Hey guys its Pryanka,
Sorry for the late update, let’s just say my blog “wasn’t working”…
Anyways, yesterday in pre-cal we learnt a new lesson which was called: Solving Systems of Linear Equations: Two Equations, Two Unknowns. This is method 3: Elimination. If you didn’t understand the lesson we learnt on December 13, then this lesson will be difficult for you. Also this lesson is very important to know because this knowledge will apply to today’s lesson. So try remembering this lesson!!
*The blog on the lesson we learnt today will be posted tomorrow.
Steps for the elimination method:
Step 1: Put the like terms (variables) together in the equation so that they are in the same column.
Ex). 5x=15=-7y ------> 5x+7y=-15
9y=2x=-6 ------> -2x+9y=-6
Step 2: Choose any variable with a coefficient and make it the same for both equations. Also remember to multiply the whole equation by the multiple you have chosen.
Ex). 5x(2)+7y(2)= -15(2) (In this case the common factor is 10 for the “x” value)
-2x(5)+9y(5)= -6(5)
Step 3: Add or subtract the equations so that we can eliminate one variable. After doing that we can solve for the other variable.
Ex). 10x+14y= -30 (In this case we will have to subtract to eliminate the "x" variable)
- 10x+45= -30
0-31y=0
-31y = 0
-31 -31
y=0
Step 4: Substitute back in the value found, into any equation to solve for the second variable.
Ex). 5x+7y= -15
5x+7(0)= -15
5x= -15
5 5
x = -3
Step 5: State your answer in an ordered pair.
Ex). (x,y) ---> (-3,0)
Step 6: Finally, check your answer.
5x+15= -7y
5(-3) +15= -7(0)
-15+15=0
0=0
9y+2x= -6
9(0) +2(-3) = -6
-6= -6
Now, let’s try to do one example with a fraction in it.
2e - 4f = 7
5f = -20+7
First, put the like terms (variables) together:
2e – 4f = 7
-7e + 5f = -20
Then, chose a variable with a coefficient and make it the same as the other equation:
2e(5) – 4f(5)= 7(5) --------> 10e -20f = 35
-7e(4) + 5f(4) = -20(4) ------> -28e + 20f = -80
After you have done that, add or subtract both equations to eliminate one variable:
+ -28e + 20f = -80
-18e +0 = -45
-18e = -45
-18 -18
e= 45 (we can reduce these numbers by 9)
18
e= 5
2
Now we can solve for "f", by substituting in the "e” value in any equation.
Now we can solve for "f", by substituting in the "e” value in any equation.
-7e + 5f = -20
-7(5) + 5f= -20
1 2
-35 + 5f = -20
2
5f = -20+35 (multiply the 20 and the 1 by 2 to get the common denominator)
1 2
5f = -40+35
2 2
5f = -5 (to get 5 by itself, divide both sides by 5)
2
5f = -5
5 2
5
f = -5 x 1
2 5
f = -5 (reduce these numbers by 5)
10
f = -1
2
Now we can put these to variables in there ordered pair:
(5,-1)
2 2
Last step is to check your answer:
Check:
2e - 4f = 7
2(5) -4(-1) = 7 (substitute the values found in the equation)
2 2
10 + 4 = 7
2 2
5 + 2 = 7
7=7
5f = -20 + 7e
5(-1) = -20 + 7(5)
2 2
-5 = -20 + 35 (multiply -20 by 2 so that we can get the same denominator)
2 1 2
-5 = -40 + 35
2 2 2
-5 = -5
2 2
Both the "x" and "y" values work out in the check, this means you did the question correct!
HOMEWORK:
For homework, we have to complete the worksheet we got in class, which was: Solving Systems of Linear Equations: Two Equations, Two Unknowns.
If you guys still dont get it, here are some videos that teach you how to do these types of equations: http://www.youtube.com/watch?v=xB-oXaCoJoc
http://www.youtube.com/watch?v=CIBb7YR0FHQ&feature=related
http://www.youtube.com/watch?v=DsKT1M0Jnk0&feature=related
The last video shows you how to solve special equations.
That's all for this lesson, see you guys tomorrow!Bye(:
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