Solve the system of equations using the Addition method?

Question

2a + 3b = -1
3a + 5b = -2

Answer 1

2a + 3b = - 1- - - - - -Equation 1
3a + 5b = - 2- - - - - -Equation 2
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Multiply equation 1 by3 and equation by - 2

2a + 3b = - 1

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

6a + 9b = - 3...New equation 1
- - - - - - -

3a + 5b = - 2

- 2(3a) + (- 2)(5b) = - 2(- 2)

- 6a + (- 10b) = 4

- 6a - 10b = 4. . .New equation 2
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New equation 1 and New equation 2

Elimination of a,

6a + 9b = - 3
- 6a - 10b = 4
- - - - - - - - - -
- b = 1

- 1(- b) = - 1(1)

b = - 1

The answer b = - 1

Insert the b value into equation 1
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Solving for a

2a + 3b = - 1

2a + 3(- 1) = - 1

2a + (- 3) = - 1

2a - 3 = - 1

2a - 3 + 3 = - 1 + 2

2a = 2

2a/2 = 2/2

a = 1

The answer is a = 1

Insert the a value into equation 1

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Check for equation 1

2a + 3b = - 1

2(1) + 3(- 1) = - 1

2 + (-3) = - 1

2 - 3 = - 1

- 1 = - 1

- - - - - - - - - - -

Check for equation 2

3a + 5b = - 2

3(1) + 5(- 1) = - 2

3 + (- 5) = - 2

3 - 5 = - 2

- 2 = - 2

- - - - - - - - -s-

Answer 2

2a + 3b = -1 multiply by -3
-6a-9b=3
3a + 5b = -2 multiply by 2
6a+10b=-4
-6a-9b=3 add
b=-1
substitute
2a + 3b = -1
2a-3=-1
2a=2
a=1
solution is (1,-1)

Answer 3

You multiply one equation by a constant and the other equation by another constant and add them so as to cancel one of the variables, then solve.

e.g.:

5x+4y=8
2x+5y=9
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10x+8y=16 multiply top by 2
-10x-25y=-45 multiply bottom by -5
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-17y=-29 then add them
y=29/17 and solve
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10x+232/17=16 substitute y into one of the equations
10x=40/17
x=4/17
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solution:(4/17,29/17)

Answer 4

multiply first eq. by 5. second eq by -3 and add them-
10 a + 15 b = -5
-9a -15 b = 6
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a = 1

put 'a' in any eq. to get b= -1