Elimination method in simontanious equation ......solve 2a +3b=24 3a +2b=26?

To solve the simultaneous equations using the elimination method, we can multiply the first equation by 2 and the second equation by 3 to make the coefficients of either 'a' or 'b' the same in both equations.

Multiplying the first equation by 2:
4a + 6b = 48

Multiplying the second equation by 3:
9a + 6b = 78

Now, we can subtract the first equation from the second equation to eliminate 'b':
(9a + 6b) - (4a + 6b) = 78 - 48
9a + 6b - 4a - 6b = 30
5a = 30
a = 30 / 5
a = 6

Now that we have found the value of 'a', we can substitute it back into one of the original equations to find the value of 'b'. Let's use the first equation:
2(6) + 3b = 24
12 + 3b = 24
3b = 24 - 12
3b = 12
b = 12 / 3
b = 4

Therefore, the solution to the simultaneous equations 2a + 3b = 24 and 3a + 2b = 26 is a = 6 and b = 4.

Step 1: Multiply the first equation by 2 and the second equation by 3 to make the coefficients of b the same.

- 2(2a + 3b) = 2(24)
- 3(3a + 2b) = 3(26)

Step 2: Simplify both equations.

- 4a + 6b = 48
- 9a + 6b = 78

Step 3: Subtract the first equation from the second equation to eliminate b.

- (9a + 6b) - (4a + 6b) = 78 - 48
- 5a = 30

Step 4: Solve for a.

- a = 30 / 5
- a = 6

Step 5: Substitute the value of a back into the first equation to solve for b.

- 2(6) + 3b = 24
- 12 + 3b = 24
- 3b = 12
- b = 4

Therefore, the solution to the system of equations is a = 6 and b = 4.