### Theory:

Algebraic expressions are used to find out the unknown values. We assign some values to those expressions to find out the values of the expression. We can see these things in our daily life.
Here we see an example to understand clearly.

You went to a shop to buy chocolates and biscuits. The total amount you have in your hand is $$100$$, and you spent all the amount on buying these two. If the cost of the biscuits is 30, then can you find out the cost of the chocolates?

Let's solve this problem!

Let the cost of the chocolates is $$x$$, and the cost of the biscuits is $$y$$, then the total amount is $$100$$.

If you write an algebraic equation with these data we get,

$\begin{array}{l}x\phantom{\rule{0.147em}{0ex}}+y\phantom{\rule{0.147em}{0ex}}=100\\ \\ \mathit{Substitute}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{knwon}\phantom{\rule{0.147em}{0ex}}\mathit{values}\phantom{\rule{0.147em}{0ex}}\mathit{we}\phantom{\rule{0.147em}{0ex}}\mathit{get}\\ \\ x\phantom{\rule{0.147em}{0ex}}+30\phantom{\rule{0.147em}{0ex}}=100\\ \\ x\phantom{\rule{0.147em}{0ex}}=100\phantom{\rule{0.147em}{0ex}}-30\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}70\\ \\ \mathit{The}\phantom{\rule{0.147em}{0ex}}\mathit{cost}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{chocolates}\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}70\end{array}$

Here we know the value of $$y$$ and total value, so we easily find out the value of $$x$$.

What you will do to find the total value of the algebraic expression if the value of the variables $$x, y, z, a, b, m, n,$$ etc are given.
Let's learn a few steps to solve these sorts of problems.

Step 1:
Understand the problem first and then fix the variable and write the algebraic expression.

Step 2:
Substitute each variable by the given numerical value to get an arithmetical expression.

Step 3:
Try to solve or simplify the arithmetical expression by BIDMAS method.

Step 4:
Now the final value you obtained is the required value of the expression.
[Note: BIDMAS is the rule for the arithmetic operation that explains in what order the arithmetic operations of a mathematical expression should be performed].

Let's see some examples to find the value of the expression clearly.

If $$x =$$ 52$$y =$$ 30 then find the values of

I) $x\phantom{\rule{0.147em}{0ex}}+y$

II) $\mathit{2x}\phantom{\rule{0.147em}{0ex}}-\mathit{2y}$.

III) $\mathit{3}{x}^{2}\phantom{\rule{0.147em}{0ex}}-\mathit{2}{y}^{2}$

Now we can solve by just substituting the given values of $$x$$ and $$y$$ in the given expression, and we have to do the arithmetic expression like addition, subtraction, multiplication according to the expression.

Question 1)

$\begin{array}{l}=x\phantom{\rule{0.147em}{0ex}}+y\\ \\ \mathit{Now}\phantom{\rule{0.147em}{0ex}}\mathit{subtitute}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{known}\phantom{\rule{0.147em}{0ex}}\mathit{values}\phantom{\rule{0.147em}{0ex}}\mathit{we}\phantom{\rule{0.147em}{0ex}}\mathit{get}\\ \\ =52+30\\ \\ =82\end{array}$

The value of this $x\phantom{\rule{0.147em}{0ex}}+y$ algebraic expression is 82.

Question 2)

$\begin{array}{l}=\mathit{2x}\phantom{\rule{0.147em}{0ex}}-\mathit{2y}\\ \\ \mathit{Now}\phantom{\rule{0.147em}{0ex}}\mathit{subtitute}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{values}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}\mathit{and}\phantom{\rule{0.147em}{0ex}}y\phantom{\rule{0.147em}{0ex}}\mathit{we}\phantom{\rule{0.147em}{0ex}}\mathit{get}\\ \\ =2\left(52\right)\phantom{\rule{0.147em}{0ex}}-2\left(30\right)\\ \\ =104\phantom{\rule{0.147em}{0ex}}-60\\ \\ =44\end{array}$

The value of this $\mathit{2x}\phantom{\rule{0.147em}{0ex}}-\mathit{2y}$ algebraic expression is 44.

Question 3)

$\begin{array}{l}=\mathit{3}{x}^{2}\phantom{\rule{0.147em}{0ex}}-\mathit{2}{y}^{2}\\ \\ \mathit{Now}\phantom{\rule{0.147em}{0ex}}\mathit{substitute}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{values}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}\mathit{and}\phantom{\rule{0.147em}{0ex}}y\phantom{\rule{0.147em}{0ex}}\mathit{we}\phantom{\rule{0.147em}{0ex}}\mathit{get}\\ \\ =3×\left({14}^{2}\right)\phantom{\rule{0.147em}{0ex}}-2×\left({3}^{2}\right)\\ \\ =3×\left(196\right)\phantom{\rule{0.147em}{0ex}}-2×\left(9\right)\\ \\ =588-18\\ \\ =570\end{array}$

The value of this  $\mathit{3}{x}^{2}\phantom{\rule{0.147em}{0ex}}-\mathit{2}{y}^{2}$ algebraic expression is 570.

So these are all the steps to find out the value of algebraic expressions if the value of $$x$$ and $$y$$ are known.