Question

The product of two numbers and the difference them are 289 and 20 respectively. find the numbers

Answer 1

x = 10 + √389, y = -10 + √389 // x = 29.7231, y = 9.72308

or

x = 10 - √389, y = -10 - √389 // x = -9.72308, y = -29.7231

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

Algebra I

• Solving systems of equations using substitution/elimination
• Standard Form: ax² + bx + c = 0
• Quadratic Formula:
  `x=(-b\pm√(b^2-4ac) )/(2a)`

Explanation:

Step 1: Define Systems

xy = 289

x - y = 20

Step 2: Rewrite Systems

x - y = 20

1. Subtract x on both sides: -y = 20 - x
2. Divide -1 on both sides: y = x - 20

Step 3: Redefine Systems

xy = 289

y = x - 20

Step 4: Solve for x

Substitution

1. Substitute in y: x(x - 20) = 289
2. Distribute x: x² - 20x = 289
3. Rewrite [SF]: x² - 20x - 289 = 0
4. Define variables: a = 1, b = -20, c = -289
5. Substitute [QF]:
   `x=(20\pm√((-20)^2-4(1)(-289)) )/(2(1))`
6. Exponents:
   `x=(20\pm√(400-4(1)(-289)) )/(2(1))`
7. Multiply:
   `x=(20\pm√(400+1156) )/(2)`
8. Add:
   `x=(20\pm√(1556) )/(2)`
9. Simplify:
   `x=(20\pm 2√(389) )/(2)`
10. Factor GCF:
    `x=(2(10\pm √(389) ))/(2)`
11. Divide:
    `x=10\pm √(389)`

Step 5: Solve for y

Possibility 1: x = 10 + √389

1. Define equation: x - y = 20
2. Substitute in x: (10 + √389) - y = 20
3. Isolate y term: -y = 20 - (10 + √389)
4. Isolate y: y = (10 + √389) - 20
5. Combine like terms: y = -10 + √389
6. Evaluate: y = 9.72308

Possibility 2: x = 10 - √389

1. Define equation: x - y = 20
2. Substitute in x: (10 - √389) - y = 20
3. Isolate y term: -y = 20 - (10 - √389)
4. Isolate y: y = (10 - √389) - 20
5. Combine like terms: y = -10 - √389
6. Evaluate: y = -29.7231

Step 6: Identify Solutions

Possibility 1:

x = 10 + √389, y = -10 + √389

x = 29.7231, y = 9.72308

Possibility 2:

x = 10 - √389, y = -10 - √389

x = -9.72308, y = -29.7231