How to Solve System of Two Linear Equations

Jason76

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How to solve the system below? :confused: What variable are we solving for?

System (a)

\(\displaystyle 2x_{1} - 4x_{2} = 6\)

\(\displaystyle 3x_{1} = 9\)
 
Ok i get it now. We solve the equation which can be solved, the one with only one variable, then we plug that value into the other equation. The goal being to find the value of the other variable. ;)
 
How to solve the system below? :confused: What variable are we solving for?

System (a)

\(\displaystyle 2x_{1} - 4x_{2} = 6\)

\(\displaystyle 3x_{1} = 9\)

You;re looking for both x1 and x2.

Addition method: Add three times the first to two times the second to solve for x2.

Substitution method (easier in this case): Solve equation (2) for x1, and substitute the value into equation(1).
 
Ok i get it now. We solve the equation which can be solved, the one with only one variable, then we plug that value into the other equation. The goal being to find the value of the other variable. ;)
Yes for this situation, but no generally.

If we have two equations with two unknowns, we first use one equation to solve for one unknown in terms of the other unknown (not a numeric solution). Then we reduce the other equation to one unknown by substituting for the unknown that is known in terms of the other unknown. Then we get numeric solutions for the remaining unknown. And finally, we get a numeric solution for the first unknown. This is abstract so here is an example.

\(\displaystyle x^2 - 2xy + y^2 = 81.\)

\(\displaystyle x + y = 5.\)

Step 1: Solve for one unknown in terms of the other.

You can use either equation to do so. In this case, it is obviously easier to use the second equation.

\(\displaystyle y = 5 - x.\)

Step 2: Reduce the other equation to an equation in one unknown by substitution.

\(\displaystyle x^2 - 2xy + y^2 = 81 \implies x^2 - 2x(5 - x) + (5 - x)^2 = 81 \implies x^2 - 10x + 2x^2 + 25 - 10x + x^2 = 81 \implies 4x^2 - 20x - 56 = 0.\)

Step 3: Solve the reduced equation numerically.

\(\displaystyle 4x^2 - 20x - 56 \implies x^2 - 5x - 14 = 0 \implies (x - 7)(x + 2) = 0 \implies x = 7\ or\ x = - 2.\)

Step 4: Use the other equation to solve for the other unknown numerically.

\(\displaystyle x = 7 \implies 7 + y = 5 \implies y = - 2.\)

\(\displaystyle x = - 2 \implies - 2 + y = 5 \implies y = 7.\)

So the solutions are (7, - 2) and (- 2, 7).
 
How to solve the system below?

We helped you solve systems of linear equations in 2012 and 2013.

Do you recall any of the following?

coefficient matrix

augmented matrix

matrix equation

row operations

reduced echelon form (ref)

row reduced echelon form (rref)

elimination method

substitution method

Cramer's Rule

:cool:
 
I remember how to do all that linear algebra stuff somewhat. I pretty much know what's going on, as far as this problem.
 
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