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Question 1190535: I'm having a really hard time understanding solving using elimination. Could somebody explain it to me in lots of detail.
x-y=3 and 2x+4y=9
Found 3 solutions by Alan3354, math_tutor2020, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! x-y=3 and 2x+4y=9
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Multiply the 1st eqn by 2
2x-2y = 6
2x+4y = 9
---------------------- Subtract
-6y = -3 ..... x is eliminated
y = 1/2
==============
x-y=3
x - (1/2) = 3
x = 3 1/2 or 3.5
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Let's multiply both sides of x-y=3 by 4
We should end up with 4x-4y=12
The original system of
is equivalent to
Notice how we have -4y and +4y from the first and second equations respectively.
They add to 0y, aka 0. The y terms go away.
Let's add the terms straight down- 4x+2x becomes 6x
- -4y+4y becomes 0y and goes away (hence the term "elimination")
- 12+9 becomes 21
We're left with this equation: 6x = 21
Divide both sides by 6 to solve for x. You should get x = 21/6 = 7/2 = 3.5
Let's plug this into any equation involving x and y, and solve for y
x-y = 3
3.5-y = 3
-y = 3-3.5
-y = -0.5
y = 0.5
y = 1/2
Or,
2x+4y = 9
2(3.5)+4y = 9
7+4y = 9
4y = 9-7
4y = 2
y = 2/4
y = 1/2
y = 0.5
Answer in fraction form: x = 7/2 and y = 1/2
Answer in decimal form: x = 3.5 and y = 0.5
Answer by ikleyn(52775)  (Show Source):
You can put this solution on YOUR website! .
I'm having a really hard time understanding solving using elimination. Could somebody explain it to me in lots of detail.
x-y=3 and 2x+4y=9
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Specially for the given problem, the procedure (the Elimination method) works as follows.
Your starting system is
x - y = 3 (1)
2x + 4y = 9 (2)
Now multiply first equation by 2 (both sides). Keep the second equation as is. You will get
2x - 2y = 6 (3)
2x + 4y = 9 (4)
Next, subtract equation (3) from equation (4).
The terms "2x" in both equations will cancel each other, and you will obtain an equation for one single unknown "y":
4y - (-2y) = 9 - 6,
or
6y = 3
y = 3/6 = 0.5.
So, one unknown is just found: y = 0.5.
To find another unknown "x", substitute the found value y= 0.5 into EITHER equation (1) or (2),
For example, substituting into equation (1) gives you
x = 3 + y = 3 + 0.5 = 3.5.
At this point, the problem is just solved, and the answer is x= 3.5, y= 0.5.
You may check (and you MUST check) the answer by substituting the found values into the original / (given) equations.
I leave this job to you, since it is elementary.
At this point, the solution is complete.
The idea of the method is to make the coefficients EQUAL in two equations for some
of the two unknowns, and then subtract equations, ELIMINATING this unknown.
Having this explained solution, please NEVER say again that you do not understand the subject:
it was explained to you in the most comprehensive and detailed way.
//////////////
To see many other similar solved problems, look into the lesson
- Solution of the linear system of two equations in two unknowns by the Elimination method
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
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