Substitution can be done any time we know the value of a variable. If all variables are substituted out, this turns the expression into an equation (as a side note, anytime an algebraic expression is equal to some value, that becomes an algebraic equation).
Let’s return to our original example and provide parameters to the variable.
20 x 2 + 4 x + 6 x − 10 {"version":"1.1","math":"20x^2 + 4x + 6x - 10"}
What is the result?
Try it for yourself, then check the box below for the answer.
Where x = − 1 2 {"version":"1.1","math":"x=-\frac{1}{2}"}
∴ 20 x 2 + 4 x + 6 x − 10 {"version":"1.1","math":"\therefore 20x^2 + 4x + 6x - 10"}
= 20 x 2 + 10 x − 10 {"version":"1.1","math":"=20x^2 + 10x - 10"}
= 20 ( − 1 2 ) 2 + 10 ( − 1 2 ) − 10 {"version":"1.1","math":"= 20\left(-\dfrac{1}{2}\right)^2 + 10\left(-\dfrac{1}{2}\right)-10"}
= 5 − 5 − 10 {"version":"1.1","math":"=5-5-10"}
= − 10 {"version":"1.1","math":"=-10"}
Remember to apply BEDMAS accordingly (Refer to the Order of Operations section in Module 1: Numbers and Whole Numbers).
Substitution is not, however, only limited to knowing the values of all variables and solving an algebraic equation. You may encounter algebraic equations where the value of the equation and values of all variables are known apart from one. From there, you’ll be ask to solve for the unknown variable.
