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Recap of what was just presented:
1) System of Linear Equations are 2 or more linear equations
2) There may be 1 solution, no solutions, or infinite solutions 3) Solutions can be seen on the graph wherever the lines touch or intersect
**If they intersect at just one point, then there is only 1 solution
If they never touch, no solution
If they are the same line (equivalent equations) then there
are infinite solutions.
1 solution
no solutions
look for this in parallel
line scenarios
infinite solutions
they are equivalent equations
so they look like the same line
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Basic idea is to add or subtract the equations together to get
a) 1 equation
b) 1 variable
THEN you can solve for the one remaining variable. After that
you simply plug in the variable solution you have to let you solve
for the other variable you need.
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It's similar to how you found the 2nd variable using the elimination method.
This causes confusion for people because it feels like you're doing the same thing. Are you? Yes, but in a different order.
Elimination method you get rid of a variable by adding or subtracting, solve for
the remaining variable AND THEN FINALLY SUBSTITUTE IT TO GET THE 2ND VARIABLE.
In the substitution method you substitute in a value to GET THE 1st VARIABLE,
then you DO IT AGAIN to get the 2nd variable.
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Graphing is the 1st way we can
find a solution for a system of equations. Can be time consuming.
Usually this is for when you already have
a graph for at least 1 of the equations.
Remember to divide by ­2
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slope is 2 and y­intercept is ­6
Note the y­intercept is ­6, and the slope is 2/1 6
Slope is 1/2 and y­intercept is ­3
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where they
cross is the solution
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y = 2x ­ 6
(­2) = 2(2) ­ 6
­2 = 4 ­ 6
­2 = ­2
x ­ 2y = 6
2 ­ 2(­2) = 6
2 + 4 = 6
6 = 6
Notice that you don't get
the same value for each equation. Instead you'll see that both equations
reach a true statement. 9
You could solve with graphing or substitution
method as well.... but the problem specifically
states to use elimination method. A lot of problems will push you into a specific type of solution. LEARN ALL THREE METHODS.
There are other numbers
you could have chosen to multiply by.... just would
have caused a little more
work. So did it have to be 2?
No. 10
Notice that we are not using 2x + y = 7 now.... we're using
the new equivalent equation we
made in step 1 by multiplying it
by 2. Now we add because it'll
knock out the y variables and leave
just x.
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Here's where people start
to think we're using the
substitution method. Remember,
substitution method does not have
you adding or subtracting the equations like we did in Step 2.
You could substitute x=4
into the other equation as well.
It'd give you y = ­1 just as
easily. My rule of thumb is substitute it
back into the "easiest" looking
equation. 12
­3x ­ 2y = ­10 2x + y = 7
­3(4) ­ 2(­1) = ­10 2(4) + (­1) = 7
­12 + 2 = ­10 8 + (­1) = 7
­10 = ­10 7 = 7
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The new first equation is y = 1,
it becomes ­y = ­1 after multiplication. Why multiply by ­1?
because you want to add the equation to the other equation....and it'll let the y's cancel out. Easy to graph these equations....
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Here's a graph
of the original
equations showing
a solution of (1,1) as well.
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Overall I think example B is to be thought of
as an additional strategy to attack these situations.
It's a slightly different way to use the elimination method.
Look back at Example A and then compare it to Example B.
You'll see the differences if you pay attention. 17
The key to substitution
is getting 1 of the variables by itself. Doesn't matter which one really.
Now we can take what x equals (­4 ­ 4y) and plug it in for the x variable in the other equation....
Why? So it'll just have 1
variable we can solve for.
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no x's left, just y's
So we just got done isolating the x value in step one.... we now go to the other equation and plug in ­4­ 4y for the x value. Now we can solve for y. 19
Choose the "easiest" equation.
Plug in 1/2 for y (we got this from
step 2). Notice now you have no y's
left so you're free to solve for x. 20
I would imagine you see that y is sitting by itself with no coefficient in front of it. Look like the easiest equation to isolate?
1)
y = ­3x + 9
Now I can plug in ­3x + 9 for the y in the 1st equation
2)
4x ­ 3(­3x+9) = ­1
4x + 9x ­ 27 = ­1
13x ­27 = ­1
13x = 26
x = 2
3)
3x + y = 9
3(2) + y = 9
6 + y = 9
y = 3
4) Solution is (2,3)
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Try your best on the Practice Problems for Lesson 15. 26
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