 Math calculations frequently involve the use of letters instead of numbers. By using algebraic expressions, we can easily solve math problems.

One of the most important areas of algebraic expressions is linear equations. Linear equations are frequently used in daily life calculations and are the basis of mathematics. Even if there is a number that we don’t know, we can find the answer with linear equations.

Equations have properties and must be understood in their definitions and outlines. If you can create linear equations by solving word problems, you will be able to use linear equations in your daily life.

How should we think about using linear equations? We will explain the properties of linear equations and how to solve them.

## Meaning and Definition of Equations

First, what is an equation? We have to understand the definition. If an equation contains numbers that we don’t know, an equation that uses the alphabet is called algebraic expressions. By using letters, we can create an expression.

If we are learning to calculate polynomials, we already understand expressions that use letters. In polynomials, we can substitute numbers into $x$ to get various answers. For example, substituting numbers into $3x+1$ gives the answer as follows. In linear equations, on the other hand, the answer can be correct or wrong by substituting numbers into the equation.

For example, given the equation $3x+1=10$, what does the answer look like? If we apply various numbers to $x$, we get the following.

• $x=1, 3×1+1=10$: wrong
• $x=2, 3×2+1=10$: wrong
• $x=3, 3×3+1=10$: correct

We can’t substitute any number. Only a specific number can be used to get the answer. Because there is an equality sign ($=$) in the equation, the number of $x$ that satisfies the equation is limited.

### Five Properties of the Equals Sign of Equations

In linear equations, there is always an equality sign ($=$). Since it means equal, we can use an equality sign if the two are equal. In addition, the following words exist in the equation.

• Left side: the equality on the left.
• Right side: the equation on the right.
• Both sides: left and right sides These words exist in linear equations.

It is important to note that the equals sign has five properties. Understanding the properties of equations containing the equals sign will enable us to solve problems with linear equations.

To understand the properties of equations, it’s easy to think of balances. In equations, we use equality signs. Therefore, the left and right sides must be equal. The equation is shown in the following figure. The left and right sides are equal by an equality sign. If $A=B$, then the following five things hold.

• Adding the same numbers, the equation holds.
• Subtracting the same numbers, the equation holds.
• Multiplying the same numbers, the equation holds.
• Dividing the same numbers, the equality still holds.
• Exchanging both sides, the equation still holds.

I will explain each of these.

-If $A=B$, then $A+C=B+C$

Why is it okay to add the same number to both sides of an equation? This is because if you add the same number to both sides, both sides will be equal. If we add the same thing, the balance will be equal as follows. -If $A=B$, then $A-C=B-C$

In the equation, you can subtract the same number from both sides of the equation. You can subtract as well as add. -If $A=B$, then $A×C=B×C$

In the equation, you can multiply both sides. This is because both sides of the equation are equal. -If $A=B$, then $\displaystyle\frac{A}{C}=\displaystyle\frac{B}{C}$

If you divide the same number on both sides, the equation still holds. In addition, algebraic expressions don’t use division, but instead convert all of the divisions into multiplication of fractions. Therefore, the formula is represented by multiplying the fractions, not by division. -If $A=B$, then $B=A$

If the equation holds, then it is okay to exchange the left and right sides. Since both sides are equal, they are equal even if they are exchanged. Some people may think that the properties of the equals sign that we have described so far are obvious. However, this is understandable because we can consider the example of balances. On the other hand, if $A=B$, then $A-C=B-C$, and so on, it is impossible to understand the formula without figures.

When learning linear equations, many people have trouble understanding the properties of equations. So, let’s try to understand the properties of these equations with figures.

### Solving Linear Equation Problems by Transposition

How do we solve the problem of linear equations? To do this, we use the property of equality sign explained earlier.

For both sides, you can add or subtract the same number and still get the same equation. For example, $x+2=11$ can be calculated in the following order.

1. $x+2=11$
2. $x+2\textcolor{red}{-2}=11\textcolor{red}{-2}$
3. $x=11\textcolor{red}{-2}$
4. $x=9$

In an equation, we can get the answer by adding or subtracting the same number on both sides.

However, in mathematics, we have to be able to calculate quickly. In fact, there is a time limit on the exam. So we use a technique called transposition. A transposition is a method of moving a number from the left side to the right (or right to left) by changing the sign.

For the previous calculation, we can omit the equation (b), as shown below. By transposition, we can eliminate the calculation process. As a result, miscalculation will be reduced. We all use transposition in algebraic expressions.

Transposition can be used not only with numbers but also with letters. We can transpose numbers and letters as follows. Note that we can multiply or divide the same number by both sides. By multiplying an integer or fraction, you get the value of $x$. For example, we get the following.

$3x=10$

$3x×\textcolor{red}{\displaystyle\frac{1}{3}}=10×\textcolor{red}{\displaystyle\frac{1}{3}}$

$x=\displaystyle\frac{10}{3}$

To answer a number we don’t know ($x$), we use a linear equation. Therefore, the final answer must be in the form of $x=□$.

Note that multiplication and division (multiplication of fractions) cannot be transposed, unlike addition and subtraction. They must be calculated by multiplying them by both sides, respectively.

### Remove the Parentheses Using the Distributive Property and Convert to Integers

There are many equations that contain parentheses in algebraic expressions. In this case, you cannot transpose them. So we need to use the distributive property (or distributive law) and remove the parentheses first.

For example, how can we calculate $3(x-2)=14$? In this case, we can remove the parentheses. We can calculate as follows.

$3(x-2)=14$ : Distributive property

$3x-6=14$ : Transpose

$3x=14\textcolor{red}{+6}$

$3x\textcolor{red}{×\displaystyle\frac{1}{3}}=20\textcolor{red}{×\displaystyle\frac{1}{3}}$ : Multiply the fractions

$x=\displaystyle\frac{20}{3}$

Because you remove the parentheses, you can add and subtract by transposition. To solve the equation, use the distributive law to remove the parentheses first.

-Change Decimal to Integer by Multiplying Both Sides

In some cases, we can multiply on both sides to simplify the calculation. In equations involving decimals or fractions, both sides of the equation are frequently multiplied.

For example, how should the following equation be calculated?

$0.04x-0.12=0.02x-0.2$

You can calculate the equation as decimals, but there is a high probability that a calculation error will occur. So let’s simplify the equation by changing it to integers. In the previous equation, we can multiply both sides by 100.

$0.04x-0.12=0.02x-0.2$

$(0.04x-0.12)\textcolor{red}{×100}=(0.02x-0.2)\textcolor{red}{×100}$

$4x-12=2x-20$

$4x\textcolor{red}{-2x}=-20\textcolor{red}{+12}$

$2x=-8$

$x=-4$

It is easier to calculate integers than decimals.

-Correcting Fractional Expressions to Integers

The same is true for fractions, which are easier to calculate if you convert them to integers. For example, how should the following be calculated?

$\displaystyle\frac{2}{3}x-\displaystyle\frac{1}{4}=\displaystyle\frac{3}{8}x$

Just like with decimals, we can convert them into integer expressions by multiplying both sides of the equation. For example the following.

$\displaystyle\frac{2}{3}x-\displaystyle\frac{1}{4}=\displaystyle\frac{3}{8}x$

$\left(\displaystyle\frac{2}{3}x-\displaystyle\frac{1}{4}\right)\textcolor{red}{×24}=\displaystyle\frac{3}{8}x\textcolor{red}{×24}$

$16x-6=9x$

$16x-9x=6$

$7x=6$

$x=\displaystyle\frac{6}{7}$

Without converting the equation, it is often difficult to calculate. In that case, use the property that multiplying both sides yields equality. By multiplying both sides by a number, you can convert it into an integer expression.

## Understand the Difference Between Calculating Polynomials and Equations

It is important to note that polynomial problems and equations are different in properties. In polynomial problems, there is no equality sign. In contrast, equations have an equality sign, and the left and right sides are connected by equals.

The reason why it does not matter if we multiply both sides of the equation is because it is an equation. Because the equation has an equality sign, the five properties we just discussed can be used. However, if there is no equality sign, it cannot be used the five properties.

For example, in the following calculation, as mentioned earlier, we can multiply both sides by 100.

• $0.04x-0.12=0.02x-0.2$

Because of the equal sign, we can add, subtract, and multiply both sides by the same number. On the other hand, in the following expression, we cannot multiply by 100.

• $0.04x-0.02x-0.12+0.2$

If the expression does not contain an equal sign, it must be calculated as it is.

In typical polynomial calculations, only one expression exists. In contrast, in linear equations, there are two expressions on the left and right sides because of the presence of an equality sign. Therefore, the calculation method is different. Make sure you understand this difference to avoid calculation mistakes.

## Tips on How to Solve Word Problems with Linear Equations

Please note that when learning linear equations, solving problems is not the most important thing. The most important thing is that you create linear equations and be able to answer them. This is because linear equations are used frequently in our daily lives.

For example, you have to use linear equations to calculate your electricity and water bills. Also, if you want to calculate how much your electricity bill has been reduced by saving on air conditioning, you can use the linear equation to calculate it.

You can also use the linear equation to know what time you need to leave your home to catch a plane or bus. If you make a mistake in your calculations, you will miss your flight or bus and not be able to travel. In junior high school and high school, parents will do all of these calculations. However, if you are an adult, you have to do all the calculations yourself. Linear equations are used in every situation. When you can solve word problems with linear equations, you won’t have any problems in your life as an adult.

-How to Solve Linear Equations

How do we make linear equations? In algebraic expressions, we use alphabets to calculate the equation. The $x$ frequently used in algebraic expressions refers to an unknown number. A number you don’t know in math means that it is a number you want to answer.

For example, if you don’t know how many pens you need to buy, set the number of pens you need to buy to $x$. If you don’t know how many kilometers to walk, set the distance to walk to $x$km. For all equations, set the number you don’t know to $x$.

Then make the equations and solve the problem. This way you will be able to get the answer.

### Represent the Same Thing Two Ways and Make Equations

When you make an equation, there is a method. That is making two equations. As you can see from the fact that there is an equal sign, you attach equations that have the same meaning.

For example, how do we make an equation for the following word problem?

• How many pens costing \$2.00 each will cost you \$12.00?

Before we learned the algebraic expressions, we can see that $12÷2=6$ and the answer is 6. Let’s express this in an algebraic expression.

Set the number you don’t know to $x$. In this problem, we don’t know how many pens we need to buy. So, set the number of pens we need to buy to $x$.

Then, as a result of buying $x$ pens that cost \$2 each, we will pay \$12. In other words, we can make the following equation.

• $\$2 ×x＝\$12$

We make the equation by setting the number we want to find to $x$. The amount of money we buy $x$ pens at \$2 each must equal the total amount of money we pay (\$12). So we make an equation by connecting the two equations with an equal sign.

We have used a very simple equation to explain the equation, but be aware of the following in the word problems of the linear equations.

• Set the unknown number to $x$.
• Create two expressions with the same meaning and connect them with an equals sign.

By doing this work, we will be able to create equations. Then, all we have to do is solve the problems in the linear equations to get the answers.

## Exercises: Solve Word Problems Using Linear Equations

Q1: Do the following calculation.

1. $5(3x-3)+2=6x+3$
2. $\displaystyle\frac{5x+2}{4}=\displaystyle\frac{-3x-3}{6}$
3. $4-\displaystyle\frac{2x+3}{5}=0.4x$

In linear equations, we solve the problem by doing transposition. However, if there are parentheses, you must remove the parentheses. And if there are decimals or fractions, you can simplify the equation by multiplying both sides of the equation.

(a)

$5(3x-3)+2=6x+3$

$15x-15+2=6x+3$

$15x-6x=3+15-2$

$9x=16$

$x=\displaystyle\frac{16}{9}$

(b)

$\displaystyle\frac{5x+2}{4}=\displaystyle\frac{-3x-3}{6}$

$\displaystyle\frac{5x+2}{4}\textcolor{red}{×12}=\displaystyle\frac{-3x-3}{6}\textcolor{red}{×12}$

$3(5x+2)=2(-3x-3)$

$15x+6=-6x-6$

$15x+6x=-6-6$

$21x=-12$

$x=-\displaystyle\frac{12}{21}$

$x=-\displaystyle\frac{4}{7}$

(c)

If both decimals and fractions are present in the equation, multiply both sides by a number that will be an integer.

$4-\displaystyle\frac{2x+3}{5}=0.4x$

$\left(4-\displaystyle\frac{2x+5}{5}\right)\textcolor{red}{×5}=0.4x\textcolor{red}{×5}$

$4×5-\displaystyle\frac{2x+3}{5}×5=0.4x×5$

$20-(2x+3)=2x$

$20-2x-3=2x$

$-2x-2x=-20+3$

$-4x=-17$

$x=\displaystyle\frac{17}{4}$

Q2: Solve the following word problem.

Currently, the father’s age is 44 and the child’s age is 14. In how many years will the father’s age be twice the age of his child?

Set $x$ for the unknown number in the word problem. In this word problem, we don’t know how many years later the father’s age will be twice the age of his child. So we set the number of years that have passed to $x$.

After $x$ years have passed, the father’s age is $44+x$ years old. In contrast, the child’s age is $14+x$ years old. And by doubling $14+x$ years, it is equal to the age of the father. Therefore, the following equation holds.

• $44+x=(14+x)×2$

Solving this linear equation will give us the correct answer.

$44+x=2(14+x)$

$44+x=28+2x$

$x-2x=28-44$

$-x=-16$

$x=16$

In word problems of linear equations, we can check if the answer is correct or not by using the answer we get. The trick to solving the problem is to check whether the answer you get is really correct or not.

In this calculation, $x=16$. So, let’s apply the number. 16 years from now, the father’s age will be $44+16=60$ years old. In contrast, the child’s age is $14+16=30$ years old. Multiplying 30 by 2 gives us 60, so we can be confident that $x=16$ is the correct answer.

In word problems of linear equations, you can check if a miscalculation is occurring by reviewing the equation in this way.

Q3: Solve the following word problem.

You are thinking of purchasing an item for \$5 a piece and selling it for \$8. However, 10% of the goods you stock are defective and have scratches on the surface that make them look bad, so you sell them for \$6. How many items do you need to purchase and sell in order to earn a profit of \$140?

In this word problem, we don’t know how many products to stock and sell. So, the number of products to stock is $x$.

It is important to note that the number of regular products is not $x$ pieces. There are also defective products with scratches in the mix. When we stock $x$ pieces, the number of defective products is $0.1x$ pieces. We need to subtract the number of defective products from the goods we stock, so the number of regular products is $(x-0.1x)$ pieces.

Also, if we sell 1 piece of the regular product, our profit is $8-5=\$3$. On the other hand, if we sell one defective item with a scratch, the profit is$6-5=\$1$. Selling all of these results in a profit of \$140. Therefore, we have the following equation. $3x×0.9x+1×0.1x=1402.7x+0.1x=1402.8x=1402.8x×10=140×1028x=1400x=\displaystyle\frac{1400}{28}x=50$-Reviewing the Answer In the same way, be sure to review the word problem by applying the answer. 50 products are purchased, the number of defective products is$50×0.1=5$pieces. On the other hand, the number of regular products is$50-5=45$pieces. As already described, if we sell one regular product, our profit is \$3. If we sell one defective product, the profit is \$1. Therefore, we can get the amount of profit by the following calculation. •$3×45+1×5$If we calculate this equation, we get \$140. So we can see that we have made the equation correctly and that we have the answer.

Q4: Solve the following word problem.

We have 200 g of 14% salt water. We want to make it 20% salt water by adding salt. How many grams of salt should be added?

An important factor in the salt problem is the fact that adding salt increases the number of grams of solution. Therefore, when adding salt, the equation must also take into account the weight of the adding salt itself.

The number we don’t know in this question is how many grams of salt to be added. So, set the amount of salt to be added to $x$g.

A 200g of 14% salt water contains $200×0.14=28$g of salt. By adding $x$g of salt to it, we make it 20% salt water.

On the other hand, how many grams of salt are in a 20% salt water? To 200 grams of salt water, $x$g is added, so the weight after adding salt is $(200+x)$g. Since the concentration of salt water is 20%, we can calculate the amount of salt in 20% salt water by the equation $(200+x)×0.2$.

As a result, we can create two different formulas. Both equations show the amount of salt in 20% salt solution. Solving this equation gives us the following.

$200×0.14+x=(200+x)×0.2$

$28+x=40+0.2x$

$x-0.2x=40-28$

$0.8x=12$

$0.8x×\displaystyle\frac{5}{4}=12×\displaystyle\frac{5}{4}$

$x=15$

After answering the equation, let’s review it. To do this, make a different equation. 200 grams of 14% salt water contains $200×0.14=28$g of salt. 15 grams of salt is added to the solution resulting in $28+15=43$g of salt.

If a 20% concentration of salt water contains 43g of salt, how many grams of total salt water is there? If we set the weight of the salt solution to $y$g, we can make the following equation.

• $y×0.2=43$

Solving this equation gives us the following.

$y×0.2=43$

$y×0.2\textcolor{red}{×5}=43\textcolor{red}{×5}$

$y=215$

A 14% salt water is 200g, and if we add 15g of salt to this, we get 215g of salt water. The numbers match $y=215$. So we can check that we have made the equation and calculated it correctly.

## It Is Important to Solve Problems with Linear Equations

The reason why we need to learn algebraic expressions in mathematics is to be able to create equations. Linear equations are used in all aspects of daily life. Therefore, if you don’t understand linear equations, you will often have trouble calculating them when you live as an adult because you can’t make the correct equations.

In linear equations, make sure you understand the properties of the equals sign. There are five properties of an equality sign, and all people use these properties in math calculations to get answers.

Another important part of linear equations is being able to solve word problems. The ability to solve word problems allows us to apply mathematics to our daily lives. Consider an unknown number as $x$ and try to make two equations.

In addition, you can review the word problems of linear equations as we have discussed so far. Reviewing them will help you to notice miscalculations, so be sure to review your answers to see if they are correct. Understanding the considerations and tips for solving these equations will help you reduce your math mistakes.