Simultaneous Equations help?

2a + 2(a - 8) = 32

a + a - 8 = 16

2a = 24

a = 12

12 - b = 8

b = 4

Answer: a = 12, b = 4

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a + 2b - 3c + 2a + b + c + 5a - 2b - c = 0 + 8 + 7

b = 15 - 8a + 3c

a + 2(15 - 8a + 3c) - 3c = 0

a + 30 - 16a + 6c - 3c = 0

3c = 15a - 30

c = 5a - 10

b = 15 - 8a + 3(5a - 10)

b = 15 - 8a + 15a - 30

b = 7a - 15

2a + 7a - 15 + 5a - 10 = 8

14a = 33

a = 33/14

b = 7(33/14) - 15

b = (33 - 30)/2

b = 3/2

c = 5(33/14) - 10

c = (165 - 140)/14

c = 25/14

Answer: a = 33/14; b = 3/2; c = 25/14

Proof—3rd equation:

5(33/14) - 2(3/2) - 25/14 = 7

(165 - 42 - 25)/14 = 7

98/14 = 7

7 = 7

a-b=8

2a + 2b = 32

get a by itself:

a = 8 + b

insert value of a into second equation:

2(8 + b) + 2b = 32

16 + 2b + 2b = 32

4b = 16

b = 4

insert b's answer back into first equation:

a - 4 = 8

a = 12

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Same concept, so just the steps:

a = 3c - 2b

b = 8 - 2a - c

b = 8 - 6c + 4b - c

b = 4b - 7c + 8

3b = 7c - 8

b = (7c - 8)/3

c = 5a - 2b - 7

c = 15c - 10b - 2b -7

14c = 12b + 7

c = (12b + 7)/14

c = (12 (7c - 8) + 7))/14

1 3

c = (28c - 32 + 7) / 14

14c = 28c - 32 + 7

25 = 14c

c = 1 11/14 !!!

b = 1.5 !!!

a = 3 (25/14) - 3

a = 75/14 - 3

a = 5 5/14 - 3

a = 2 5/14 !!!

Proof:

2 5/14 + 3 - 5 5/14

0 = 0

4 10/14 + 1.5 + 25/14 = 8

5 17/14 + 25/14

87/14 + 25/14

112/14

8 = 8

10 25/14 - 3 - 1 11/14 = 7

8 11/14 - 1 11/14 = 7

7 = 7

Edit: there was an error in my first answer. This one is complete and correct.

For the first question:

multiply the top equation by two (so that you can add the equations and cancel out the b terms):

2a - 2b = 16

2a + 2b = 32

Now, simply add the equations together:

4a = 48

a = 12

substitute this value back into the very first equation:

24 + 2b = 32

2b = 8

b = 4 (their answer isn't right)

You can always check if your answer's right by subbing your answers back into the equations - keep in mind, your answer has to work for both to be right.

Try the second problem by multiplying the second equation by 3, then adding it to the first equation - that'll leave you with an equation with only a and b (and a constant) - (call this equation four).

Then add the second and third equation to get another equation with only a and b (call this equation five).

See how you go from there - it becomes a lot like the first problem after that.

Hope that at least kind of helped =)

Use the multiplication method to multiply the all terms of the 1st equation in order to knock out one of the variables when you add both equations.

2(a-b=8) = 2a-2b=16

2a+2b=32

Now add both 4a =48

equations a=12

substitute a=12 into the first equation to solve for b: 12-b=8; 12-8=b; 4=b

Ans: a=12; b=4

2nd Question:

Step1: Multiply the entire 1st equation by -2 so that when you add it to the entire 2nd equation you will be able to knock out the a's

-2(a+2b-3c=0) -2a-4b+6c=0

+ 2a +b +c =8

Result 1: -3b+7c=8

Step2 Now multiply the 2nd equation by -5 and the 3rd equation by 2 so that you will be able to knock out the a's when you add both equations

-5(2a+b+c=8) -10a -5b-5c=-40

2(5a-2b-c=7) + 10a -4b-2c=14

Result 2: -9b-7c=26

Now add both Result 1 & Result 2 in order to knock out the c's & solve for b

-3b +7c = 8

+ -9b -7c = -26

-12b =-18

b=3/2

Now substitute b=3/2 for b in equations 1&3 since each b has a coefficient of 2 and you will be able to get rid of it so that it will make your task of solving for the other variables much easier.

a+2(3/2) -3c= 0 a+3-3c=0 a-3c=-3

5a-2(3/2)-c=7 5a-3-c=7 5a-c =10

-5(a-3c=-3) -5a+15c=15

5a-c=10

14c= 25

c=25/14

Now use Equation 1 in order to substitute for b&c and solve for a:

a+2(3/2) -3(25/14)=0; a+3 -(75/14)=0; a =-3+(75/14) ; a= 33/14

Ans: a= 33/14 b=3/2 c= 25/14

For the first one, multiply the first equation by 2, then add the two equations, to get 4a = 16+32, solve for a, then put that value in one of the equations and solve for b.

Second is a lot more complicated.

1. Work with 1 and 2, and eliminate one of the variables, say c. Multiply 2 by 3 and add to do that.

2. Work with 2 and 3, and eliminate the same variable. Just add them to do that.

3. now you have two equations in 2 unknowns, solve them the same way, and get values for a and b.

4. Put those values back in one of the equations, and solve for c.

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a - b = 8 -------------(1)

2a + 2b = 32 -------(2)

Follow the steps carefully

(1) * 2 ==> 2a - 2b = 16 ------------(3)

(2) + ( 3) ==> 4a = 48

so a = 12 now plug in a = 12 in (1)

12 - b = 8 ==> b = 20

{a = 12}

{b = 20}

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a + 2b - 3c = 0-------------(1)

2a + b + c = 8 -------------(2)

5a - 2b - c = 7 -------------(3)

(1) + (3) ==> 6a - 4c = 7-------(4)

(2) * 2 ==> 4a + 2b + 2c = 16--------------(5)

(3) + (5) ==> 9a + c = 23 ---------(6)

(4) ==> 6a - 4c = 7 ------------(4)

(6)*4==> 36a + 4c = 92 ------(7)

(4) + (7) ==> 42 a = 99 ==> a = 99/42 = 33//14 pulg in onto (4)

6*33/14 - 4c = 7

66 - 28c = 49

-28c = - 17

c = 17/28

Now plug in a , c in (1)

a + 2b - 3c = 0

33/14 + 2b - 3*17/28 = 0

2b = 51/28 - 33/14 = -15/14

b = -15/28

{a = 33/14}

{b = -15/28}

{c = 17/28}

a=b+8

2(b+8)+2b=32

2b+16+2b-32=0

4b=16

b=4

a=b+8

a=4+8

a=12

a+2b-3c=0

a=3c-2b

2(3c-2b)+b+c=8

6c-4b+b+c=8

7c-3b=8-----equation 1

a=3c-2b

5(3c-2b)-2b - c = 7

15c-10b-2b - c = 7

14c-12b=7------equation 2

7c-3b=8

14c-12b=7

2(7c-3b=8)

14c-12b=7

14c-6b=16

-(14c-12b=7)

6b=9

b=9/6

b=3/2

Put b=3/2 in equation 1 or 2

7c-3(3/2)=8

7c-9/2=8

Take L.C.M

14c/2-9/2=8

14c-9=16

14c=25

c=25/14

a=3c-2b

a=3(25/14)-2(3/2)

a=75/14-3

a=75/3-42/3

a=33/3

a=11

If i did any step wrong then i'm sorry!

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