## Tuesday, June 21, 2011

### The Problem of the Week Day 2: Week 6/19 - 6/25

Here is the second part of our first problem of the week! Make sure you remember your answers from yesterday to plug into today's equations!

Easy Problem: To do this problem, you need to do some simple Algebra. Remember to do the opposite operation of what is there. For instance, if you see multiplication, you would do division to both sides to get rid of it. For example, if you had 6 = 2k (a letter with a number to its left without a symbol means multiplication), you would divide both sides of the equation by two to get 3 = k.

Step 1: Plug c into this equation: (2 x c^2) + 1 = ?
Step 2: Make the answer to that equal 17z
In other words, solve (2 x c^2) + 1 = 17z

z = ___

Hard Problem: Since the easy people get one equation to solve, we can handle two! This is a system of linear equations, where you have to solve for two variables, requiring a second equation. In order to solve it, you will multiply through one or both equations by a number that will make two coefficients the same number, but one negative and one positive. Then, you add both equations together, which eliminates a variable and allows you to use simple Algebra to determine the other. After finding that variable, plug its answer into the easiest of the original equations and solve for the other variable. Since the math will be difficult, you may use a calculator for today's work. You will be plugging yesterday's answers into this equation where you see the letter a or b.

-3ax + 156y = 519 - 6b
3bx + 48.2y = 6^3 + .9

x = ____
y = ____

Also, the problem of the week is only going to occur once every month. Yesterday, I said it would be weekly. My mistake. On the first day of the next problem of the week, I will inform you of who got the answer right.