Explainer: Two-Step Equations

In this explainer, we will learn how to solve two-step equations.

Equations are like an enigmaβ€”they give us information about an unknown number, denoted with a letter (often π‘₯) that we can use to find the value of this number. A simple equation tells us that when the unknown number has undergone some operations, it led to a certain result. The equation can be solved by reversing all the operations and in the contrary order, to finally find the value of the starting number. More complex equations cannot be solved in that way; however, we are going to derive with this first approach a general method that can be applied to solve any equation.

Let us first imagine that when a given number is doubled and the result is added to 5, we get 19. Let π‘₯ be this unknown number. The corresponding equation is 2π‘₯+5=19. How can we find the value of this unknown number? We can use diverse models to visualize what is happening. First, we can use a bar diagram.

The bar diagram suggests that if we can split the 19 in 19βˆ’5 and 5, we find that 2π‘₯=19βˆ’5.

This shows us that the first step to solve the equation is to subtract 5 from both its sides. We find then that 2π‘₯=19βˆ’5=14. The second step is to split both sides in two, that is, dividing by two.

We find that π‘₯=142=7.

We can also write the operations that have been performed on π‘₯ as given in the question.

And, from there, we simply go the other way around and perform the reverse operations on 19 to get to the value of π‘₯.

Both models show us that the first step to solve the equation 2π‘₯+5=19 is to subtract 5 from each side, which can be formally written as

that is, 2π‘₯=14.

The second step is to divide both sides by two:

which gives the solution of π‘₯=7.

We have illustrated here a general method that will be applied to solve all equations, not only the type that we have just taken as example. This method is called the balance method and is based on the additive and multiplicative properties of equalities.

Balance Method: Additive and Multiplicative Properties of Equalities

The balance method is a general method for solving equations based on the additive and multiplicative properties of equalities. These properties are very intuitive. The first states that if the same amount is added to (or subtracted from) both sides of the equality, then the equality is still true. The second states that if both sides of the equation are multiplied by the same number (or both sides are divided by the same number), then the equality is still true.

The additive and multiplicative properties of equalities can be written in symbols as follows: ifthenifthenπ‘Ž=𝑏,π‘Ž+𝑐=𝑏+𝑐;π‘Ž=𝑏,π‘Žβ‹…π‘=𝑏⋅𝑐.

This balance method can be illustrated with a balance scale.

We have 2π‘₯+5=19; we can see numbers as weights put on a balance scale. We place each side of the equation on either side of the scale. Since 2π‘₯+5=19, the scale is balanced. It is clear that if we add or remove a weight from only one side of the scale, it will not be balanced any more. It is exactly the same with the equation. For it to remain balanced, we need to do exactly the same operation on each side.

Let us now look at some examples.

Example 1: Solving an Equation from a Description of the Operations on an Unknown Number

A number is tripled and 7 is subtracted from the result. If the answer is 17, what is the number?

Answer

Let us call the unknown number π‘₯. When this number is tripled, we get three times π‘₯, that is, 3π‘₯. Then, 7 is subtracted from 3π‘₯, which is written as

And this number is 17, so we have 3π‘₯βˆ’7=17.

To solve this equation, we apply the balance method, which consists in adding (subtracting) a number to (from) both sides of the equation and/or multiplying (dividing) both sides by a number. We first want to isolate the π‘₯ term on one side of the equation, so we need to add 7 to both sides to get rid of the β€œβˆ’7” on the left-hand side. We have

which gives 3π‘₯=24.

Then, we need to divide both sides by three (which is the reverse operation of tripling) to get the value of π‘₯:

which is π‘₯=8.

Hence, we find that the number is 8.

We can check our result by performing the operations as given in the question. First, the number is tripled, so we get 24. Then, seven is subtracted from the result, so we get 17. It is indeed the result given in the question, so our answer is correct.

Example 2: Solving a Two-Step Equation

Find the solution set of 3π‘₯βˆ’9=βˆ’6 in β„•.

Answer

Here, we are asked to find the solution set of an equation in the set of natural numbers. It means that we need to find the value(s) of π‘₯ so that 3π‘₯βˆ’9=βˆ’6 is true and so that π‘₯ is a natural number.

Let us first find the value of π‘₯ that is the solution of 3π‘₯βˆ’9=βˆ’6. For this, we want to get rid of the β€œβˆ’9” on the left-hand side to have the π‘₯ term on its own. So, we need to add 9 to each side of the equation. We have

that is, 3π‘₯=3.

It is quite obvious that if 3 times a number equals 3, then this number is 1. We can, however, write this step formally as dividing both sides by three:

that is, π‘₯=1.

We can check our answer by plugging 1 into the equation 3π‘₯βˆ’9=βˆ’6. We find 3β‹…1βˆ’9=βˆ’6.

This equality is true, so our answer is correct.

Next, we need to check that our solution is a natural number. The number 1 is a natural number; therefore, the solution set of 3π‘₯βˆ’9=βˆ’6 is {1}.

Example 3: Solving a Two-Step Equation

Find the solution set of the equation βˆ’4π‘₯+3=4.

Answer

To solve this equation, we apply the balance method, which consists in adding (subtracting) a number to (from) both sides of the equation and/or multiplying (dividing) both sides by a number. We first want to isolate the π‘₯ term on one side of the equation, so we start by subtracting 3 from each side. We have

which gives βˆ’4π‘₯=1.

The second step is to divide both sides of the equation by βˆ’4:

which gives π‘₯=βˆ’14.

We can check our answer by plugging it into βˆ’4π‘₯+3, evaluating it, and comparing with 4. We find βˆ’4β‹…ο€Όβˆ’14+3=1+3=4.

Therefore, the value of βˆ’14 is the solution to the equation βˆ’4π‘₯+3=4; our answer is correct.

The solution set of βˆ’4π‘₯+3=4 is ο¬βˆ’14.

Example 4: Solving a Two-Step Equation with a Fractional Leading Coefficient

Solve 𝑧3βˆ’13=11.

Answer

To solve this equation, we apply the balance method, which consists in adding (subtracting) a number to (from) both sides of the equation and/or multiplying (dividing) both sides by a number. We first want to isolate the 𝑧 term on one side of the equation, so our first step is to add 13 to both sides. We have

which gives 𝑧3=24.

We now need to multiply by 3 both sides of the equation to get the value of 𝑧. We find

hence, 𝑧=72.

We can check our answer by plugging it into 𝑧3βˆ’13 and evaluating it. We find 723βˆ’13=24βˆ’13=11.

Therefore, the value of 72 is the solution to the equation 𝑧3βˆ’13=11; our answer is correct.

So, the solution set of the equation 𝑧3βˆ’13=11 is {72}.

Example 5: Solving a Two-Step Equation Involving a Fraction

Find the solution set of π‘₯βˆ’54=5 in β„•.

Answer

To solve this equation, we apply the balance method, which consists in adding (subtracting) a number to (from) both sides of the equation and/or multiplying (dividing) both sides by a number. We want to isolate the π‘₯ term on one side of the equation. For this, our first step is to multiply both sides by 4. We have

which gives π‘₯βˆ’5=20.

We now need to add 5 to both sides of the equation.

We find

hence, π‘₯=25.

We can check our answer by plugging it into π‘₯βˆ’54 and evaluating this expression. We find 25βˆ’54=204=5.

Therefore, the value of 25 is the solution to the equation π‘₯βˆ’54=5. We now need to check that 25 is a natural number: it is. Therefore, the solution set to π‘₯βˆ’54=5 is {25}.

Key Points

  1. A simple equation such as 2π‘₯+5=19 can be solved by reversing all the operations and in the contrary order, to finally find the value of the starting number.
  2. More complex equations cannot be solved in that way.
  3. The balance method is a general method for solving equations based on the additive and multiplicative properties of equalities.
  4. These properties state the following:
    1. If the same amount is added to (or subtracted from) both sides of the equality, then the equality is still true.
    2. If both sides of the equation are multiplied by the same number (or if both sides are divided by the same number), then the equality is still true.
  5. It is written in symbols as follows: ifthenifthenπ‘Ž=𝑏,π‘Ž+𝑐=𝑏+𝑐;π‘Ž=𝑏,π‘Žβ‹…π‘=𝑏⋅𝑐.

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