Question

Solve each equation algebraically. a) \(\frac{5 x}{3 x+4}=7\) b) \(2=\frac{20-3 x}{x}\) c) \(\frac{x^{2}}{x-2}=x-6\) d) \(1+\frac{2}{x}=\frac{x}{x+3}\)

Short answer

a) \( x = -\frac{7}{4} \); b) \( x = 4 \); c) \( x = \frac{3}{2} \); d) \( x = -\frac{6}{5} \).

Step-by-step solution

Solve equation (a)

Start by isolating the variable. The equation is \ \( \frac{5 x}{3 x + 4} = 7 \). Multiply both sides by \( 3 x + 4 \) to get: \ \( 5 x = 7 (3 x + 4) \). Simplify and solve for \( x \): \ \( 5 x = 21 x + 28 \) \ Subtract \( 21 x \) from both sides: \ \( 5 x - 21 x = 28 \) \ \( -16 x = 28 \) \ Divide both sides by \( -16 \) to get: \ \( x = -\frac{28}{16} = -\frac{7}{4} \).

Solve equation (b)

Rearrange the given equation \( 2 = \frac{20 - 3 x}{x} \) by multiplying both sides by \( x \): \ \( 2 x = 20 - 3 x \). Combine like terms: \ \( 2 x + 3 x = 20 \) \ \( 5 x = 20 \). Divide both sides by \( 5 \): \ \( x = \frac{20}{5} = 4 \).

Solve equation (c)

Begin with \( \frac{x^2}{x - 2} = x - 6 \). Multiply both sides by \( x - 2 \) to clear the fraction: \ \( x^2 = (x - 6)(x - 2) \). Expand the right-hand side: \ \( x^2 = x^2 - 8 x + 12 \). Set the equation to zero: \ \( 0 = -8 x + 12 \). Solve for \( x \): \ \( 8 x = 12 \) \ \( x = \frac{12}{8} = \frac{3}{2} \).

Solve equation (d)

Starting from \( 1 + \frac{2}{x} = \frac{x}{x + 3} \), first subtract \( 1 \) from both sides: \ \( \frac{2}{x} = \frac{x}{x + 3} - 1 \). Express \( 1 \) as a fraction with a common denominator: \ \( \frac{x}{x + 3} - 1 = \frac{x - (x + 3)}{x + 3} = \frac{x - x - 3}{x + 3} = \frac{-3}{x + 3} \). This simplifies the equation to: \ \( \frac{2}{x} = \frac{-3}{x + 3} \). Cross-multiply to solve for \( x \): \ \( 2(x + 3) = -3 x \). Simplify: \ \( 2 x + 6 = -3 x \) \ \( 2 x + 3 x = -6 \) \ \( 5 x = -6 \) \ \( x = -\frac{6}{5} \).

Key concepts

Linear Equations

Linear equations are the simplest type of algebraic equations. They describe a straight line when plotted on a graph. The structure of a linear equation is usually in the form of \(ax + b = 0\).The goal is to solve for the variable, x.
To solve a linear equation, follow these steps:

• Combine like terms on each side of the equation.
• Isolate the variable on one side by using basic arithmetic operations such as addition, subtraction, multiplication, or division.

For example, consider solving the equation \(5x = 21x + 28\):

• Subtract 21x from both sides: \(5x - 21x = 28\) forms a simpler equation, \(-16x = 28\).
• Divide both sides by \(-16\) to isolate x: \(x = -\frac{28}{16} = -\frac{7}{4}\).

Remember, the key to solving linear equations is to isolate the variable by performing inverse operations.

Rational Equations

Rational equations involve expressions that are fractions containing polynomials in the numerator and denominator. To solve a rational equation:

• Identify the least common denominator (LCD) of all rational expressions in the equation.
• Multiply each term by the LCD to clear the fractions.

For example, consider the equation \(2 = \frac{20 - 3x}{x}\):

• Multiply both sides by x: \(2x = 20 - 3x\).
• Combine like terms: \(2x + 3x = 20\) simplifies to \(5x = 20\).
• Divide both sides by 5 to isolate x: \(x = 4\).

The primary goal while solving rational equations is to simplify and eliminate fractions, making it easier to solve for the variable.

Cross-Multiplication

Cross-multiplication is a useful technique when dealing with an equation containing two fractions set equal to each other. This method simplifies the equation by eliminating the denominators. Here’s how to do it:

• If you have \(\frac{a}{b} = \frac{c}{d}\), you multiply `a` by `d` and `b` by `c`:
  \(a \times d = b \times c\).

For instance, consider solving \(\frac{2}{x} = \frac{-3}{x + 3}\) through cross-multiplication:

• Cross-multiply to get \(2(x + 3) = -3x\).
• Simplify to obtain \(2x + 6 = -3x\).
Combine like terms: \(2x + 3x = -6\) gives you \(5x = -6\).
• Divide both sides by 5 to solve for x: \(x = -\frac{6}{5}\).

Cross-multiplication helps to quickly eliminate fractions, making it easier to solve the equation.

Variable Isolation

Variable isolation is the process of manipulating an equation to get the variable of interest by itself on one side. This step is crucial in solving all types of equations, whether linear, rational, or others. Follow these basic steps:

• Use inverse operations (addition/subtraction, multiplication/division) on both sides of the equation to simplify.
• Perform the same operation on both sides to keep the equation balanced.

For instance, given \(\frac{x^2}{x - 2} = x - 6\), isolated the variable by:

• Multiply both sides by \(x - 2\) to clear the fraction: \(x^2 = (x - 6)(x - 2)\).
• Expand the right-hand side: \(x^2 = x^2 - 8x + 12\).
• Set the equation to zero: \(0 = -8x + 12\).
• Solve for x: \(8x = 12\) gives \(x = \frac{12}{8} = \frac{3}{2}\).

The primary goal is to have the variable on one side and a number or an expression on the other. This simplification makes it easier to see the solution.