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## #1 2012-11-26 22:20:39

demha
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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

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

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

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

5.
y = 2x
x + y = 3
x + y = 3
x + (2x) = 3
3x = 3
x = 1

y = 2x
y = 2(1)
y = 2

6.
x = 3 - 3y
x + 3y = -6

x + 3y = -6
(3 – 3y) + 3y = -6
3 + 0 = -6
3 = -6

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

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

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

11.
x + y = 0
3x + y = -4

2x + 0 = -4
2x = -4
x = -2

x + y = 0
-2 + y = 0
+2         +2
y = 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

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

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

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

17.
y = (1/2)x - 3
y = (3/2)x – 1

0 = -x – 4
x = 2

y = x – 5
y = (-2) – 5
y = 7

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

"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

## #2 2012-11-26 22:37:40

bob bundy
Moderator

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### 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

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

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

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

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

CORRECT

6.
x = 3 - 3y
x + 3y = -6

x + 3y = -6
(3 – 3y) + 3y = -6
3 + 0 = -6
3 = -6

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

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

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

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

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

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

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

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

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

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

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
Full 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

---

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

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

---

(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

---

16.
y = -2x + 1
y = x – 5

0 = 3x + 6
3x = 6
x = 2

y = x – 5
y = 2 – 5
y = -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

---

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

"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

## #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

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

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

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

CORRECT

---

16.
y = -2x + 1
y = x – 5

0 = 3x + 6
3x = 6
x = 2

y = x – 5
y = 2 – 5
y = -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

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
Full 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.

"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

## #6 2012-11-27 21:38:23

bob bundy
Moderator

Offline

### Re: Substitution and Elimination Method

You are welcome.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei