Week 12 – Solving systems of equations algebraically

This week in pre-calc, we learned about substitution and elimination. Using substitution or elimination can help us find the points of intersection(s) for lines or parabola’s if they intersect. Doing this gives us the co-ordinates of where they intersect and how many there are.

For 2 lines, there can only be 0, 1 or infinite. Parabola’s can have 0, 1, 2 or infinite. Parabola’s and lines can have 0, 1 or 2 solutions.

Using substitution is taking one of your two equations and getting one variable all by itself so you can input the solution for that variable and replace it with any other variable that has the same value.

Ex. x – 2y = -10

3x – y = 0

this becomes 3(2y-10) – y = 0

Then you want to gather like terms

6 y – 30 – y = 0

5y = 30

then you divide the ‘y’ value by it’s co-efficient on both sides to get the value of ‘y’

y=6

Now that you know your x-value, you can input this value into the first equation.

x – 2y = -10

x – 2(6) = 10

x – 12 = 10

x = 2

Now you know your co-ordinates for this system is (2,6)

Ex. x – 2y = -10

3x – y = 0

With elimination, you want to take your 2 equations, and make one variable equal to the other.

x – 2y = -10

6x -2y = 0

Now you can subtract the equations to make 1 equation.

-5x = -10

Once again, you just divide the ‘x’ term with its co-efficient on both sides and you get your ‘x’ value.

x= 2

Now you can input your ‘x’ value into the on of the equations and then you can simplify for ‘y’

6x -2y = 0

6(2) – 2y = 0

12 – 2y = 0

-2y = -12

y = 6

Co-ordinates = (2,6)

Both ways work for finding the point(s) of intersections it just depends on which method you prefer to use.