2 variable systems. Solve using any method

Question

x-3y=-9

5x-3y=3

Answer 1

The other tutor mentioned solving by using elimination or by substitution. There are many other ways to solve systems of linear equations such as these. However, elimination and substution are the most commonly used in algebra classes.

You can always solve this type of equations with either method. Picking the right method can make the solution process much easier. That's why you should know both methods: so that you will be able to pick the best method for the problem.

And that's exactly what the other tutor did. Substution would have been more time consuming in this case.

AND one more point: always check your final answer. It is very easy to make a mistake with all these calculations.

Go back to the original system:

x-3y=-9

5x-3y=3

Plug in (substitute) the soluion values for x and y, respectively:

(3) - 3(4) = 3 -12 = -9

So the solution found solves that equation.

5(3) -3(4) = 15 - 12 = 3

So that solution checks out as well.

The solution found must work in both equations, or it is not a valid solution.

We are done. We have found the correct solution and checked it.

Answer 2

There are two methods for solving systems of equations: substitution and elimination.

For this problem, I see that we have a 3y in both equations, this means it will be simple to get these y's to cancel out, we just need to make one of them positive and one of them negative.

x-3y=-9 equation (1)

5x-3y=3 equation (2)

Let's multiply equation (1) by -1

-x+3y=9 equation (1)

5x-3y=3 equation (2)

Now let's add the two equations together:

-x + 3y = 9 equation (1)

+ 5x - 3y = 3 equation (2)

4x + 0 = 12

Now we solve for x by dividing both sides by 4

x = 3

Now we plug in x=3 into one of our equations (doesn't matter which one) - how about equation (1)

-3 + 3y = 9

3y = 12

y = 4

Our solution is the point (3, 4)