1. Linear equations

A linear equation consists of variables raised only to the power of 1, with no higher exponents or roots. Consider the following table.

Equation	Linear/Non-linear
$2x = 10$	One variable linear
$5x + 1 = 3$	One variable linear
$x^2 + 2x + 1 = 0$	Non-linear (Power of $x$ is $2$ )
$2x - 3y = 1$	Two variable linear
$y = x^2 + 3$	Non-linear (Power of $x$ is $2$ )
$\sqrt{x} + y = 5$	Non-linear ( $x$ is under square root which means it's raised to the power of $\frac{1}{2}$ )
$2x = y^2 + 1$	Two-variable linear

According to the table, in the equations, $2x = 10$ , $5x + 1 = 3$ , each consists of only one variable raised to the power of $1$ , thus can be categorized as one-variable linear equations.
The equations, $x^2 + 2x + 1 = 0$ , $y = x^2 + 3$ , and $\sqrt{x} + y = 5$ are nonlinear because the variables of these equations are not raised to the power of $1$ .
The equations, $2x - 3y = 1$ and $2x = y^2 + 1$ can be categorized as two variable linear equations since they consist of two variables.

Now, we can distinguish between one variable linear equations and others. By solving equations in one variable, we find the value of the variable.

Let’s consider a few examples of two variable linear equations:

$5x = 15$
$2a - 3 = 5a - 21$
$5p - q - 4 = 0$
$\frac{x}{2} + \frac{2x}{3} = \frac{5}{4}$
$a - \frac{a}{2}- 7 = 0$

Consider the one-variable linear equation $\boldsymbol{x + 1 = 5}$ . Let’s solve this equation.

The left side of the equation consists of an algebraic expression, while the right side contains only the integer $5$ . Solving an equation means isolating the variable on one side and finding its value on the other side.

As we want to isolate $x$ on the left side, we need to eliminate the $+1$ from the left side. We perform the inverse operation of $+1$ to both sides to eliminate it from the left side. The inverse of addition is subtraction, so we subtract $1$ from both sides.

$x + 1 = 5$

$x + 1 - 1 = 5 - 1$

Now, $+1$ and $-1$ are getting canceled on the left side while $5 - 1$ simplifies to $+4$ on the right side. Therefore, the solution for the equation $x + 1 = 5$ is $x = 4$

Consider the one-variable linear equation $\boldsymbol{x - 3 = 4}$ . Let’s solve this equation.

As we want to isolate x on the left side, we need to eliminate the $-3$ from the left side. We perform the inverse operation of $-3$ on both sides to eliminate it from the left side. The inverse of subtraction is addition, so we add $3$ to both sides.

x - 3 = 4

x - 3 + 3 = 4 +3

Now, $-3$ and $+3$ are getting canceled on the left side while $4 +3$ simplifies to $+7$ on the right side.

Therefore, the solution for the equation $\boldsymbol{x - 3 = 4}$ is $\boldsymbol{x = 7}$ .

Consider the one-variable linear equation $\boldsymbol{5x = 20}$ . Let’s solve this equation.

The inverse of multiplication is division. When simplifying this equation, since $5 \times x$ leads to $5x$ , we divide both sides by $5$ .

5x = 20

\frac{5x}{5} = \frac{20}{5}

Now, $5x$ is divided by $5$ to obtain $x$ on the left side while $20$ divided by $5$ simplifies to $4$ on the right side.

Therefore, the solution for the equation $\boldsymbol{5x = 20}$ is $\boldsymbol{ x = 4}$ .

Consider the one variable linear equation $\boldsymbol{\frac{x}{3} = 2}$ . Let’s solve this equation.

The inverse of division is multiplication. When simplifying this equation, since $x \div 3$ leads to $\frac{x}{3}$ , we multiply both sides by $3$ .

\frac{x}{3} = 2

\frac{x}{3} \times 3 = 2 \times 3

Now, $\frac{x}{3}$ is multiplied by $3$ to obtain x on the left side while $2$ multiplied by $3$ simplifies to $6$ on the right side.

Therefore, the solution for the equation $\boldsymbol{\frac{x}{3} = 2}$ is $\boldsymbol{x = 6}$ .

John’s brother is $\boldsymbol{5}$ years older than John. If John’s brother is $\boldsymbol{21}$ years old, how old is John?

Let John’s age be $x$ . Since John is $5$ years older than John, his age in terms of $x$ should be $x + 5$ . But, his actual age is $21$ . Now we can construct an equation by setting the expression $x + 5$ equal to $21$ . Therefore, we can write the equation as $x + 5 = 21$ .

Now, we can solve the equation $x + 5 = 21$ .

x + 5 = 21

x + 5 - 5 = 21 - 5

x = 16

Therefore, John is $\boldsymbol{x = 16}$ years old.

Let’s consider a few more examples.

$y + 10 = 3$
$= y + 10 - 10 = 3 - 10$ (Subtracting $10$ from both sides)
$y = -7$
$p - 2.5 = -2$
$p - 2.5 + 2.5 = -2 + 2.5$ (Adding $2.5$ to both sides)
$p = 0.5$
$-x - 5 = -1$
$-x - 5 + 5 = -1 + 5$ (Adding $5$ to both sides)
$-x = 4$
$-x-1 = 4-1$ (Dividing both sides by $-1$ )
$x = -4$
$-5.2 + 4 = 21 - 4.3 + x + 2.4$
$-1.2 = 21 - 4.3 + 2.4 + x$
$-1.2 = 19.1 + x$
$x + 19.3 = -1.2$
$x + 19.3 - 19.3 = -1.2 - 19.3$ (Subtracting $19.3$ from both sides)
$x = -20.5$
$y - \frac{2}{3} = \frac{1}{2}$
$y - \frac{2}{3} + \frac{2}{3} = \frac{1}{2} + \frac{2}{3}$ (Adding $\frac{2}{3}$ to both sides)
$y = \frac{1 \times 3}{2 \times 3} + \frac{2 \times 2}{3 \times 2}$ (Common denominator is $6$ )
$y = \frac{3}{6} + \frac{4}{6}$
$y = \frac{7}{6}$
$\frac{b}{6} = \frac{2}{3}$
$\frac{b}{6} \times 6 = \frac{2}{3} \times 6$ (Common denominator is $6$ , so both sides are multiplied by $6$ )
$b = 2 \times 2$
$b = 4$
$2a = -7$
$\frac{2a}{2} = \frac{-7}{2}$ (Dividing both sides by $2$ )
$a = \frac{-7}{2}$
$\frac{3}{5} - 2 = \frac{1}{5} + y$
$\frac{1}{5} + y = \frac{3}{5} - 2$
$\frac{1}{5} + y - \frac{1}{5} = \frac{3}{5} - 2 - \frac{1}{5}$ (Subtracting $\frac{1}{5}$ from both sides)
$y = \frac{3}{5} - \frac{1}{5} - 2$
$y = \frac{2}{5} - 2$
$y = \frac{2}{5} - \frac{2 \times 5}{1 \times 5}$ (Common denominator is 5)
$y = \frac{2}{5} - \frac{10}{5}$
$y = \frac{-8}{5}$
The width of a rectangle is $5$ meters less than its length. If the width of the rectangle is $21$ meters, what is the length of the rectangle?

Let $x$ be the length of the rectangle. Let’s construct the equation in terms of $x$ and solve it.

$x - 5 = 21$

$x - 5 + 5 = 21 + 5$

$x = 26$

Therefore, the length of the rectangle is $\boldsymbol{26}$ meters.

Reminder!
When solving an equation, we can check for the accuracy of our answer by substituting the obtained value of the variable back into the original equation. After working out in the substitution, If the same number appears on both sides of the equation, then the accuracy is established. If not, we have made a mistake.
For example, to check the accuracy of the solution, let’s consider the equation $x - 1 = 14$ and solve it.

x - 1 = 14

x - 1 + 1 = 14 + 1

x = 15

Our solution is $x = 15$ . Now let’s substitute this value to both the left side and right side and check.

Substituting $x = 15$ to the left side,

$x - 1$

$= 15 - 1$

$= 14$

Already we have $14$ only on the right side, thus there is nothing to work out. We obtain the same answer which is $15$ on both sides when we do the substitution.

Therefore, we can conclude that our solution is correct.

1. Solve each of the following equations.

$x + 7 = 13$
$y - 10 = 15$
$p + 2.7 = -7$
$-1.3 + 7 = 12 - 2.3 + x - 3.6$
$-x - 2 = -4$
$y - \frac{2}{7} = \frac{1}{3}$
$\frac{t}{4} = \frac{1}{3}$
$3a = -16$
$\frac{4}{5} - 3 = \frac{1}{5} + y$
$\frac{4}{5} - 3 = \frac{1}{5} + 1 - x$

Answer

a. $x = 6$

b. $y = 25$

c. $p = -9.7$

d. $x = -0.4$

e. $x = 2$

f. $y = \frac{13}{21}$

g. $t = \frac{4}{3}$

h. $a = -\frac{16}{3}$

i. $y = -\frac{12}{5}$

j. $x = \frac{17}{5}$

2. Form an equation and solve each of the following.

A rope is 10 meters longer than a metal rod. If the rope is 28 meters long, how long is the metal rod?

The price of a tablet is $100 less than the price of a laptop. If the laptop costs $350, what is the price of the tablet?

c. Sarah is 3 years younger than her sister. If Sarah’s sister is 18 years old, how old is Sarah?

Answer

a. $x = 18 \text{ m}$

b. $t = \$250$

c. $s = 15 \text{ years}$

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