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.