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#1 2012-11-26 22:20:39
- demha
- Member
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Substitution and Elimination Method
This is for my school work. I am working online. Before I send it in I would like to know if it's all correct.
Solving Systems of Equations Using Substitution Method and Elimination Method:
SUBSTITUTION METHOD
1.
y = (2/3)x - 1
y = -x + 4
y = (2/3) - 1
(-x + 4) = (2/3)x – 1
(-1 1/2)x + 4 = -1
(-1 1/2)x = -5
x = 3
y = -x + 4
y = -3 + 4
y = 1
Answer: (3, 1)
2.
x + y = 0
3x + y = -4
x + y = 0
-x -x
y = 0 – x
3x + (0 – x) = -4
2x + 0 = -4
2x = -4
x = -2
x + y = 0
(-2) + y = 0
+2 +2
y = 0 + 2
y = 2
Answer: (-2, 2)
3.
4x + 3y = -15
y = x + 2
4x + 3(x + 2) = -15
4x + (3x + 6) = -15
7x + 6 = -15
-6 -6
7x = -21
x = -3
y = x + 2
y = (-3) + 2
y = -1
Answer: (-3, -1)
4.
x + 2y = -4
4y = 3x + 12
x + 2y = -4
x = -4 - 2y
4y = 3(-4 - 2y) + 12
4y = (-12 – 6y) + 12
10y = -12 + 12
10y = 0
y = 0
4y = 3x + 12
4(0) = 3x + 12
3x = 12
x = 4
Answer: (4, 0)
5.
y = 2x
x + y = 3
x + y = 3
x + (2x) = 3
3x = 3
x = 1
y = 2x
y = 2(1)
y = 2
Answer: (1, 2)
6.
x = 3 - 3y
x + 3y = -6
x + 3y = -6
(3 – 3y) + 3y = -6
3 + 0 = -6
3 = -6
Answer: There is no solution.
7.
y = -2x + 1
y = x – 5
y = -2x + 1
(x – 5) = -2x + 1
+5 +5
x = -2x + 6
+2x +2x
3x = 6
x = 2
y = x – 5
y = (2) – 5
y = 3
Answer: (2, 3)
8.
y = (1/2)x - 3
y = (3/2)x – 1
y = (1/2)x – 3
(3/2)x – 1 = (1/2)x – 3
+1 +1
(3/2)x = (1/2)x – 4
(4/2)x = -4
2x = -4
x = -2
y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = 4
Answer: (-2, 4)
9.
x + y = 2
4y = -4x + 8
x + y = 2
y = 2 – x
4y = -4x + 8
4(2 – x) = -4x + 8
8 – 4x = -4x + 8
+4x +4x
8 – 8x = 8
-8 -8
8x = 0
0 = 0
Answer: This equation has infinite solutions.
ELIMINATION METHOD
10.
y = (2/3)x - 1
y = -x + 4
0 = (-5/3)x + 5
(-5/3)x = 5
x = -3
y = -x + 4
y = 3 + 4
-3 -3
y = 1
Answer: (-3, 1)
11.
x + y = 0
3x + y = -4
2x + 0 = -4
2x = -4
x = -2
x + y = 0
-2 + y = 0
+2 +2
y = 2
Answer: (-2, 2)
12.
4x + 3y = -15
y = x + 2
y = x + 2 (times equation by 3)
3y = 3x + 6
4x + 3y = -15
7x = -21
x = -3
y = x + 2
y = -3 + 2
y = -1
Answer: (-3, -1)
13.
x + 2y = -4
4y = 3x + 12
x + 2y = -4 (times equation by 3)
3x + 6y = -12
4y = 3x + 12
3x - 4y = 3x – 3x + 12
4y – 3x = 12
3x + 6y = -12
3x + 6y + 4y – 3x = -12 + 12
6y + 4y = -12 + 12
10y = 0
y = 0
x + 2y = -4
x + 2(0) = -4
x = -4
Answer: (-4, 0)
14.
y = 2x
x + y = 3
x + y = 3 (times equation by 2)
2x + 2y = 6
y = 2x
y – 2x = 2x – 2x
y – 2x = 0
y – 2x = 0
2x + 2y = 6
2x + 2y + y – 2x = 6 + 0
2y + y = 6 + 0
3y = 6
y = 2
y = 2x
(2) = 2x
2x = 2
x = 1
Answer: (1, 2)
15.
x = 3 - 3y
x + 3y = -6
x + 3y = 3 – 3y + 3y
x + 3y = 3
x + 3y = 3
x + 3y = -6
Answer: Both equations are the same, but both equal two different things. Therefore the answer is no solution.
16.
y = -2x + 1
y = x – 5
0 = -x + -4
x = -4
y = -2x + 1
y = -2(-4) + 1
y = 8 + 1
y = 9
Answer: (-4, 9)
17.
y = (1/2)x - 3
y = (3/2)x – 1
0 = -x – 4
x = 2
y = x – 5
y = (-2) – 5
y = 7
Answer: (2, -3)
18.
x + y = 2
4y = -4x + 8
4y = -4x + 8
4y – 4y = -4x – 4y + 8
-4x – 4y = 8
x + y = 2 (times equation by 4)
4x + 4y = 8
-4x + 4y = 8
4x + 4y = 8
8y = 16
y = 2
x + y = 2
x + (2) = 2
-2 -2
x = 0
Answer: (0, 2)
If you dream about beating me, you better wake up and apologize.
-Muhammad Ali
#2 2012-11-26 22:37:40
- bob bundy
- Moderator
Offline
Re: Substitution and Elimination Method
hi demha
Welcome to the forum.
LATER EDIT: As I checked these I began to think I'd seen them before but I was rushing to finish while you were still on-line.
Now I've had time to look carefully, I see the second 9 are the same as the first 9, just a different method. You should, of course, get the same answers either way, which may help you to sort out the corrections.
SUBSTITUTION METHOD
1.
y = (2/3)x - 1
y = -x + 4
y = (2/3) - 1
(-x + 4) = (2/3)x – 1
(-1 1/2)x + 4 = -1
(-1 1/2)x = -5
x = 3
y = -x + 4
y = -3 + 4
y = 1
Answer: (3, 1)
CORRECT
2.
x + y = 0
3x + y = -4
x + y = 0
-x -x
y = 0 – x
3x + (0 – x) = -4
2x + 0 = -4
2x = -4
x = -2
x + y = 0
(-2) + y = 0
+2 +2
y = 0 + 2
y = 2
Answer: (-2, 2)
CORRECT
3.
4x + 3y = -15
y = x + 2
4x + 3(x + 2) = -15
4x + (3x + 6) = -15
7x + 6 = -15
-6 -6
7x = -21
x = -3
y = x + 2
y = (-3) + 2
y = -1
Answer: (-3, -1)
CORRECT
4.
x + 2y = -4
4y = 3x + 12
x + 2y = -4
x = -4 - 2y
4y = 3(-4 - 2y) + 12
4y = (-12 – 6y) + 12
10y = -12 + 12
10y = 0
y = 0
4y = 3x + 12
4(0) = 3x + 12
3x = 12
x = 4
Answer: (4, 0)
SOMETHING HAS GONE WRONG HERE. YOUR VALUES DON'T FIT THE EQUATIONS.
5.
y = 2x
x + y = 3
x + y = 3
x + (2x) = 3
3x = 3
x = 1
y = 2x
y = 2(1)
y = 2
Answer: (1, 2)
CORRECT
6.
x = 3 - 3y
x + 3y = -6
x + 3y = -6
(3 – 3y) + 3y = -6
3 + 0 = -6
3 = -6
Answer: There is no solution.
CORRECT
7.
y = -2x + 1
y = x – 5
y = -2x + 1
(x – 5) = -2x + 1
+5 +5
x = -2x + 6
+2x +2x
3x = 6
x = 2
y = x – 5
y = (2) – 5
y = 3
Answer: (2, 3)
CHECK Y AGAIN
8.
y = (1/2)x - 3
y = (3/2)x – 1
y = (1/2)x – 3
(3/2)x – 1 = (1/2)x – 3
+1 +1
(3/2)x = (1/2)x – 4
(4/2)x = -4
2x = -4
x = -2
y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = 4
Answer: (-2, 4)
CHECK Y AGAIN
9.
x + y = 2
4y = -4x + 8
x + y = 2
y = 2 – x
4y = -4x + 8
4(2 – x) = -4x + 8
8 – 4x = -4x + 8
+4x +4x
8 – 8x = 8
-8 -8
8x = 0
0 = 0
Answer: This equation has infinite solutions.
CORRECT
ELIMINATION METHOD
10.
y = (2/3)x - 1
y = -x + 4
0 = (-5/3)x + 5
(-5/3)x = 5
x = -3
y = -x + 4
y = 3 + 4
-3 -3
y = 1
Answer: (-3, 1)
THESE VALUES DON'T FIT THE EQUATIONS
11.
x + y = 0
3x + y = -4
2x + 0 = -4
2x = -4
x = -2
x + y = 0
-2 + y = 0
+2 +2
y = 2
Answer: (-2, 2)
CORRECT
12.
4x + 3y = -15
y = x + 2
y = x + 2 (times equation by 3)
3y = 3x + 6
4x + 3y = -15
7x = -21
x = -3
y = x + 2
y = -3 + 2
y = -1
Answer: (-3, -1)
CORRECT
13.
x + 2y = -4
4y = 3x + 12
x + 2y = -4 (times equation by 3)
3x + 6y = -12
4y = 3x + 12
3x - 4y = 3x – 3x + 12
4y – 3x = 12
3x + 6y = -12
3x + 6y + 4y – 3x = -12 + 12
6y + 4y = -12 + 12
10y = 0
y = 0
x + 2y = -4
x + 2(0) = -4
x = -4
Answer: (-4, 0)
CORRECT
14.
y = 2x
x + y = 3
x + y = 3 (times equation by 2)
2x + 2y = 6
y = 2x
y – 2x = 2x – 2x
y – 2x = 0
y – 2x = 0
2x + 2y = 6
2x + 2y + y – 2x = 6 + 0
2y + y = 6 + 0
3y = 6
y = 2
y = 2x
(2) = 2x
2x = 2
x = 1
Answer: (1, 2)
CORRECT
15.
x = 3 - 3y
x + 3y = -6
x + 3y = 3 – 3y + 3y
x + 3y = 3
x + 3y = 3
x + 3y = -6
Answer: Both equations are the same, but both equal two different things. Therefore the answer is no solution.
CORRECT
16.
y = -2x + 1
y = x – 5
0 = -x + -4
x = -4
y = -2x + 1
y = -2(-4) + 1
y = 8 + 1
y = 9
Answer: (-4, 9)
CHECK AGAIN
17.
y = (1/2)x - 3
y = (3/2)x – 1
0 = -x – 4
x = 2
y = x – 5
y = (-2) – 5
y = 7
Answer: (2, -3)
CHECK AGAIN
18.
x + y = 2
4y = -4x + 8
4y = -4x + 8
4y – 4y = -4x – 4y + 8
-4x – 4y = 8
x + y = 2 (times equation by 4)
4x + 4y = 8
-4x + 4y = 8
4x + 4y = 8
8y = 16
y = 2
x + y = 2
x + (2) = 2
-2 -2
x = 0
Answer: (0, 2)
CANNOT BE RIGHT. COMPARE WITH Q9
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#3 2012-11-27 18:57:38
- demha
- Member
Offline
Re: Substitution and Elimination Method
Thanks for the welcome and super thanks for answering back! I knew that 1 - 9 & 10 - 18 were the same equations... but now I feel kind of stupid ._.
I should have known that I would get the same answers! =P
Here are the mistakes I redid:
4.
x + 2y = -4
4y = 3x + 12
x + 2y = -4
x = -4 - 2y
4y = 3(-4 - 2y) + 12
4y = (-12 – 6y) + 12
10y = -12 + 12
10y = 0
y = 0
x + 2y = -4
x + 2(0) = -4
x = -4
Answer: (4, 0)
---
7.
y = -2x + 1
y = x – 5
y = -2x + 1
(x – 5) = -2x + 1
+5 +5
x = -2x + 6
+2x +2x
3x = 6
x = 2
y = x – 5
y = (2) – 5
y = -3
Answer: (2, -3)
8.
y = (1/2)x - 3
y = (3/2)x – 1
y = (1/2)x – 3
(3/2)x – 1 = (1/2)x – 3
+1 +1
(3/2)x = (1/2)x – 4
(4/2)x = -4
2x = -4
x = -2
y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = -4
Answer: (-2, -4)
---
(Elimination Method)
10.
y = (2/3)x - 1
y = -x + 4
0 = (-5/3)x + 5
(- 5/3)x = 5
x = 3
y = 2/3x – 1
y = 2/3(3) – 1
y = 2 – 1
y = 1
Answer: (3, 1)
---
16.
y = -2x + 1
y = x – 5
0 = 3x + 6
3x = 6
x = 2
y = x – 5
y = 2 – 5
y = -3
Answer: (2, -3)
17.
y = (1/2)x - 3
y = (3/2)x – 1
0 = -x – 2
x = -2
y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = -4
Answer: (-2, -4)
---
Now for number 18, I understand I am supposed to get the same answer as number 9 which is a "word solution." The answer should be infinate (0 = 0 for example). I have tried a few times but I just can't seem to get it. Would you mind helping me? Here is the equation yet again:
(Elimination Method)
18.
x + y = 2
4y = -4x + 8
If you dream about beating me, you better wake up and apologize.
-Muhammad Ali
#4 2012-11-27 19:14:24
- bob bundy
- Moderator
Offline
Re: Substitution and Elimination Method
hi demha
Here are the mistakes I redid:
4.
x + 2y = -4
4y = 3x + 12
x + 2y = -4
x = -4 - 2y
4y = 3(-4 - 2y) + 12
4y = (-12 – 6y) + 12
10y = -12 + 12
10y = 0
y = 0
x + 2y = -4
x + 2(0) = -4
x = -4
Answer: (4, 0)
YOU MEANT (-4,0) OF COURSE !
---
7.
y = -2x + 1
y = x – 5
y = -2x + 1
(x – 5) = -2x + 1
+5 +5
x = -2x + 6
+2x +2x
3x = 6
x = 2
y = x – 5
y = (2) – 5
y = -3
Answer: (2, -3)
CORRECT
8.
y = (1/2)x - 3
y = (3/2)x – 1
y = (1/2)x – 3
(3/2)x – 1 = (1/2)x – 3
+1 +1
(3/2)x = (1/2)x – 4
(4/2)x = -4
2x = -4
x = -2
y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = -4
Answer: (-2, -4)
CORRECT
---
(Elimination Method)
10.
y = (2/3)x - 1
y = -x + 4
0 = (-5/3)x + 5
(- 5/3)x = 5
x = 3
y = 2/3x – 1
y = 2/3(3) – 1
y = 2 – 1
y = 1
Answer: (3, 1)
CORRECT
---
16.
y = -2x + 1
y = x – 5
0 = 3x + 6
3x = 6
x = 2
y = x – 5
y = 2 – 5
y = -3
Answer: (2, -3)
CORRECT
17.
y = (1/2)x - 3
y = (3/2)x – 1
0 = -x – 2
x = -2
y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = -4
Answer: (-2, -4)
CORRECT
---
Now for number 18, I understand I am supposed to get the same answer as number 9 which is a "word solution." The answer should be infinate (0 = 0 for example). I have tried a few times but I just can't seem to get it. Would you mind helping me? Here is the equation yet again:
(Elimination Method)
18.
x + y = 2
4y = -4x + 8
4y = -4x + 8
4y – 4y = -4x – 4y + 8
-4x – 4y = 8
This is where you went wrong.
line 2 is ok
4y – 4y = -4x – 4y + 8
so
0 = -4x – 4y + 8
Now what are you doing? If you add 4x and add 4y to each side you'll have
0 +4x + 4y = -4x+4x – 4y + 4y + 8
4x + 4y = 8
Now the other equation was
x + y = 2
so multiply by 4
4x + 4y = 8
So now eliminating gives
0 = 0 as you wanted.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#5 2012-11-27 21:27:50
- demha
- Member
Offline
Re: Substitution and Elimination Method
Hello Bob! 
So it was adding 4x & 4y to each side that would give me the correct answer. I see now. Then we have both similar equations then start the elimination to get 0 + 0 = 0 which in the end is 0 = 0: there is an infinate solution.
Thanks you very much Bob! I appreciate you taking your time helping me with this.
If you dream about beating me, you better wake up and apologize.
-Muhammad Ali