Problem

Solve for x (2x+3)/3=6/(x-1)

The problem consists of a rational equation where an algebraic fraction involving the variable x is set equal to another algebraic fraction containing x. The objective is to find the value(s) of x that make the equation true. This typically involves finding a common denominator, multiplying both sides of the equation to clear the fractions, and then solving for x using algebraic methods such as combining like terms, isolating x, and checking for any potential restrictions on the solution set (for example, x cannot be equal to values that would make any denominator zero).

2x+33=6x1

Answer

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Solution:

Step 1

Cross-multiply the terms of the given equation 2x+33=6x1.

Step 2

Isolate the variable x.

Step 2.1

First, expand the expression (2x+3)(x1).

Step 2.1.1

Utilize the FOIL (First, Outer, Inner, Last) method to expand (2x+3)(x1).

Step 2.1.1.1

Distribute 2x across (x1) and 3 across (x1), resulting in 2x(x1)+3(x1)=36.

Step 2.1.1.2

Continue distribution to obtain 2xx2x1+3(x1)=36.

Step 2.1.1.3

Complete distribution to get 2x22x+3x3=36.

Step 2.1.2

Combine like terms and simplify.

Step 2.1.2.1

Handle each term individually.

Step 2.1.2.1.1

Combine powers of x by adding exponents.

Step 2.1.2.1.1.1

Rewrite 2(xx)2x+3x3=36.

Step 2.1.2.1.1.2

Multiply x by x to get 2x22x+3x3=36.

Step 2.1.2.1.2

Multiply 1 by 2 to get 2x22x+3x3=36.

Step 2.1.2.1.3

Multiply 3 by 1 to get 2x22x+3x3=36.

Step 2.1.2.2

Combine 2x and 3x to get 2x2+x3=36.

Step 2.2

Calculate 36 to get 2x2+x3=18.

Step 2.3

Subtract 18 from both sides to get 2x2+x21=0.

Step 2.4

Factor the quadratic equation.

Step 2.4.1

Identify two numbers that multiply to 221=42 and add to 1.

Step 2.4.1.1

Express 1x as the sum of 6x+7x.

Step 2.4.1.2

Expand to 2x26x+7x21=0.

Step 2.4.2

Group and factor by grouping.

Step 2.4.2.1

Group as (2x26x)+(7x21)=0.

Step 2.4.2.2

Factor out the GCF from each group to get 2x(x3)+7(x3)=0.

Step 2.4.3

Factor out the common binomial (x3) to get (x3)(2x+7)=0.

Step 2.5

Solve for x by setting each factor equal to zero.

Step 2.5.1

Set x3=0 and solve for x to get x=3.

Step 2.5.2

Set 2x+7=0 and solve for x.

Step 2.5.2.1

Subtract 7 from both sides to get 2x=7.

Step 2.5.2.2

Divide by 2 to get x=72.

Step 2.6

Combine the solutions to get x=3,72.

Step 3

Present the solution in various forms.

Exact Form: x=3,72 Decimal Form: x=3,3.5 Mixed Number Form: x=3,312

Knowledge Notes:

  1. Cross-Multiplication: This technique involves multiplying the numerator of one fraction by the denominator of the other fraction and setting the products equal to each other. It is used to solve equations involving two fractions.

  2. FOIL Method: A mnemonic for remembering the steps to multiply two binomials: First, Outer, Inner, Last. It is a specific case of the more general method of distribution.

  3. Distributive Property: This property states that a(b+c)=ab+ac. It is used to expand expressions and to simplify equations.

  4. Combining Like Terms: This involves adding or subtracting terms that have the same variable raised to the same power.

  5. Factoring: The process of breaking down an expression into a product of simpler expressions. It can be used to solve quadratic equations by setting each factor equal to zero and solving for the variable.

  6. Quadratic Equations: These are polynomial equations of the second degree, generally in the form ax2+bx+c=0. They can be solved by factoring, completing the square, or using the quadratic formula.

  7. Factoring by Grouping: A factoring technique used when an expression has four or more terms. It involves grouping terms with common factors and then factoring out the greatest common factor from each group.

  8. Zero Product Property: This property states that if the product of two factors is zero, then at least one of the factors must be zero. It is used to find the roots of polynomial equations.

  9. Solving Linear Equations: The process of finding the value of the variable that makes the equation true. This often involves isolating the variable on one side of the equation using inverse operations.

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