Lesson Video: Solving Linear Equations over the Real Numbers | Nagwa Lesson Video: Solving Linear Equations over the Real Numbers | Nagwa

Lesson Video: Solving Linear Equations over the Real Numbers Mathematics • Second Year of Preparatory School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

In this video, we will learn how to solve linear equations over the real numbers and how to represent their solutions on a number line.

14:02

Video Transcript

In this video, we will learn how to solve linear equations over the real numbers and represent their solutions on a number line. We will be focusing on linear equations primarily of the form 𝑎𝑥 plus 𝑏 equals 𝑐, where 𝑎, 𝑏, and 𝑐 are real constants and 𝑎 is nonzero.

We recall that when solving equations, we perform the same operations to both sides of the equation so that the equation remains balanced. These operations will be the inverses of the operations performed when forming the equation. As such, for equations of the form 𝑎𝑥 plus 𝑏 equals 𝑐, 𝑥 will be equal to 𝑐 minus 𝑏 divided by 𝑎. To solve the two-step equation, we first need to isolate the 𝑥-term and then divide by its coefficient. This is equivalent to adding the additive inverse of 𝑏 to both sides and then multiplying both sides by the multiplicative inverse of 𝑎. We will recap this formal method in our first example.

Find, in the set of real numbers, the solution set of the equation two 𝑥 plus three is equal to five. Which of the following represents the solution of the equation on a number line? Is it option (A), (B), (C), (D), or (E)?

The first part of this question wants us to solve a linear equation in the form 𝑎𝑥 plus 𝑏 is equal to 𝑐, where 𝑎, 𝑏, and 𝑐 are constants. In order to solve the equation two 𝑥 plus three equals five, we begin by subtracting three from both sides. And as five minus three is equal to two, we are left with two 𝑥 equals two. Our next step is to divide through by the coefficient of 𝑥, in this case two. As two 𝑥 divided by two is 𝑥 and two divided by two is one, we have 𝑥 is equal to one. The solution set of the equation two 𝑥 plus three equals five contains the solitary element one.

The second part of this question asks us to identify the solution to the equation on a number line. Option (E) represents the solution 𝑥 is equal to eight. Option (D) is the solution 𝑥 is equal to three. Option (C) is the solution 𝑥 is equal to four. And option (B) represents the solution 𝑥 is equal to two. Since the solution to our equation was 𝑥 is equal to one, none of these options are correct. The number line that represents the solution of the equation two 𝑥 plus three equals five is option (A). We have a solid dot representing the single value one.

In this first example, solving the equation was reasonably straightforward as the coefficients and constants involved were all integers. We will now consider more complicated equations which involve coefficients or constant terms which are radicals or surds. These will sometimes lead to solutions themselves which involve radicals.

Which of the following is the solution set of the equation two 𝑥 plus two root three equals two in the real numbers? Is it (A) one plus root three, (B) one minus root three, (C) two plus root three, (D) two minus root three, or (E) four minus root three?

In this question, we need to solve a linear equation in the form 𝑎𝑥 plus 𝑏 equals 𝑐, where 𝑎, 𝑏, and 𝑐 are constants. We solve the two-step equation two 𝑥 plus two root three equals two by firstly isolating the 𝑥-term. To do this, we subtract two root three from both sides. This is the same as adding the additive inverse of 𝑏 to both sides of the general equation. When we do this, the left-hand side becomes two 𝑥. And on the right-hand side, we have two minus two root three.

Our next step is to divide both sides of the equation by two. This is the coefficient of 𝑥. On the left-hand side, the twos cancel, leaving us with 𝑥. And on the right-hand side, we are left with one minus root three. We can therefore conclude that the solution set of the equation two 𝑥 plus two root three equals two is option (B) one minus root three.

In our next example, the coefficient of the unknown variable 𝑥 is a radical. As such, we will need to divide by this radical, which will lead to an answer involving an irrational denominator. We will then need to recall the process of rationalizing the denominator of a fraction.

Find, in the set of real numbers, the solution set of the equation root three 𝑥 plus two is equal to five. Which of the following shows the solution of the equation on a number line? Is it option (A), option (B), option (C), option (D), or option (E)?

The first part of this question involves solving a linear equation of the form 𝑎𝑥 plus 𝑏 equals 𝑐, where 𝑎, 𝑏, and 𝑐 are nonzero constants. To solve the equation root three 𝑥 plus two is equal to five, we begin by isolating the 𝑥-term. We do this by subtracting two from both sides. Root three 𝑥 is therefore equal to three. Next, we divide through by root three. On the left-hand side, the root threes cancel and we are left with 𝑥. Our expression on the right-hand side, three over root three, is a fraction, and its denominator is irrational. We therefore need to rationalize the denominator by multiplying both the numerator and denominator by root three. And since root three multiplied by root three is three, we have 𝑥 is equal to three root three over three. And this simplifies to 𝑥 is equal to root three.

The solution set of the equation root three 𝑥 plus two equals five is the single value root three.

The second part of this question asks us to identify the number line that represents this solution. Since the square root of one is equal to one and the square root of four is equal to two, we know that the square root of three must lie between one and two. This means that we can rule out options (B), (C), (D), and (E), as (B) has a solution between three and four, (C) and (D) have a solution between zero and one, and (E) has solution 𝑥 is equal to three. The correct number line is therefore option (A). Whilst it is not required in this question, it is worth noting that the square root of three is equal to 1.732 and so on. And since the solid dot lies closer to two than one, this backs up the answer of option (A).

Before looking at one final example, we will summarize the steps involved in solving two-step linear equations. In order to solve two-step linear equations of the form 𝑎𝑥 plus 𝑏 equals 𝑐, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is nonzero, we follow the following steps. Firstly, we isolate the unknown variable by adding the additive inverse of 𝑏 to both sides of the equation. Secondly, we divide both sides of the equation by the coefficient of the unknown variable. This is equivalent to multiplying both sides by the multiplicative inverse of 𝑎. Finally, if this leads to a fraction with an irrational denominator, we rationalize the denominator by multiplying both the numerator and denominator of the fraction by the radical in the denominator.

We will now consider one final example where we need to follow these three steps.

Find, in the set of real numbers, the solution set of the equation root five multiplied by root three 𝑥 minus two is equal to four root five.

There are a couple of approaches we could use to answer this question. For example, we could begin by distributing the parentheses on the left-hand side. However, as the unknown variable 𝑥 appears inside the parentheses, it will be simpler to divide both sides of the equation by root five first. When we do this, we have root five multiplied by root three 𝑥 minus two all divided by root five is equal to four root five over root five. On both sides of the equation, the root fives cancel. And as such, our equation becomes root three 𝑥 minus two is equal to four. We can then add two to both sides of this equation to isolate the 𝑥-term. As four plus two is equal to six, we have root three 𝑥 equals six.

Next, we divide through by the coefficient of 𝑥, in this case root three. And 𝑥 is therefore equal to six over root three. Since the denominator of our fraction is a radical, we need to rationalize the denominator by multiplying the numerator and denominator by root three. Recalling that root three multiplied by root three is three, we have 𝑥 is equal to six root three over three, which in turn simplifies to 𝑥 is equal to two root three. The solution set of the equation root five multiplied by root three 𝑥 minus two is equal to four root five is the single value two root three. We could check this answer by substituting our value of 𝑥 back in to the original equation.

We will now finish this video by recapping the key points. When solving equations, we always perform the same operations to both sides. We saw that a two-step linear equation of the form 𝑎𝑥 plus 𝑏 equals 𝑐 can be solved by first isolating the unknown variable and then dividing by its coefficient. If we have a fractional answer with an irrational denominator, we need to rationalize this. And we do so by multiplying both the numerator and denominator of the fraction by the radical in the denominator. We then need to simplify this answer. Finally, we saw that the solution to a linear equation can be represented using a solid dot on a number line. In the case of irrational solutions, the position of the dot will be approximate.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy