Solving Simultaneous Linear Equations
This lesson covers how to solve simultaneous linear equations using substitution and elimination methods. It also explains how to identify linear equations and includes examples for practice.
- 01:00 “I have two equations on the board, X plus 4 equals to 7, X squared plus 4 equals to 7. The power of the unknown here, that is X, is 1. Any time you have an equation, which asks a power of 1. So, we'll call such equation...”
- 01:51 “Example 1. Use the substitution method to solve the following PS of simultaneous equations. A. Q plus 2R equals to 8. R minus Q over 2 plus 1 equals to 0. I'm to use substitution. Don't forget I told you, we are going to...”
- 13:06 “The value of y is 5. Put that into the equation. Then that will give us 23 minus 15. 23 minus 15 will give you what? It. So with this, you have been able to substitute using substitution method, the value of x and the va...”
Simplification of Algebraic Expressions
This lesson covers the definition of algebraic expressions and their components, including terms and coefficients. It explains how to identify like terms, perform arithmetic operations on similar terms, and simplify expressions by inserting or removing brackets.
- 20:52 “Like it says here. Multiply 3 by the second term, like I explained earlier, and you get 3b. At the end of the day, you get 6a plus 3b. More examples. You can solve this on your own. I'll just solve just one of it. Expand...”
- 25:31 “is the next one, or division. In the expression, you do multiplication next. So you use the distributive law to multiply the brackets by its coefficient. Remember that 4 times 3x means 4 times 3 times x. 16x plus 2 plus...”
- 16:03 “8x plus the tin y minus x minus 4y. Just as I did earlier, you rearrange, gathering those ones with x together, 8x minus x plus 13y minus 4y. So 8x minus x. 8x is minus 1x. Remember, there's an invisible one. So you have...”