Solve by substitution:
6a - b = -5
4a - 3b = -8
Im trying to do my work, but idk where to start..
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So you want to be able to cancel out one of the variables and be left with just one variable in an equation. To do this, we can subtract equations. But, we need to make sure the variables will cancel out first.
We have 6a - b = -5 and 4a - 3b = -8
We can multiply the entire first equation by 3 so that b will have a coefficient or -3, just like in the second equation, so it will be easy to cancel out.
3(6a - b) = 3(-5)
18a-3b = -15
Now, we can subtract the two equations. We have (18a-3b=-15) - (4a-3b=-8)
First, we subtract the left side. 18a-4a=14a and -3b-(-3b)=0.
The left side is 14a+0, or just 14a.
The right side: -15-(-8) = -7
So, 14a = -7. Dividing by 14 on both sides, we get a=-1/2
You can now solve for b by plugging a back into one of the equations.
Please let me know if any part of this is confusing
6a - b = -5
... adding b and 5 ... 6a + 5 = b
... SUBSTITUTE this value for b into the second equation
4a - 3(6a + 5) = -8
... solve for a ... 4a - 18a - 15 = -8
substitute the value for a back into one of the original equations
... solve for b
To solve this system of equations by substitution, follow these steps:
Step 1: Choose one equation to solve for one variable in terms of the other variable. Let's choose the first equation:
6a - b = -5
Solve for 'b':
b = 6a + 5
Step 2: Substitute the expression you found for the variable in the other equation.
Replace 'b' in the second equation with '6a + 5':
4a - 3(6a + 5) = -8
Step 3: Simplify and solve for 'a':
4a - 18a - 15 = -8
-14a - 15 = -8
-14a = 7
a = 7 / -14
a = -1/2
Step 4: Substitute the value of 'a' back into one of the original equations. Let's use the first equation:
6a - b = -5
Replace 'a' with '-1/2':
6(-1/2) - b = -5
-3 - b = -5
Step 5: Solve for 'b':
-b = -5 + 3
-b = -2
b = 2
So the solution to the system of equations is a = -1/2 and b = 2.
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