Algebra 4/7 - 4/11

Algebra Scholars,

Welcome back! Hopefully many of you will continue to view the work that students create during class as a resource for learning how to solve systems. This week we will continue to prepare for the MCA test as well as learn more strategies for solving systems.

Have a great week! --Mr. Hanson

Geometry 4/7 - 4/11

Geo students,

We begin chapter 11:  Indirect and Coordinate Proofs.  We are also in MCA month.  We will be spending significant time preparing during our warm-ups for the MCA test. 

Have a great week.  --Mr. Hanson

Solving Linear Systems by Multiplication

Math Scholars, welcome back from Spring Break. We dig right back in to Linear Systems by solving them with multiplication. At this point, all of you have learned many methods to solve systems and multiplication is yet another strategy to add to your skill set.

Emma helped us out today and has clearly demonstrated how to use multiplication to arrange a linear system so it can be solved by addition. The goal is to multiply the entire equation by a number that will create a situation that will cancel either the 'x' or the 'y' when doing addition. If you have any question, be sure to ask during class. Welcome back!

Solving Systems with Addition!

Wow, just when you thought systems were the most difficult, complex problem you have ever solved...you got it! Check out this work by one of my students.

This student is demonstrating how to solve linear systems with addition. As long as there is an x in each system that is the opposite of one another (-8x & 8x), they will cancel and become zero. Therefore, we are then able to solve for y. Once we find y, then -3 is substituted back in for y to solve for x. What a great thing!

Remember, up to this point, this method will only work if you have a situation that allows the cancellation of one of the two variables.

Algebra & Geometry 3/31 - 4-4

Spring Break Folks. Enjoy!

More examples of using substitution to solve linear systems

Math Scholars, we are fortunate once again to have some some student work to post to the blog. In this post, you will see that it is not necessary to have both equations in a system in slope intercept form (y=mx+b) to use substitution.

Notice that in example (a), (2x-6) is substituted in for 'y' in the first equation. This allows us to solve for x. Once 'x' is found to be zero, we substitute zero back in for 'x' and solve for 'y'. What a beautiful thing, eh?

If you find that you are still having trouble, please ask me questions in class. There still isn't a better place to get help. :-)

Geometry 3/24 -3/28

Geometry scholars,

This week is the last week of quarter 3. Our test is Friday! If you know you will be gone, your test is on Wednesday as well all of your work and stamp sheet. Be sure you have everything done and ready to be stamped off on Wednesday.

Mr. Hanson