Solve the system by the elimination method.?

Question

{2a + 3b=6}
{5a + 2b - 4=0}

Answer 1

2a + 3b = 6- - - - - - - - -Equation 1
5a + 2b - 4 = 0- - - - - - -Equation 2
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Convert equation 2 to standard form

5a + 2b - 4 = 0

5a + 2b - 4 + 4 = 0 + 4

5a + 2b = 4

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2a + 3b = 6
5a + 2b = 4
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Multiply equqation 1 by 2 and equation 2 by - 3

2a + 3b = 6

2(2a) + 2(3b) = 2(6)

4a + 6b = 12
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5a + 2b = 4

- 3(5a) + (-3)(3a) = - 3(4)

- 15a+ (- 6b = - 12

- 15a - 6b = - 12

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4a + 6b = 12
15a - 6b = - 12
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19a = 0

19a / 19 = 0 / 19

a = 0 / 19

a = 0

Insert the a value into equation 1
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2a + 3b = 6

2(0) + 3b = 6

0 + 3b = 6

3b = 6

3b / 3 = 6 / 3

b = 6 / 3

b = 2

Insert the b value into equation 1

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Check for equation 1

2a + 3b = 6

2(0) + 3(2) = 6

0 + 6 = 6

6 = 6

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Check for equation 2

5a + 2b - 4 = 0

5(0) + 2(2) - 4 = 0

0 + 4 - 4 = 0

0 = 0

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Both equations balance

a = 0

B = 2

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Answer 2

{2a + 3b=6}
{5a + 2b - 4=0}

The elimination method is also the combination or addition method...

2a + 3b = 6
5a + 2b = 4 ... moved the 4 to right side

(2a + 3b = 6)(-2)
(5a + 2b = 4)(3) .... multiplying each equation so that we can eliminate the b term

-4a - 6b = -12
15a + 6b = 12 ...... now add the two together

11a = 0
a = 0 ...... substitute a=0 into either equation to solve for
2(0) + 3b = 6
3b = 6
b = 2

a = 0
b = 2

plug into both equations to check

2(0) + 3(2) = 6.... 6 = 6 ... check
5(0) + 2(2) - 4 = 0 ..... 4 - 4 = 0 .... check

Answer 3

you can simultaneously solve the equation,

but the elimination method goes as follows,

a = (6-3b)/2 (eliminating a)

5((6-3b/2)) +2b - 4 = 0

therefore b = 2 and a = 0

Answer 4

Write each equation in standard form

2a + 3b = 6
5a + 2b = 4

Suppose we choose to eliminate the variable b. Write both equations so that the coefficients of b are opposites.
Multiply the first equation by -2 and the second by 3.

-4a - 6b = -12
15a + 6b = 12

Add the eqautions:

9a = 0
a = 0

To solve for b, you may substitute a = 0 into either of the original equations. Suppose we choose the first equation.

2(0) + 3b = 6
0 + 3b = 6
3b = 6
b = 2

So, the solution for this system is (a, b) = (0, 2).

Answer 5

Very simple.1)Firstly write 2a + 3b=6. 2)Then u shift the 2a to the RHS of the equation.U should get 3b=6-2a. 3)Then divide 3 throughout the equation.U should get:b=(6/3)-(2/3)a.From here u can substitute the value of b which is b=(6/3)-(2/3)a into the second equation. 4)After subsituting the value u should get this:5a+[2(6/3)-(2/3)a]-4=0. 5)Then u simplify the equation.U should get this:5a+4-(4/3)a -4=0. 6)Then write this:5a+4-4-(4/3)a=0 7)Then simplify it to get this:5a-(4/3)a=0. 8)Then solve for a.U should get 0 for a.From here u can find b by subsituting 0 to every a into any equation here.