3a-2b=12
2a+b=1
multiply eq 2 x 2
4a+2b=2
eliminate b by subtracting eq (1) from (3)
4a+2b=2
3a-2b=12
a=-10 am i going right so far??
Yes, that looks correct.
Continue by substituting in a=-10 into one of the equations to find b.
Let's go through each step and check if you're going in the right direction.
1. The given system of equations is:
3a - 2b = 12 ...(Equation 1)
2a + b = 1 ...(Equation 2)
2. To eliminate the variable "b", we can multiply Equation 2 by 2:
2(2a + b) = 2(1)
4a + 2b = 2 ...(Equation 3)
3. To eliminate "b" from Equations 1 and 3, we can subtract Equation 1 from Equation 3:
(4a + 2b) - (3a - 2b) = 2 - 12
4a + 2b - 3a + 2b = -10
a + 4b = -10 ...(Equation 4)
4. However, notice that we still haven't eliminated "b" entirely. To do so, we need to eliminate "a" as well.
To solve the system of equations, we need to simultaneously solve Equations 4 and 2. We can do this by using the method of substitution or the method of elimination.
Let's use the method of substitution:
We have Equation 4 as:
a + 4b = -10
From Equation 2, we can express "a" in terms of "b":
2a + b = 1
2a = 1 - b
a = (1 - b)/2
Now, substitute this value of "a" into Equation 4:
(1 - b)/2 + 4b = -10
Expanding and solving the equation:
(1 - b + 8b)/2 = -10
(1 + 7b)/2 = -10
Multiply both sides by 2 to eliminate the denominator:
1 + 7b = -20
7b = -21
b = -3
Now substitute the value of "b" back into Equation 2 to find "a":
2a + (-3) = 1
2a - 3 = 1
2a = 4
a = 2
Therefore, the solution to the system of equations is a = 2 and b = -3.
Hence, after checking the steps, we can conclude that you are going in the right direction. The solution is a = 2 and b = -3.