Linear equations in one variable (LEOV) are mathematical equations that contain only one variable with the highest exponent equal to one. The general form of a linear equation in one variable is $$ax + b = 0,$$where:
- $a$ and $b$ are real numbers,
- $a \neq 0$, and
- $x$ is the variable.
The value of $x$ that satisfies the equation is called the solution or the root of the equation. Note that the coefficient of $x,$ namely $a,$ cannot be equal to $0$ because if this occurs, the variable $x$ will disappear.
The following are some examples of equations that are classified as LEOV:
- $x + 4 = 10$
- $2x -5 = 7$
- $3x = 15$
- $5 -2x = 1$
Meanwhile, the following equations are not classified as linear equations in one variable:
- $x^2 + 3 = 7$ (the exponent of the variable is greater than one)
- $2x + y = 5$ (contains more than one variable)
Basic Principles of Solving LEOV
The main objective in solving LEOV is to determine the value of the variable $x$ so that the equation becomes true. The basic principle used is maintaining the balance of both sides of the equation, namely:
- If a number is added to or subtracted from one side, then the other side must also be added to or subtracted from by the same number.
- If a number is multiplied or divided on one side, then the other side must also be multiplied or divided by the same number (as long as it is not zero).
Using this principle, the equation can be simplified until the value of $x$ is obtained.
General Steps for Solving LEOV
The general steps for solving linear equations in one variable are as follows:
- Simplify both sides of the equation if there are still parentheses or like terms.
- Move all terms containing the variable to one side (usually the left side).
- Move all constant numbers to the other side.
- Simplify the equation until the form $ax = b$ is obtained.
- Determine the value of the variable by dividing both sides by $a$, so that $x = \dfrac{b}{a}$ is obtained.
Suppose the following equation is given $$x + 4 = 10.$$The solution steps are as follows:
$$\begin{aligned} x + 4 & = 10 \\ x & = 10 -4 \\ x & = 6 \end{aligned}$$Therefore, the value of $x$ that satisfies the equation is $6$.
Below are several problems and discussions regarding LEOV. The problems presented are arranged gradually, starting from direct questions to contextual modeling problems. Through these exercises, readers are expected to understand the concept of linear equations in one variable more deeply and be able to apply it in various everyday life situations. Each problem is equipped with detailed and systematic solution steps to help readers follow the mathematical thinking process correctly.
Quote by Jim Rohn
Multiple Choice Section
The following equation that is classified as a linear equation in one variable is $\cdots \cdot$
A. $3x + 5 = 2$
B. $2x+y =-3$
C. $x^2-y^2=9$
D. $\frac{2}{x} + 9 = 12$
A linear equation in one variable is a mathematical equation that contains only one variable with the highest exponent equal to one.
Check option A:
The equation $3x+5=2$ contains one variable, namely $x,$ with exponent one. Therefore, $3x+5=2$ is classified as a linear equation in one variable.
Check option B:
The equation $2x+y=-3$ contains two variables, namely $x$ and $y.$ Therefore, $2x+y=-3$ is not classified as a linear equation in one variable.
Check option C:
The equation $x^2-y^2 = 9$ contains two variables, namely $x$ and $y,$ each with exponent two. Therefore, $x^2-y^2=9$ is not classified as a linear equation in one variable.
Check option D:
The equation $\frac{2}{x} + 9 = 12$ can be rewritten as $2x^{-1} + 9 = 12.$ It can be seen that this equation has one variable, namely $x,$ but its exponent is $-1,$ not $1.$ Therefore, the equation $\frac{2}{x} + 9 = 12$ is not classified as a linear equation in one variable.
(Answer A)
The solution of the equation $2x- 8 =12$ is $\cdots \cdot$
A. $x = 2$
B. $x = 4$
C. $x = 6$
D. $x = 10$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 2x-8 & = 12 \\ 2x & = 12+8 \\ 2x & = 20 \\ x & = \dfrac{20}{2} = 10. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = 10}.$
(Answer D)
Based on the following illustration, the price of a pencil is $\cdots \cdot$
A. Rp2.000
B. Rp3.500
C. Rp4.500
D. Rp7.000
Suppose $p$ represents the price of a pencil. This means that the linear equation corresponding to the illustration is $$4p+20.000 = 41.000-2p.$$Using simple algebraic calculations, we obtain
$$\begin{aligned} 4p+20.000 & = 41.000-2p \\ 4p+2p & = 41.000-20.000 \\ 6p & = 21.000 \\ p & = \dfrac{21.000}{6} = 3.500 . \end{aligned}$$Therefore, the price of a pencil is Rp3.500.
(Answer B)
It is known that the equation $$5(2x-3) + 4 = 2(3x + 1)-(-3)$$has the solution $m.$ The value of $3m + 4$ is $\cdots \cdot$
A. $3$ C. $13$
B. $7$ D. $16$
From the given equation, we obtain
$$\begin{aligned} (10x-15) + 4 & = (6x+2) + 3 \\ 10x-11 & = 6x+5 \\ 10x-6x & = 5+11 \\ 4x & = 16 \\ x & = 4. \end{aligned}$$This means that $m = 4$ so that $3m + 4 = 3(4) + 4 = 16.$
Therefore, the value of $3m+4$ is $\boxed{16}.$
(Answer D)
The value of $x$ that satisfies $\dfrac{4x+5}{2x+1} = \dfrac{16}{5}$ is $\cdots \cdot$
A. $\dfrac23$ C. $\dfrac34$
B. $\dfrac32$ D. $\dfrac43$
By performing cross multiplication, we obtain
$$\begin{aligned} 5(4x + 5) & = 16(2x + 1) \\ 20x + 25 & = 32x + 16 \\ 25-16 & = 32x-20x \\ 9 & = 12x \\ x & = \dfrac{9}{12} = \dfrac34. \end{aligned}$$Therefore, the value of $x$ that satisfies the equation is $\boxed{\dfrac34}.$
(Answer C)
It is known that $n$ is the solution of the equation $2\dfrac12x + \dfrac34 = 2x-1\dfrac12.$ The value of $n+4$ is $\cdots \cdot$
A. $-\dfrac32$ C. $\dfrac12$
B. $-\dfrac12$ D. $\dfrac32$
It is known that $2\dfrac12x + \dfrac34 = 2x-1\dfrac12.$ First, multiply both sides by $4$ so that a new equation $10x + 3 = 8x-6$ is obtained. Using simple algebraic calculations, we obtain
$$\begin{aligned} 10x-8x & = -6-3 \\ 2x & = -9 \\ x & = -\dfrac92. \end{aligned}$$This means that the solution of the equation is $n = -\dfrac92$ so that $n+4 = -\dfrac92 + 4 = -\dfrac12.$
Therefore, the value of $n+4$ is $\boxed{-\dfrac12}.$
(Answer B)
The value of $x$ that satisfies the equation $\dfrac{x+5}{4} = \dfrac{2x+7}{2}$ is $\cdots \cdot$
A. $-9$ C. $3$
B. $-3$ D. $9$
It is known that the equation $\dfrac{x+5}{4} = \dfrac{2x+7}{2}.$ Multiply both sides by $4$ to avoid fractions, then simplify. We will obtain
$$\begin{aligned} \cancel{4} \cdot \dfrac{x+5}{\cancel{4}} & = \cancelto{2}{4} \cdot \dfrac{2x+7}{\cancel{2}} \\ x+5 & = 2(2x + 7) \\ x + 5 & = 4x + 14 \\ x-4x & = 14-5 \\ -3x & = 9 \\ x & = -3. \end{aligned}$$Therefore, the value of $x$ that satisfies the equation is $\boxed{x=-3}.$
(Answer B)
I am a positive integer. If four more than me is multiplied by five, the result is me plus twenty-four. What number am I?
A. $1$ C. $3$
B. $2$ D. $4$
Suppose I am represented by $x.$ Thus, the mathematical model corresponding to this case is $5(x + 4) = x + 24.$ Using simple algebraic calculations, we obtain
$$\begin{aligned} 5(x + 4) & = x + 24 \\ 5x + 20 & = x + 24 \\ 5x-x & = 24-20 \\ 4x & = 4 \\ x & = 1. \end{aligned}$$Therefore, I am $1.$
(Answer A)
It is known that a square has diagonal lengths of $(6x + 3)$ m and $(4x + 15)$ m. What is the actual length of the diagonal of the square?
A. $24$ m C. $36$ m
B. $33$ m D. $39$ m
A square has two diagonals of equal length. Therefore, we obtain
$$\begin{aligned} 6x + 3 & = 4x + 15 \\ 6x-4x & = 15-3 \\ 2x & = 12 \\ x & = 6. \end{aligned}$$Substituting the value $x = 6$ into the expression $6x+3$ gives $6(6) + 3 = 39.$
Therefore, the actual length of the diagonal of the square is $\boxed{39}$ m.
(Answer D)
It is known that the perimeter of a rectangle is $64$ cm. If its length is $(3x + 7)$ cm and its width is $(2x+5)$ cm, the actual length and width of the figure respectively are $\cdots \cdot$
A. $26$ cm and $6$ cm
B. $22$ cm and $10$ cm
C. $19$ cm and $13$ cm
D. $18$ cm and $14$ cm
Suppose the rectangle has length $p$ cm and width $\ell$ cm.
Since the perimeter of the rectangle is $64$ cm, we obtain $2(p + \ell) = 64,$ or simplified into $p + \ell = 32$ after dividing both sides by $2.$ Next, substitute $p = 3x + 7$ and $\ell = 2x + 5.$
$$\begin{aligned} (3x+7) + (2x+5) & = 32 \\ 5x + 12 & = 32 \\ 5x & = 20 \\ x & = 4. \end{aligned}$$This means that $p = 3x + 7 = 3(4) + 7 = 19$ cm and $\ell = 2x + 5 = 2(4) + 5 = 13$ cm.
Therefore, the actual length and width of the figure respectively are $19$ cm and $13$ cm.
(Answer C)
It is known that Ardi’s age is $p$ years, while Lili is $6$ years younger than Ardi. If the sum of their ages is $54$ years, then the correct mathematical model for this case is $\cdots \cdot$
A. $p + 6 = 54$
B. $2p + 6 = 54$
C. $2p -6= 54$
D. $4p -6 = 54$
It is known that Ardi’s age is $p$ years. Since Lili is $6$ years younger than Ardi, Lili’s age is $p-6$ years. The sum of their ages is $54$ years, so it can be written as
$$p + (p-6) = 54 \Rightarrow 2p-6 = 54.$$Therefore, the correct mathematical model for this case is $\boxed{2p-6 = 54}.$
(Answer C)
The sum of three consecutive odd numbers is $63.$ The sum of the largest and smallest numbers among them is $\cdots \cdot$
A. $38$
B. $42$
C. $46$
D. $54$
Suppose the three consecutive odd numbers are denoted by $n, n+2,$ and $n+4.$ Thus, we obtain
$$\begin{aligned} n + (n+2) + (n+4) & = 63 \\ 3n + 6 & = 63 \\ 3n & = 57 \\ n & = \dfrac{57}{3} = 19. \end{aligned}$$This means that the three odd numbers are $19, 21,$ and $23.$
Therefore, the sum of the largest and smallest numbers among them is $\boxed{19 + 23 = 42}.$
(Answer B)
Essay Section
Solve the following linear equation.
$$x + 4 = 1$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} x + 4 & = 1 \\ x & = 1-4 \\ x & = -3. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = -3}.$
Solve the following linear equation.
$$x -5=4$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} x-5 & = 4 \\ x & = 4+5 \\ x & = 9. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = 9}.$
Solve the following linear equation.
$$x-3 = -4$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} x-3 & = -4 \\ x & = -4+3 \\ x & = -1. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = -1}.$
Solve the following linear equation.
$$2x + 1 = 5$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 2x + 1 & = 5 \\ 2x & = 5-1 \\ 2x & = 4 \\ x & = \dfrac{4}{2} = 2. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = 2}.$
Solve the following linear equation.
$$3x-5=7$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 3x-5 & = 7 \\ 3x & = 7+5 \\ 3x & = 12 \\ x & = \dfrac{12}{3} = 4. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = 4}.$
Solve the following linear equation.
$$4x + 10 = 30$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 4x + 10 & = 30 \\ 4x & = 30-10 \\ 4x & = 20 \\ x & = \dfrac{20}{4} = 5. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = 5}.$
Solve the following linear equation.
$$5x-10=10$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 5x-10 & = 10 \\ 5x & = 10+10 \\ 5x & = 20 \\ x & = \dfrac{20}{5} = 4. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = 4}.$
Solve the following linear equation.
$$6x-5 = -2$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 6x-5 & = -2 \\ 6x & = -2+5 \\ 6x & = 3 \\ x & = \dfrac{3}{6} = \dfrac12. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = \dfrac12}.$
Solve the following linear equation.
$$7x-4 = -5$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 7x-4 & = -5 \\ 7x & = -5+4 \\ 7x & = -1 \\ x & = -\dfrac{1}{7}. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = -\dfrac17}.$
Solve the following linear equation.
$$4-3x=-5$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 4-3x & = -5 \\ -3x & = -5-4 \\ -3x & = -9 \\ x & = \dfrac{-9}{-3} = 3. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = 3}.$
Solve the following linear equation.
$$5-4x =1 $$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 5-4x & = 1 \\ -4x & = 1-5 \\ -4x & = -4 \\ x & = \dfrac{-4}{-4} = 1. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = 1}.$
Solve the following linear equation.
$$-4 + 4x = 9$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} -4 + 4x & = 9 \\ 4x & = 9+4 \\ 4x & = 13 \\ x & = \dfrac{13}{4}. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = \dfrac{13}{4}}.$
Solve the following linear equation.
$$2 + 3x = 5 + x$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 2 + 3x & = 5 + x \\ 3x-x & = 5-2 \\ 2x & = 3 \\ x & = \dfrac{3}{2}. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = \dfrac32}.$
Solve the following linear equation.
$$2-3x = 5-2x$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 2-3x & = 5-2x \\ -3x + 2x & = 5-2 \\ -x & = 3 \\ x & = -3. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = -3}.$
Solve the following linear equation.
$$-4x + 5 = 7x + 10$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} -4x + 5 & = 7x + 10 \\ -4x-7x & = 10-5 \\ -11x & = 5 \\ x & = -\dfrac{5}{11}. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = -\dfrac{5}{11}}.$
Solve the following linear equation.
$$6x-1=x-10$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 6x-1 & = x-10 \\ 6x-x & = -10+1 \\ 5x & = -9 \\ x & = -\dfrac95. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = -\dfrac95}.$
Solve the following linear equation.
$$5-3x = 10-7x$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 5-3x & = 10-7x \\ -3x + 7x & = 10-5 \\ 4x & = 5 \\ x & = \dfrac54. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = \dfrac54}.$
Solve the following linear equation.
$$4-10x = 5(x-3)$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 4-10x & = 5(x-3) \\ 4-10x & = 5x-15 \\ -10x-5x & = -15-4 \\ -15x & = -19 \\ x & = \dfrac{-19}{-15} = \dfrac{19}{15}. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = \dfrac{19}{15}}.$
Solve the following linear equation.
$$2(4-x) = 10x + 7$$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 2(4-x) & = 10x+7 \\ 8-2x & = 10x+7 \\ -2x-10x & = 7-8 \\ -12x & = -1 \\ x & = \dfrac{-1}{-12} = \dfrac{1}{12}. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = \dfrac{1}{12}}.$
Solve the following linear equation.
$$4(5-7x) = 5(3-2x) $$
Using simple algebraic calculations, we obtain
$$\begin{aligned} 4(5-7x) & = 5(3-2x) \\ 20-28x & = 15-10x \\ -28x + 10x & = 15-20 \\ -18x & = -5 \\ x & = \dfrac{-5}{-18} = \dfrac{5}{18}. \end{aligned}$$Therefore, the solution of the linear equation is $\boxed{x = \dfrac{5}{18}}.$
Model the following statement into a linear equation, then solve the equation by finding the appropriate value of the variable.
Twice Skibidi’s current age is equal to Skibidi’s age 8 years from now.
Suppose $x$ represents Skibidi’s current age so that the appropriate linear equation is $2x = x + 8.$ Using simple algebraic calculations, we obtain
$$\begin{aligned} 2x & = x+8 \\ 2x -x & =8 \\ x & = 8. \end{aligned}$$Therefore, the solution of the linear equation is $x = 8.$
This means that Skibidi’s current age is $8$ years old.
Model the following sentence into a linear equation, then solve the equation by finding the appropriate value of the variable.
The price of three mangoes is 16 thousand rupiah less than a one hundred thousand rupiah banknote.
Suppose $m$ represents the price of one mango so that the appropriate linear equation is $3m = 100.000-16.000.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} 3m & = 100.000-16.000 \\ 3m & = 84.000 \\ m & = \dfrac{84.000}{3} = 28.000. \end{aligned}$$Therefore, the solution to the linear equation is $m = 28.000.$
This means that the price of one mango is Rp28.000.
Model the following sentence into a linear equation, then solve the equation by finding the appropriate value of the variable.
The price of four apples is equal to the price of two apples plus twenty thousand rupiah.
Suppose $a$ represents the price of one apple so that the appropriate linear equation is $4a = 2a + 20.000.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} 4a & = 2a+20.000 \\ 4a-2a & = 20.000 \\ 2a & = 20.000 \\ a & = \dfrac{20.000}{2} = 10.000. \end{aligned}$$Therefore, the solution to the linear equation is $a = 10.000.$
This means that the price of one apple is Rp10.000.
Model the following sentence into a linear equation, then solve the equation by finding the appropriate value of the variable.
Five years from now, Sigma Boy’s age will be twice his current age.
Suppose $S$ represents Sigma Boy’s current age so that the appropriate linear equation is $S + 5 = 2S.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} S + 5 & = 2S \\ 2S-S & = 5\\ S & = 5. \end{aligned}$$Therefore, the solution to the linear equation is $S = 5.$
This means that Sigma Boy’s current age is $5$ years old.
Model the following sentence into a linear equation, then solve the equation by finding the appropriate value of the variable.
Skibidi buys 3 notebooks and 3 pens with a total price of Rp24.000. The price of 1 pen is Rp3.000. (Hint: Suppose $b$ = the price of $1$ notebook)
Suppose $b$ represents the price of $1$ notebook. Since the price of $1$ pen is Rp3.000, then $3$ pens cost Rp9.000. Therefore, the appropriate linear equation is $3b + 9.000 = 24.000.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} 3b + 9.000 & = 24.000 \\ 3b & = 24.000-9.000 \\ 3b & = 15.000 \\ b & = \dfrac{15.000}{3} = 5.000. \end{aligned}$$Therefore, the solution to the linear equation is $b = 5.000.$
This means that the price of $1$ notebook is Rp5.000.
Model the following sentence into a linear equation, then solve the equation by finding the appropriate value of the variable.
Four times a number plus eight is equal to twice the number plus twenty.
Suppose $x$ represents the number referred to so that the appropriate linear equation is $4x + 8 = 2x + 20.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} 4x+8 & = 2x+20 \\ 4x-2x & = 20-8 \\ 2x & = 12 \\ x & = \dfrac{12}{2} = 6. \end{aligned}$$Therefore, the solution to the linear equation is $x = 6.$
This means that the number referred to is $6.$
Model the following sentence into a linear equation, then solve the equation by finding the appropriate value of the variable.
Three times a number minus four is equal to twice the number plus seven.
Suppose $x$ represents the number referred to so that the appropriate linear equation is $3x-4 = 2x+7.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} 3x-4 & = 2x+7 \\ 3x-2x & = 7+4 \\ x & = 11. \end{aligned}$$Therefore, the solution to the linear equation is $x = 11.$
This means that the number referred to is $11.$
Model the following sentence into a linear equation, then solve the equation by finding the appropriate value of the variable.
Three less than a number is equal to four times the number plus eight.
Suppose $x$ represents the number referred to so that the appropriate linear equation is $x-3 = 4x + 8.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} x-3 & = 4x+8 \\ -3-8 & = 4x-x \\ -11 & = 3x \\ x & = -\dfrac{11}{3}. \end{aligned}$$Therefore, the solution to the linear equation is $x = -\dfrac{11}{3}.$
This means that the number referred to is $-\dfrac{11}{3}.$
Write the linear equation that matches the following illustration, then solve it.
Suppose $d$ represents the price of one durian. This means that the appropriate linear equation for the illustration is $3d + 25.000 = 85.000.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} 3d+25.000 & = 85.000 \\ 3d & = 85.000-25.000 \\ 3d & = 60.000 \\ d & = 20.000. \end{aligned}$$Therefore, the solution to the linear equation is $d = 20.000.$
This means that the price of one durian is Rp20.000.
Write the linear equation that matches the following illustration, then solve it.
Suppose $k$ represents the price of one chair. This means that the appropriate linear equation for the illustration is $250.000-3k = k + 10.000.$ By using simple algebraic calculations, we obtain
$$\begin{aligned} 250.000-3k & = k+10.000 \\ 250.000-10.000 & = k+3k \\ 240.000 & = 4k \\ k & = \dfrac{240.000}{4} = 60.000 \end{aligned}$$Therefore, the solution to the linear equation is $k = 60.000.$
This means that the price of one chair is Rp60.000.