Math Problem Statement

Solve the following algebraic equations: 1. 2x + 8 = 20 2. (3/2)x + 5 = 14 3. (x - 9)/(-4) = 7

Solution

I see the image contains three algebraic equations. Let me list and solve each one for you:

Equation 1:

2x+8=202x + 8 = 20

Steps to solve:

  1. Subtract 8 from both sides: 2x=208    2x=122x = 20 - 8 \implies 2x = 12
  2. Divide by 2: x=122    x=6x = \frac{12}{2} \implies x = 6

Equation 2:

32x+5=14\frac{3}{2}x + 5 = 14

Steps to solve:

  1. Subtract 5 from both sides: 32x=145    32x=9\frac{3}{2}x = 14 - 5 \implies \frac{3}{2}x = 9
  2. Multiply both sides by the reciprocal of 32\frac{3}{2} (which is 23\frac{2}{3}): x=9×23    x=6x = 9 \times \frac{2}{3} \implies x = 6

Equation 3:

x94=7\frac{x - 9}{-4} = 7

Steps to solve:

  1. Multiply both sides by 4-4 to eliminate the denominator: x9=7×(4)    x9=28x - 9 = 7 \times (-4) \implies x - 9 = -28
  2. Add 9 to both sides: x=28+9    x=19x = -28 + 9 \implies x = -19

Summary of Solutions:

  1. x=6x = 6
  2. x=6x = 6
  3. x=19x = -19

Would you like more details or further clarification?

Here are 5 related questions:

  1. How do you solve equations with fractions?
  2. What is the method to solve equations with variables on both sides?
  3. How do you deal with negative coefficients in equations?
  4. How does multiplying both sides of an equation affect the solution?
  5. Can we apply the same approach for solving systems of equations?

Tip: Always check your solution by substituting the value of xx back into the original equation to verify correctness.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Fractions in Equations

Formulas

Basic linear equation form: ax + b = c
Solving for x with fractional coefficients

Theorems

Multiplication Property of Equality
Addition Property of Equality

Suitable Grade Level

Grades 7-9