Multiplication of Algebraic Expressions - Examples, Exercises and Solutions

To solve the exercise, we will reorder the numbers using the substitution property.

$18x-8x+4x-7-9=$

To continue, let's remember an important rule:

1. It is impossible to add or subtract numbers with variables.

That is, we cannot subtract 7 from 8X, for example...

We solve according to the order of arithmetic operations, from left to right:

$18x-8x=10x$$10x+4x=14x$$-7-9=-16$Remember, these two numbers cannot be added or subtracted, so the result is:

$14x-16$

To solve this question, we need to remember that we can perform addition and subtraction operations when we have the same variable,
but we are limited when we have several different variables.
 

We can see in this exercise that we have three variables:
$45$ which has no variable
$8y$ and $34y$ which both have the variable $y$
and $45z$ with the variable $z$

Therefore, we can only operate with the y variable, since it's the only one that exists in more than one term.

Let's rearrange the exercise:

$45-34y+8y-45z$

Let's combine the relevant terms with $y$

$45-26y-45z$

We can see that this is similar to one of the other answers, with a small rearrangement of the terms:

$-26y+45-45z$

And since we have no possibility to perform additional operations - this is the solution!

It is important to remember that when we have numbers and variables, it is impossible to add or subtract them from each other.

We group the elements:

$7.3×4a + 2.3 + 8a =$

29.2a + 2.3 + 8a = 

$37.2a + 2.3$

And in this exercise, this is the solution!

You can continue looking for the value of a.

But in this case, there is no need.

Let's break down the fraction's numerator into an expression:

$8x^2=4\times2\times x\times x$

And now the expression will be:

$\frac{4\times2\times x\times x}{4x}+3x=$

Let's reduce and get:

$2x+3x=5x$

We'll use the law of exponents for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

We'll apply this law to the expression in the problem:

$2x^3\cdot x^2-3x\cdot x^4+6x\cdot x^2-7x^3\cdot 5=2x^{3+2}-3x^{1+4}+6x^{1+2}-35x^3$

When we apply the above law to the first three terms from the left, while remembering that any number can always be considered as that number raised to the power of 1:

$a=a^1$

And in the last term we performed the numerical multiplication,

We'll continue and simplify the expression we got in the last step:

$2x^{3+2}-3x^{1+4}+6x^{1+2}-35x^3=2x^5-3x^5+6x^3-35x^3=-x^5-29x^3$

Where in the first stage we simplified the expressions in the exponents of the terms in the expression and in the second stage we combined like terms,

Therefore the correct answer is answer A.

According to the laws of multiplication, we must first simplify everything into one exercise:

$\frac{9m\times3m}{3m^2\times6}=$

We will simplify and get:

$\frac{9m^2}{m^2\times6}=$

We will simplify and get:

$\frac{9}{6}=$

We will factor the expression into a multiplication:

$\frac{3\times3}{3\times2}=$

We will simplify and get:

$\frac{3}{2}=1.5$