Solve the system by the elimination method. check your work. 2a + 3b = 6 5a + 2b - 4 = 0 make sure there are no spaces in your answer. include a comma in your answer. {( a0 )}

From the computation, a will be 0 and b will be 2.

The equations given are:

2a + 3b = 6

5a + 2b = 4

Multiply equation i by 2

Multiply equation ii by 3

4a + 6b = 12

15a + 6b = 12

Subtract the equations

a = 0

Since 2a + 3b = 6

2(0) + 3b = 6

3b = 6

b = 6/3 = 2

Therefore, a = 0, b = 2.

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2a + 3b = 6multiply by 2
5a + 2b = 4 ...multiply by -3

4a + 6b = 12 (result of multiplying by 2)
-15a - 6b = -12 (result of multiplying by -3)
add
- 11a = 0
a = 0

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

solution is (0,2) < :)

check...
2a + 3b = 6                    5a + 2b = 4
2(0) + 3(2) = 6               5(0) + 2(2) = 4
6 = 6 (correct)                4 = 4 (correct)

The answer to this is D