always easy to find multiplicative factors using LCMs of one variable or the other to allow substitution into the other equation:

Example:

4X + 5Y = 9

5X -  4Y = 1

Notice that 20 is a LCM of either the X or Y variable.  So multiply the first by 4 and the second by 5 and then adding the two (Y's will drop out allowing for substitution)

4(4X + 5Y = 9)

5(5X -  4Y = 1)  

Multiplying to produce the LCM factors:

16X + 20Y = 36

25X -  20Y = 5

Adding the equations

41X = 41

X = 1

Substitution into either equation yields

Y = 1

Elimination techniques are preferred for Rank-n>3

a. -0.3 > -0.33

answer: -1

step-by-step explanation: