Simple one-variable linear equations II

Consider the following equations.

$2x - 3 = 5$
$5 + 3x = x - 1$
$\frac{y}{2} - 10 = 1$
$x + 2 = 15 + \frac{x}{2}$

As we can see, the above equations can be identified as a combination of arithmetic operations, addition, subtraction, multiplication, and division. For example, the equation $2x - 3 = 5$ , $3$ is subtracted from $x$ multiplied by $2$ on the left side. So, multiplication and subtraction are involved. In this unit, we will discuss how we can solve such equations which have several arithmetic operations combined in the same equation.

Let’s discuss the solutions of the above equations.

Consider the equation $\boldsymbol{2x - 3 = 5}$ .

The left side of the equation consists of an algebraic expression $2x - 3$ while the right side contains only the integer $5$ .

As we want to isolate $x$ on the left side, first, we need to eliminate the $-3$ from the left side. We add $3$ to both sides to eliminate $-3$ from the left side.

2x - 3 = 5

2x - 3 + 3 = 5 + 3

2x = 8

Now, we have obtained $2x = 8$ as the simplified equation. To make the coefficient of $x$ equal to $1$ , we divide both sides of the equation $2x = 8$ by $2$ .

2x = 8

\frac{2x}{2} = \frac{8}{2}

x = 4

Therefore, the solution for the equation $\boldsymbol{2x - 3 = 5}$ is $\boldsymbol{x = 4}$ .

Consider the equation $\boldsymbol{5 + 3x = x - 1}$ .

In the above example, even though we had an integer on the right side, in this equation $5 + 3x = x - 1$ , we have expressions on both sides. Therefore, firstly, we isolate all the variable terms on the left while we move all the constants to the right side.

Isolating variable terms on the left side of the equation refers to eliminating $x$ from the right side. To do that, we subtract $x$ from both sides. Then, on the left, $x$ variable terms are simplified since they are like terms, and on the right side, $+x$ and $-x$ are getting canceled.

5 + 3x = x - 1

5 + 3x - x = x - 1 - x

5 + 2x = -1

Now we can eliminate $5$ from the left side by subtracting $5$ from both sides.

5 + 2x = -1

5 + 2x - 5 = -1 - 5

2x = -6

Now, we have obtained $2x = -6$ as the simplified equation. To make the coefficient of $x$ equal to $1$ , we divide both sides of the equation $2x = -6$ by $2$ .

2x = -6

\frac{2x}{2} = \frac{-6}{2}

x = -3

Therefore, the solution for the equation $\boldsymbol{5 + 3x = x - 1}$ is $\boldsymbol{x = -3}$ .

Consider the equation $\boldsymbol{\frac{y}{2} - 10 = 1}$ .

The left side of the equation consists of an algebraic expression $\frac{y}{2} - 10$ while the right side contains only the integer $1$ .

First, we eliminate $-10$ from the left side by adding $10$ to both sides.

\frac{y}{2} - 10 = 1

\frac{y}{2} - 10 + 10 = 1 + 10

\frac{y}{2} = 11

Now, we have obtained $\frac{y}{2} = 11$ as the simplified equation. To make the coefficient of $y$ equal to $1$ , we multiply both sides of the equation $\frac{y}{2} = 11$ by $2$ .

\frac{y}{2} = 11

\frac{y}{2} \times 2 = 11 \times 2

y = 22

Therefore, the solution for the equation $\boldsymbol{\frac{y}{2} - 10 = 1}$ is $\boldsymbol{y = 22}$ .

Consider the equation $\boldsymbol{x + 2 = 15 + \frac{x}{2}}$ .

In the equation $x + 2 = 15 + \frac{x}{2}$ , we have expressions on both sides. First, we isolate all the variable terms on the left while we move all the constants to the right. We eliminate $\frac{x}{2}$ from the right side by subtracting $\frac{x}{2}$ from both sides.

x + 2 = 15 + \frac{x}{2}

x + 2 - \frac{x}{2} = 15 + \frac{x}{2} - \frac{x}{2}

x - \frac{x}{2} + 2 = 15

Now, on the left, $x$ variable terms are simplified since they are like terms.

x - \frac{x}{2} + 2 = 15

\frac{2x}{2} - \frac{x}{2} + 2 = 15

\frac{x}{2} + 2 = 15

Now we can eliminate $2$ from the left side by subtracting $2$ from both sides.

\frac{x}{2} + 2 = 15

\frac{x}{2} + 2 - 2 = 15 - 2

\frac{x}{2} = 13

Now, we have obtained $\frac{x}{2} = 13$ as the simplified equation. To make the coefficient of $x$ equal to $1$ , we multiply both sides of the equation $\frac{x}{2} = 13$ by $2$ .

\frac{x}{2} = 13

\frac{x}{2} \times 2 = 13 \times 2

x = 26

Therefore, the solution for the equation $\boldsymbol{x + 2 = 15 + \frac{x}{2}}$ is $\boldsymbol{x = 26}$ .

Let’s consider a few more examples.

$5x + 2 = 27$
$5x + 2 - 2 = 27 - 2$ (Subtracting $2$ from both sides)
$5x = 25$
$\frac{5x}{5} = \frac{25}{5}$ (Dividing both sides by $5$ )
$x = 5$

$7 - 3y = -4 + 3y$
$7 -3y -3y = -4 + 3y -3y$ (Subtracting $3y$ from both sides)
$7 -6y = -4$
$7 -6y - 7 = -4 -7$ (Subtracting $7$ from both sides)
$-6y = -11$
$\frac{-6y}{-6} = \frac{-11}{-6}$ (Dividing both sides by $-6$ )
$y = \frac{11}{6}$

$4 + \frac{t}{5} = 3$
$4 + \frac{t}{5} -4 = 3 -4$ (Subtracting $4$ from both sides)
$\frac{t}{5} = -1$
$\frac{t}{5} \times 5 = -1 \times 5$ (Multiplying both sides by $5$ )
$t = -5$

$1 - x + 9 + x2 = 3x + 2$
$- x + \frac{x}{2} + 1 + 9 = 3x + 2$
$- x + \frac{x}{2} + 10 = 3x + 2$
$- x + \frac{x}{2} + 10 -3x = 3x + 2 -3x$ (Subtracting $3x$ from both sides)
$-4x + \frac{x}{2} + 10 = 2$
$\frac{-4x \times 2}{1 \times 2} + \frac{x}{2} + 10 = 2$
$\frac{-8x}{2} + \frac{x}{2} + 10 = 2$
$\frac{-7x}{2} + 10 = 2$
$\frac{-7x}{2} + 10 - 10 = 2 - 10$ (Subtracting $10$ from both sides)
$\frac{-7x}{2} = -8$
$\frac{-7x}{2} \times 2 = -8 \times 2$ (Multiplying both sides by $2$ )
$-7x = -16$
$\frac{-7x}{7} = \frac{-16}{7}$ (Dividing both sides by $7$ )
$x = \frac{-16}{7}$

Tom’s age is $5$ years less than twice his brother's age. If Tom is $11$ years old now, how is Tom’s brother?

Let Tom’s brother’s age be $x$ . Twice his brother's age is $2x$ . Therefore, $5$ less than twice his brother's age is $2x - 5$ . Now, we can construct the equation for Tom’s age as $2x - 5 = 11$ .

Let’s solve the equation and find the value.

$2x - 5 = 11$

$2x - 5 + 5 = 11 + 5$

$2x = 16$

$\frac{2x}{2} = \frac{16}{2}$

$x = 8$

Therefore, Tom’s brother is $\boldsymbol{8}$ years old.

1. Solve each of the following equations.

$3x + 4 = 19$
$9y - 7 = 47$
$5 - 2t = 3 + 4t$
$9 - 3y = 1 + 2y$
$6 + \frac{t}{2} = 7$
$- 2x + 9 + \frac{x}{3} = 2$
$5 - x + \frac{2x}{3} = 2x + 1$

Answer

a. $x = 5$

b. $y = 6$

c. $t = \frac{1}{3}$

d. $y = \frac{8}{5}$

e. $t = 2$

f. $x = \frac{21}{5}$

g. $x = \frac{12}{7}$

2. Find the solution for each of the following prolem.

a. Emily’s age is $3$ years more than four times her sister’s age. If Emily is $27$ years old now, how old is her sister?

b. The length of a rectangle is $3$ meters less than twice its width. If the rectangle's length is $9$ meters, what is the width of the rectangle?

Answer

a. Sister's age = $6$

b. Width of the rectangle = $6$

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Simple one-variable linear equations II

1. Solve each of the following equations.

2. Find the solution for each of the following prolem.