Linear Equations with One Varible

The topic of linear equations is one of the most basic and important for any student who is beginning with the study of algebra. This article will help you to learn about linear equations or reinforce your previous knowledge. Specifically, we will look at linear equations with one variable along with several exercises to facilitate understanding the process used to solve the equations.

ALGEBRA
linear equations with one variable

Relevant for…

Learning to solve linear equations with one variable.

See equations

ALGEBRA
linear equations with one variable

Relevant for…

Learning to solve linear equations with one variable.

See equations

Definition of linear equations

Equations are of linear when they can be written in the form ax+b=c, where x is a variable and a,~ b and c are known constants. When a linear equation is reduced to its simplest form, it contains only variables raised to the first power.

Thus, the equations x+3=-6 and 2x+4=x-2 are linear equations. The equation 3{{x}^2}+4x-2-3{{x}^2}=5 does not appear to be linear at first glance.

However, when we write it in its simplest form, combining like terms, the terms {{x}^2} disappear and the equation reduces to 4x=7. Therefore, this equation is linear.

There is exactly one solution to a linear equation with one variable.

EXAMPLES

The following are linear equations:

3x+5x+2=18

2x+5=5x-20

2{{x}^{2}}+3x-1=2{{x}^{2}}+10

5x+2{{x}^{3}}-2{{x}^{3}}=19


Rule of equations

We can perform any operation (addition, subtraction, multiplication, division) on one side of the equation, as long as we perform the same operation on the other side of the equation.

For example, suppose we have the equation 2x+5=5x-20.

If we wanted to, we could subtract 10 from both sides of the equation: 2x+5-10=5x-20-10.

Note that this time, it doesn’t help us much to have subtracted 10. When we solve equations, we want to perform operations in a way that simplifies the current equation and helps us solve the equation.

Now we are going to look at techniques for solving linear equations with one variable.


Method for solving linear equations with one variable

1. Simplify each side:

a. Remove the parentheses (using the distributive property).

b. Eliminate the fractions (multiplying both sides by the least common multiple).

c. Combine like terms.

2. Solve for the variable:

This means adding/subtracting variables to get the variables to be on only one side of the equation and adding/subtracting numbers to get the numbers to be on the other side of the equation.

3. Perform operations so that the x is alone.

4. Check your answer.


Linear equations with one variable – Examples with answers

EXAMPLE 1

  • Find the value of x in the equation 4x+5=17.

Solution: 

1. We do not have parentheses or fractions. There are also no like terms to combine.

2. Isolate the variable: move the 5 to the right:

4x+5-5=17-5

4x=12

3. We perform operations so that the x is alone: we divide both sides by 4: 

\frac{4}{4}x=\frac{{12}}{4}

x=\frac{{12}}{4}=3

4. Check your answer: we substitute the value in the original equation:

4\left( 3 \right)+5=17

12+5=17

17=17

This is true

Answer: x=3

EXAMPLE 2

  • Find the value of x in the equation 2x+6-2=-3x+14.

Solution: 

1. There are no parentheses or fractions. We can combine like terms: 

2x+4=-3x+14

2. Solve for the variable: we move the 4 to the right and the -3x to the left:  

2x+4-4=-3x+14-4

2x=-3x+10

2x+3x=-3x+10+3x

5x=10

3. We perform operations so that the x is alone: we divide both sides by 5: 

\frac{5}{5}x=\frac{{10}}{5}

x=\frac{{10}}{5}=2

4. Check your answer: we substitute the value in the original equation:

2(2)+6-2=-3(2)+14

4+6-2=-6+14

8=8

This is true

Answer: x=2

EXAMPLE 3

  • Find the value of x in the equation 2x+5(x+1)=26.

Solution: 

1. Simplify: 

We remove parentheses: 2x+5x+5=26.

There are no fractions.

We combine like terms: 7x+5=26.

2. Solve for the variable: we move the 5 to the right: 

7x+5-5=26-5

7x=21

3. We perform operations so that the x is alone: we divide both sides by 7: 

\frac{7}{7}x=\frac{{21}}{7}

x=\frac{{21}}{7}=3

4. Check your answer: we replace the value in the original equation:

2(3)+5(3+1)=26

6+5(4)=26

6+20=26

This is true

Answer: x=3

EXAMPLE 4

  • Find the value of x in the equation \frac{1}{2}x+4(x+2)=14-2x+20.

Solution: 

1. Simplify: 

We simplify the parentheses: \frac{1}{2}x+4x+8=14-2x+20.

We simplify fractions: x+8x+16=28-4x+40.

We combine like terms: 9x+16=68-4x.

2. Solve for the variable: we move the 16 to the right and the -4x to the left: 

9x+16-16=68-4x-16

9x=52-4x

9x+4x=52-4x-4x

13x=52

3. We perform operations so that the x is alone: we divide both sides by 13: 

\frac{13}{13}x=\frac{{52}}{13}

x=\frac{{52}}{13}=4

4. Check your answer: we replace the value in the original equation:

\frac{1}{2}(4)+4(4)+8=14-2(4)+20

2+16+8=14-8+20

26=26

This is true

Answer: x=4


Linear equations with one variable – Practice problems

Solve the equation 4x+10=2.

Choose an answer






Solve the equation 2(3x-4)=2x.

Choose an answer






Find the value of x in the equation 2(x+2)=5(x-1).

Choose an answer






Find the value of x in the equation 4(-2x-5)=-2(x+4).

Choose an answer







See also

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