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Some equations have variables on both sides of the equation. In such cases, the equations have different types of solutions. Learn how to use the properties of the equations to determine the nature of the solution of the equation and to find the solution. ...Read MoreRead Less

- Solving equations with variables on both sides of the equation
- Two sides of an equation
- Solution of an equation
- Solving an equation using properties of equalities
- Using the distributive property to solve an equation
- Equations with no solution
- Equations with infinitely many solutions
- Frequently Asked Questions

An equation is a mathematical statement that connects two equal expressions with an “equal to” sign. 2x + 5 = 25 – 2x, y – 6 = 10 – 3y, and 8z – 6 = 29 – z are some examples of equations having variables on both sides. The unknown values like x, y, and z found in these equations are known as variables. We can find the values of variables of an equation by rearranging the terms using some special properties.

Every equation has two sides: the expression on the left-hand side (LHS) of the equation and the expression on the right-hand side of the equation (RHS). In every true equation, the left-hand side will be equal to the right-hand side (LHS = RHS).

A solution of an equation is a value that can be used in the place of the variable to make a true statement. For example, the solution of x in the equation x – 2 = 0 is 2. If we use 2 in the place of x, we get (2) – 2 = 0, which is a true statement. Some equations have only one solution. Certain equations have no solution. And certain equations have infinitely many solutions.

While solving an equation, we are essentially finding the value of the unknown variable. So, the solution of an equation is the value of the variable. All equations can be solved by simplifying the terms on both sides and isolating the variable to find its solution. We need to use inverse math operations to isolate the variable. The general procedure for solving multi-step equations is as follows:

First, we use the distributive property to expand the terms inside the parentheses. Then, we combine the like terms and simplify the equation. Finally, we isolate the variable using inverse operations.

**For example**, let’s solve the equation y – 6 = 10 – 3y.

y – 6 = 10 – 3y Write the equation

Add 6 on both sides using the addition property of equality to simplify the equation.

y – 6 + 6 = 10 – 3y + 6 Subtraction property of equality

y = 16 – 3y Subtract

Add 3y on both sides using the addition property of equality to simplify the equation.

y + 3y = 16 – 3y + 3y Subtraction property of equality

4y = 16 Subtract

Divide both sides by 4 using the division property of equality.

\(\frac{4y}{4}\) = \(\frac{16}{4}\) Division property of equality

y = 4 Simplify

So, the value of y in this equation is 4.

The distributive property states that multiplying the sum of two or more addends by a number is the same as multiplying each addend individually by the number and then adding the products together. We can use distributive property to expand the terms inside the parentheses.

Suppose you want to solve the equation 2 ( 2x + 5 ) = 25 – x. The first step is to expand the terms inside the parentheses using distributive property.

2 ( 2x + 5 ) = 25 -x Write the equation

4x + 10 = 25 – x Distributive property

Now, we combine the like terms together. We can add x on both sides.

4x + 10 + x = 25 -x + x

5x + 10 = 25 Add

We can subtract 10 on both sides using the subtraction property of equality.

5x + 10 – 10 = 25 – 10

5x = 15 Subtract

Now, divide both sides by 5 using the division property of equality to isolate the variable.

\(\frac{5x}{5}\) = \(\frac{15}{5}\) Division property of equality

x = 3 Simplify

So, the solution of this equation is x = 3.

Unlike the examples we solved before, certain equations may not have any solution. When solving such equations, we will get an equivalent equation that is not a true statement for any value of the variable. For example, the final equation or statement might be 5 = 0, which is not true.

5x + 3 = 5x + 5 is an example of an equation that has no solution.

The first step in solving the equation is to group the like terms.

5x + 3 = 5x + 5 Write the equation

5x + 3 – 5x = 5x + 5 – 5x Subtraction property of equality

3 = 5

We write the final equation as 3 = 5, which is not a true statement. Hence, the equation 5x + 3 = 5x + 5 does not have any solution.

When solving equations with infinitely many solutions, we will get an equivalent solution that is true for all values of the variable. 8x + 3 = 8x + 3 is an example of an equation with infinitely many solutions. To solve this equation, we can group the common terms.

8x + 3 = 8x + 3 Write the equation

-8x -8x Subtraction property of equality

3 = 3

The solution 3 = 3 is true for all values of the variable. Hence, the equation has infinitely many solutions.

**Examples:**

**Example 1:** Solve 5 ( x – 1 ) = -4

**Solution:**

5 ( x – 1 ) = -4 Write the equation

5x – 5 = -4 Distributive property

5x – 5 + 5 = -4 + 5 Addition property of equality

5x = 1 Add

\(\frac{5x}{5}\) = \(\frac{1}{5}\) Division property of equality

\(x=\frac{1}{5}\) Simplify

Therefore, the solution of the equation is \(x=\frac{1}{5}\).

**Example 2:** Solve 4 ( x – 3 ) = 4x – 12.

**Solution:**

4 ( x – 3 ) = 4x – 12 Write the equation

4x – 12 = 4x – 12 Distributive property

4x – 12 + 12 = 4x – 12 + 12 Addition property of equality

4x = 4x Add

The equation has an infinite number of solutions as it is true for all values of x.

**Example 3:** Solve 8x – 3 = 8x + 5 .

**Solution:**

8x – 3 = 8x + 5 Write the equation

8x – 3 – 8x = 8x + 5 – 8x Subtraction property of equality

-3 = 5 Subtract

The final equation is not a true statement. Hence, the equation has no solution.

**Example 4:** A train starts its journey from New York with 250 passengers, stops at Philadelphia, and then reaches its destination at Washington. The number of passengers who boarded the train at Philadelphia is 3 times the number of passengers who got down at Philadelphia. Find the number of passengers who boarded the train from Philadelphia if 320 passengers got down at the final destination.

**Solution:**

Number of passengers at the beginning of the journey = 250

Let the number of passengers who got down at Philadelphia be x.

So, the number of passengers who boarded from Philadelphia is 3x.

Number of passengers who got down at the final destination = 320

An equation relating the number of passengers can be formed as follows:

250 – x + 3x = 320

Here, we are subtracting x from 250 because x number of passengers got down at Philadelphia. At the same time, 3x passengers boarded the train from the same station. So, we add 3x to the left hand side of the equation.

250 – x + 3x = 320 Write the equation

250 + 2x = 320 Group the like terms

250 + 2x – 250 = 320 – 250 Subtraction property of equality

2x = 70 Subtract

\(\frac{2x}{2}\) = \(\frac{70}{2}\) Division property of equality

x = 35 Simplify

The solution of the equation 250 – x + 3x = 320 is 35.

So, 35 passengers got down at Philadelphia, and 3 x 35 = 105 passengers boarded the train from the same station.

Frequently Asked Questions

An expression is a mathematical phrase that contains numbers and variables. Expressions will not have an “equal to” sign. An equation is a mathematical statement that connects two equal expressions with an “equal to” sign

No, a linear equation has either a unique solution, no solution, or infinitely many solutions.