Title: 6. Simplifying algebraic expressions
Course: Demonstrating STACK (v4.4.7)

Simplify the following expression by collecting together like terms:
\[[value]x+[value]y-[value]x-[value]y+5.\]
Feedback: In order to collect like terms simply pull out the common factors (the x's and y's in this example) like so: \[ = [value]x+[value]y-[value]x-[value]y+5,\] \[ = \left([value]-[value]\right)x+\left([value]-[value]\right)y+5,\] \[ = [value].\]

Gather the like terms in follow expression:
\[[value]x-[value]y+[value]z-4+[value]x-[value]y+[value]z-3\] =
Feedback: To gather like terms first factorise out the common factors (in this case \(x\),\(y\) and \(z\)) and then simply carry out the arithmetic. Don't forget that gathering like terms also includes gathering constants (in this example \(4+3 = 7\)). \[[value]x-[value]y+[value]z-4+[value]x-[value]y+[value]z-3,\] \[ = \left([value]+[value]\right)x-\left([value]+[value]\right)y+\left([value]+[value]\right)z-7,\] \[ =[value].\]

Simplify the algebraic expression: \[[value].\] Be sure to leave your answer in the form \(x^y\).
Feedback: Here we can use the rules of indices, \(c^a+c^b=c^{a+b}\), which apply to algebraic expressions as well as numeric values.

Express \({ \left([value]^[value] \right)}^[value]\) in the form \(c^x.\)
Feedback: Remember the rule: \((c^a)^b=c^{ab}\). Therefore, in this example: \[{\left( [value]^[value] \right)}^[value] = [value].\]

\({\left([value]\right)}^0\) = . You may assume that \( [value] \neq 0 \).
Feedback: Remember that \(x^0=1\) for all \(x\neq0\). Therefore \({\left([value]\right)}^0=1\).

Write the follwing expression, \(\left( [value]\cdot [value] \right)^[value]\), without brackets.
Feedback: Remember that \[\left( {a}\cdot{b} \right)^n = {a^n}\cdot{b^n}. \] Therefore in this example: \[\left([value]\cdot [value] \right)^[value] = \left([value]^[value] \right)\cdot \left([value]^[value] \right) = [value]\cdot[value]. \]

Write the follwing expression, \(\left( [value]\cdot [value]\cdot [value] \right)^[value]\), without brackets.
Feedback: Remember that \[ \left({a}\cdot{b}\cdot{c} \right)^n = {a^n}\cdot{b^n}\cdot{c^n}. \] Therefore in this example: \[ \left([value]\cdot [value]\cdot [value] \right)^[value] = {[value]^[value]}\cdot{[value]^{[value]\cdot [value]}}\cdot{[value]^[value]} = [value]\cdot[value]\cdot[value]. \]