Question

Solve the equation \(2(\mathrm{x}+3)=(3 \mathrm{x}+5)-(\mathrm{x}-5)\).

Short answer

The given equation is \( 2(x + 3) = (3x + 5) - (x - 5) \). After simplifying and combining like terms, we get the false statement \( 6 = 10 \). Therefore, there is no solution for x, and the equation is inconsistent.

Step-by-step solution

Distribute constants

First, distribute the constant 2 on the left side of the equation: \( 2(x + 3) = 2x + 6 \). Now, the equation becomes: \[ 2x + 6 = (3x + 5) - (x - 5). \]

Simplify the parentheses

Simplify the equation by performing the subtraction on the right side: \[ (3x + 5) - (x - 5) = 3x + 5 - x + 5 = 2x + 10. \] Now, the equation becomes: \[ 2x + 6 = 2x + 10. \]

Combine like terms

Move the variable terms to one side by subtracting 2x from both sides: \[ 2x - 2x + 6 = 2x - 2x + 10. \] This simplifies down to: \[ 6 = 10. \]

Determine the solution

Since the resulting equation, \( 6 = 10 \), is false, there is no solution for x in the given original equation. This means that the equation is inconsistent, and there is no value of x that will make it true.

Key concepts

Equations

Equations are mathematical statements that assert the equality of two expressions. They serve as fundamental tools in algebra to express relationships between variables and constants. The basic goal when working with equations is to find the value of the unknown variable that makes the equality true. To solve an equation, you can perform operations such as addition, subtraction, multiplication, and division.
It is essential to perform any operation to both sides of the equation to maintain balance. For example, if you want to isolate a variable on one side, you might add or subtract terms from both sides or use multiplication and division. Solving equations becomes easier as you master these operations and understand the properties of different types of equations.

Distributive Property

The distributive property is a key algebraic rule that helps simplify expressions. It states that for any numbers or variables, multiplying a sum by a number is the same as multiplying each addend by the number and then adding the results. It looks like this:

• For any numbers or variables, a(b + c) = ab + ac.

In the context of equations, the distributive property allows us to eliminate parentheses by distributing the multiplication over addition or subtraction inside the parentheses.
In the original problem, you used the distributive property to expand \(2(x + 3)\)... to \(2x + 6\).
This step is crucial because it simplifies the equation and prepares it for further operations, like combining like terms.

Like Terms

Like terms in algebra are terms that have the same variable raised to the same power. When solving equations, it is essential to combine like terms to simplify expressions. Doing so reduces the complexity of the equation and makes the solving process straightforward.
For instance, terms like \(2x\) and \(3x\) are like terms because they both contain the variable \(x\) raised to the same power. You can combine these to get \(5x\). However, \(2x\) and \(x^2\) are not like terms; \(x\) in the first term is not raised to the same power as in the second term.
In the example exercise, you combined like terms on the right side to simplify \((3x + 5) - (x - 5)\) into \(2x + 10\). This simplification was crucial for identifying the type of equation you were dealing with.

Inconsistent Equations

Inconsistent equations have no solutions because no value for the variable will satisfy the equality. This usually occurs when, after simplifying the equation, you arrive at a statement that is mathematically false.
For example, ending with an equation like \(6 = 10\) shows inconsistency because 6 will never equal 10. This indicates that there was no solution possible for the original variable in the equation.
Inconsistent equations typically result from subtracting identical terms from both sides. After doing this in the exercise, you ended with the false statement \(6 = 10\), confirming the absence of solutions. Recognizing such cases is vital in algebra to understand when a problem has no feasible solution.