Grade 11 Pre-Cal E

Systems of Linear Equations

Today in class we learned how to do a 2 variable equation question. Don't worry about 3 variables we never got to that point yet.

There are 3 ways to solve 2 variable equations:

1. Graph
Basically you just have to graph the 2 equations, but to graph it in your graphing calculator it must be in a format such as this: y = [input equation here]. After both equations are graphed look for the intersect of the two, so look for where the lines meet. If they don't intersect at all, then it's not valid.


2. Elimination
5x + 4y = 6
-3y - 2x =-1

1. First you must make sure the x-values, y-values and numbers are in the same column.
5x + 4y = 6
2x - 3y = -1

2. Now you can choose a value to get rid of, x-value or y-value? I choose y. It's time to eliminate the y-value. You can do so by multiplying all the terms in each equation by the y-value of the opposing equation.
3(5x + 4y = 6)
4(2x - 3y = -1)
==========
15x + 12y = 18
-8x - 12y = -4

3. It's time to eliminate. Add or subtract to do so.
15x + 12y = 18
-8x - 12y = -4
===========
7x = 14
x = 2

4. Substitute the value back into the original equation to solve for the second variable.
5(2) + 4y = 6
10+ 4y = 6
4y = -4
y = -1

Solution: (2, -1)

Time to see if the equation works:
-3(-1) - 2(-2) = -1
3 - 4 = -1
-1 = -1 Correct.

If you end up getting an end result of 0 = 0 then it is always true.
Any (x, y) point that solves the first line also makes the second line true.


3. Substitution
4x + y = 1
2x - 3y = 4

1. Solve the equation with 1 variable.
4x + y = 1
y = -4x + 1

2. Substitute that new equation into the second equation's y-value, and solve.
2x - 3y = 4
2x - 3(-4x + 1) = 4
2x + 12x - 3 = 4
14x = 7
x = 1/2

Check:
y = -4(1/2) + 1
y = -2 + 1
y = 1
Solution: ((1/2), 1)

Oh yes, next scribe. Since I'm pretty sure everybody already went once, I'll cycle it through again. Which means Richard is the next scribe. Just nobody pick me halfway through the cycle. Since I went just now.

posted by Anonymous @ 4:05 PM