SOLUTION: How do you know what to use to solve linear systems of equations? I've been doing problems online and they say what to use (addition, subtraction, make the left sides equal to each

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: How do you know what to use to solve linear systems of equations? I've been doing problems online and they say what to use (addition, subtraction, make the left sides equal to each      Log On


   

Question 877415: How do you know what to use to solve linear systems of equations? I've been doing problems online and they say what to use (addition, subtraction, make the left sides equal to each other etc.) but I haven't seen anything about why each way works for each problem.
For example:
Create a new equation by setting the left-hand sides of the equations equal to each other. y = 2x - 1 and 2y + x = 3.
How would you know that you can solve this, by making the left sides equal to each other (y = 2y + x), without being told to solve it that way?
Create a new equation by setting the left-hand side of the first equation equal to the right-hand side of the second equation. 2y = 2/5x - 3 and -24 = 4x + y.
How would you know that you can solve this by setting the left side of the first = to the right side of the second (2y = 4x + y)?
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
Maybe you are studying from the wrong source. Use a good Algebra 1 or Introductory Algebra textbook, and stop relying on something you find online. Good Algebra textbooks will develop the properties of numbers and the skills to use, and following those in the sequence presented in the books will give you much better, clearer development.

The techniques you use are all basically forms of substitution. In a linear system of equations with a solution, substitution works, because two expressions or numbers are stated as equal; so you can use either expression in any of the equations of the system.

Everything will make sense, if you study carefully and thoroughly from a good source of instruction.

Create a new equation by setting the left-hand sides of the equations equal to each other. y = 2x - 1 and 2y + x = 3.
Is this to solve the system?
Work first with the second equation----
2y%2Bx=3
2y%2Bx-x=3-x
2y=3-x
2y%281%2F2%29=%283-x%29%281%2F2%29
y=%283-x%29%2F2, meaning y has two expressions which are found to be equal.
-
y=2x-1, first equation
y=%283-x%29%2F2, second equation now transformed
Both expressions equal to y:
2x-1=%283-x%29%2F2----Solve this for x.
.
.
You know what to do for this one?