Solve:
4a+9b=8
8a+6c=-1
6b+6c=-1
subtract the last two, that will give you an equation in only a and b
Then use that with the first equation
To solve this system of equations, we can use the method of substitution or elimination. Let's go with the elimination method.
Step 1: Multiply the first equation by 2, the second equation by 3, and the third equation by 2 to balance the coefficients of variable 'a'.
2(4a + 9b) = 2(8)
3(8a + 6c) = 3(-1)
2(6b + 6c) = 2(-1)
This simplifies to:
8a + 18b = 16
24a + 18c = -3
12b + 12c = -2
Step 2: Cancel out 'c' terms by multiplying the third equation by 3.
8a + 18b = 16
24a + 18c = -3
36b + 36c = -6
Step 3: We now have two equations with only 'a' and 'b' terms. Multiply the first equation by -3 and the second equation by 1 to equalize the 'a' terms.
-3(8a + 18b) = -3(16)
24a + 18c = -3
-24a - 54b = -48
24a + 18c = -3
This simplifies to:
-54b = -51
Step 4: Solve for 'b'.
Divide both sides of the equation by -54:
b = -51 / -54
b = 17 / 18
Step 5: Substitute the value of 'b' back into the original second equation to solve for 'c'.
8a + 6c = -1
8a + 6(c) = -1
8a + 6(-2/3) = -1
8a - 4 = -1
8a = -1 + 4
8a = 3
a = 3 / 8
Step 6: Substitute the values of 'a' and 'b' into the first equation to solve for 'c'.
4a + 9b = 8
4(3/8) + 9(17/18) = 8
3/2 + 17/2 = 8
20/2 = 8
10 = 8
Since the equation 10 = 8 is false, it means there is no solution to this system of equations.