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Introduction
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15 resolved linear equations step by step
In mathematics, linear equations are the introduction to algebra. Their understanding is vital for any kind of equations: quadratic equations or bigger, exponentials, irrational, etc. and for equation systems.
In real life, although it may not be though of straight away, but equations are a very useful tool that allows us to resolve problems that we face on a day to day basis. We can see this by going to the linear equations problems section.
These equations are known as linear, because the monomial literal part does not have an exponent (for example, \(3x\) can be part of a linear equation, but \(3x^2\) cannot because it is quadratic), so represented in a chart appears as a straight line. This fact assures us that if there is a solution, there can only be one solution (except in special cases where there are infinite solutions).
We say if there is a solution because sometimes equations don not any solution. For example, the equation \(x = x + 1\) (which means a number equals the consecutive number) does not have a solution, because this is never true. Actually, this equation is reduced to 1 = 0, which is impossible.
In this section we resolve linear equations with an increasing difficulty: simple equations, equations with fractions (where we will use the least common multiple of the denominators), equations with parenthesis and equations with nested parenthesis (parenthesis within parenthesis).
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If we obtain an impossible equality, there is no solution, like 1 = 0.
If we obtain an equality that is always true, whatever value gives a solution. Hence the solution is all the real numbers. For example, if we obtain 0 = 0.
When there are denominators and we want to avoid them, we multiply the full equation by the same least common multiple of the denominators. Hence the denominators disappear.
To remove the parenthesis, we multiply the coefficient previous to the parenthesis by all the elements it contains. This coefficient can be a negative sign (like –1, the content changes sign), a positive sign (like +1, the content does not change) or a positive or negative number or fractions (this number multiplies everything in the parentheses, changing the signs whenever it is negative).
When we have a nested parenthesis, a parenthesis inside another parenthesis, we begging removing from outside to the inside. First, we remove the exterior parenthesis (multiplying their content by the coefficients) and afterwards, we remove the rest the same way: from the most exterior to the most interior.
Realistically, there is not any need to follow an order when removing the parentheses, but it is recommended to follow one when we are learning.
Equation 1
We add the monomials on the literal part (the x’s with x’s and numbers with numbers). What is adding on one side will change to the other side subtracting and vice versa.
Afterwards, we put the x’s to one side of the equality and the numbers to the other.
Then, the solution is \( x = 2 \).
Equation 2
Because the x has a coefficient (10) that is multiplying, it moves to the other side dividing.
Then, the solution is \( x = 1/10 \).
Equation 3
First, we remove the parenthesis: since it has a negative sign in front of it, we change the sign of all the elements in the inside.
Afterwards, all we need is to group the x’s at one side and the numbers at the other.
As the x has a coefficient, 2, that is multiplying, it moves to the other side dividing.
Equation 4
First we remove the parentheses: the one on the right has a negative sign, so it changes the sign of all the elements inside it; the one on the right has a 3 multiplying it, so it multiplies every element inside it.
Since the x has a coefficient, 32, that is multiplying, it moves to the other side dividing.
Equation 5
We have fractions. We can proceed in different ways:
multiply all the elements of the equation by the least common multiple of the denominators
or go multiplying by each denominator
We are going to multiply everything by the least common multiple of the denominators, which is 6.
Be careful with the parenthesis:
We resolve the equation:
We have 0 = 2 which is a false mathematical equality. Therefore the equation does not have a solution because whatever the value of x is, we obtain a contradiction (a false mathematical equation).
Equation 6
First we remove the parentheses: the one on the right has a 3 multiplying it, so it multiplies every element inside it (changing the signs); the one on the right has a 2 multiplying it:
Then, the solution is \( x = 3 \).
Equation 7
We are going to multiply the whole equation by the least common multiple of the denominators (3 and 2), which is 6:
We resolve the equation:
Then, the solution is \( x = 11/10 \).
Equation 8
We are going to multiply the equation by the least common multiple of the denominators (5, 2 and 3), which is 30:
We only have one parenthesis that is multiplied by 15. To remove it, we multiply the content by 15:
Therefore the solution is \( x = 23/30 \).
Equation 9
In this equation, we have a nested parenthesis (parenthesis within a parenthesis) and it is multiplied by fractions. But before we worry about this, we will multiply all the equation by the least common multiple of the denominators, 6:
Now we are going to the parentheses: on the left there are two, but we will treat it like one. We will multiply all the content of the exterior by –2. At the same time, on the right, we multiply all the content by 9.
We are left with one parenthesis, that is multiplied by 6.
Then, the solution is \( x = 11/5 \).
Equation 10
Since we have nested parenthesis (one within another), we are going to start removing them. The first parenthesis (the exterior one) is multiplied by –2. To remove it, we multiply all the content of the parenthesis by 2:
Now, the parenthesis on the exterior is multiplied by 6. To remove it we multiply the content by 6:
Finally, the parenthesis that is left is multiplied by –12, so to remove it we multiply the content by –12:
Now we are going to remove the fractions, but before this, we add some of the elements we have to reduce the length of.
We multiply the whole equation by the least common multiple of the denominators, which is 12:
Therefore the solution is \( x = 79/33 \).
Equation 11
We only have one parenthesis that is multiplied by 3. To remove it, we multiply the content by 3:
So, the solution is \( x = 11 \).
Equation 12
To remove the fractions, we multiply the whole equation by the least common multiple of the denominators, which is 6:
We resolve the equation:
Then, the solution is \( x = 17/7 \).
Equation 13
To remove the fractions, we multiply the whole equation by the least common multiple of the denominators, which is
$$LCM(12, 3, 4, 2) = 12$$
The negative coefficient in front of the fraction multiplies the numerator (changing the signs).
Then, the solution is \( x = 39/50 \).
Equation 14
We multiply the whole equation by 3 and when we reduce, all fractions disappear:
We have reached a relation that is always true and it does not depend on x. This means that whatever the value of x, the equations is always going to be true. Therefore the solution is all the real numbers (infinite solutions):
$$ x \in \mathbb{R} $$
Equation 15
Therefore the solution is \( x = 212/169 \).
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