To solve a system, there are two ways, substitution and elimination.
Today I will talk about substitution.
Let’s take two equations, 8x – y = 10 and 4x + y = 14
To find the solution, we can tackle this two different ways, we will start with substitution.
To find a solution using substitution, you first want to isolate a variable, one with preferably no coefficients.
For this example, we will be putting the y by itself in the second equation (4x + y = 14)
Moving the 4x to the other side will give you: y = -4x + 14
Now you take the other equation and now you know that y = -4x + 14, and put it in the equation: 8x -1(-4x + 14) = 10
Next you solve for the equation: 8x +4x -14 = 10
12x -14 + 14 = 10 + 14
12x = 24
12x/12 = 24/12
x = 2
Now to find y, you put x into the other equation (y = -4x + 14)
Now it equals, y = -4(2) + 14
y = -8 + 14
y = 6
Now your answer is x = 2 and y = 6
If you wish to check your answer, you just put the x and y values into one of the equations.
That’s all for this week, next week we will be talking about… probably elimination, no guarantees.