Solving Systems of Equations by Elimination
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Solving Systems of Equations by Elimination

[From: ] [author: ] [Date: 12-04-30] [Hit: ]
-1. 3x+y=14 (a),In the first question, try to eliminate x by combining (a)+(b), which gives 0x+3y=15,2.......
http://i46.tinypic.com/rvxfmx.jpg

^ picture of 5 questions that i need help with? lol, i haven't done with stuff in forever so forgive me.

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1. 3x+y=14 (a), -3x+2y=1 (b)

In the first question, try to eliminate "x" by combining (a)+(b), which gives 0x+3y=15, or y = 5 (x = 3)

2. 2x-y=6 (a), 4x-2y = 12 (b)

In this equation, if you use careful observation, you would find that (b) equation comes from 2 times of (a) equation, therefore both of the equation are the same, and in real number plain, there are infinite ordered pairs (x,y) which satisfy the equation

3. Let number of "1-step" be "x", and "2-step" be "y", the question suggested that x+y=1150 (a) and 3x-2y=1300 (b)

To eliminate the variable, you have to make coefficient in front of the variables matched each other (find the LCD). In this equation, we will eliminate "y"

Multiply (a) with 2 gives 2x+2y=2300 (c)
3x-2y=1300 (b)
Using (b)+(c) gives 5x=3600, or x = 720, and this is the answer, as the question wants us to find the number of "1-step" (or "x"). To find "y", replace "x" in (a) (y=430)

4. 3x-6y=3 (a), 7x-5y=-11 (b). In this equation, we will eliminate "y". As LCD of 6 and 5 be 30, we will make the coefficient of y be 30. In the following solution, I would prefer to do it another way, but basically, it's similar to each other

Multiply (a) with 5 gives 15x-30y=15 (c)
Multiply (b) with -6 gives -42x+30y=66 (d)
(c)+(d) gives -27x=81 or x=-3 (y=-2)

5. 4x+2y=12 (a), x-y=3 (b)
Dividing (a) with 2 gives 2x+y=6 (c)
(b)+(c) gives 3x=9 or x=3 (y=0)
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