Russell's Algebra II Website
OK, today I'll be telling you how to do three Algebra II level problems. I have listed how to do the problem below on the left side, and its answer on the handy table.
OK, today I'll be telling you how to do three Algebra II level problems. I have listed how to do the problem below on the left side, and its answer on the handy table.
PROBLEM 
SOLUTION
| 6(2+ Y) = 3(3 - Y)
12 + 6Y = 9 - 3Y 12 + 6Y + 3Y = 9 - 3Y + 3Y 12 + 9Y = 9 12 - 12 + 9Y = 9 - 12 9Y = -3 Y = -1/3 |
OK, for this problem first I got rid of the parentheses by use of the distributive property. Then I decided to "move" the variables to one side of the problem. So I added 3Y to both sides, which canceled out the Y on the right side of the equation. Then I isolated the variables on the left side by subtracting 12 from both sides. Finally, I removed the 9 from 9Y by dividing both sides of the equation, thus giving me -1/3 as the solution. |
| Give an equation for a line containing:
(-2,4) and having a Y-intercept of 8.
8-4/0 -(-2) = Slope 4/2 = Slope 2 = Slope Y = 2x + 8 |
This one seemed hard, but in actuality was quite easy. It gave the Y-intercept of 8, and a point. So in essence, you have been given two points to start [ (-2,4) (0,8) ]. So you then solve it with the formula y1-y2/x1-x2. That gives you the slope. By now you should know how to put it in the equation Y = mx + b, with the slope as m (french for montar, means to climb) and b as the Y-intercept. |
| Y = mx + b ; solve for X
Y - b = mx + b - b Y - b/m = mx/m Y - b/m = x |
This was perhaps the easiest of all. All you needed to do was carry over the variables without having to worry about mere arithmetic =) . I subtracted b from both sides, and then I divided both sides by m to get the solution. |
OK, this should help
you with your Algebra. If you need more problems then click the link below.
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to see more Algebra problems!