Math Problem Statement

Solve

Solve the following problems: - Problem #28: Solve the system of equations: 1.3x - 0.2y = 12, 0.4x + 17y = 89 - Problem #18: Cole's two student loans totaled $31,000. One loan was at 2.8% interest and the other at 4.5%. After one year, the total interest was $1024.40. Find the amount of each loan. - Problem #76: Subtract: (10xy - 4x^2y^2 - 3y^3) - (-9x^2y^2 + 4y^3 - 7xy) - Problem #80: Multiply: (2x - y)(2x + y)(4x^2 - y^2)

Solution

Let's break down the math problems one by one:


Problem #28:

Solve the system of equations:

1.3x0.2y=121.3x - 0.2y = 12 0.4x+17y=890.4x + 17y = 89

Step 1: Solve for xx or yy First, multiply the first equation by 5 to eliminate the decimal:

(5)(1.3x0.2y)=5(12)(5)(1.3x - 0.2y) = 5(12) 6.5xy=606.5x - y = 60 So, the system becomes: 6.5xy=60(1)6.5x - y = 60 \quad \text{(1)} 0.4x+17y=89(2)0.4x + 17y = 89 \quad \text{(2)}

Step 2: Solve the system using substitution or elimination. We can solve for yy from equation (1): y=6.5x60y = 6.5x - 60

Substitute this expression for yy in equation (2): 0.4x+17(6.5x60)=890.4x + 17(6.5x - 60) = 89 Expand and solve for xx: 0.4x+110.5x1020=890.4x + 110.5x - 1020 = 89 110.9x1020=89110.9x - 1020 = 89 110.9x=1109110.9x = 1109 x=1109110.9=10x = \frac{1109}{110.9} = 10

Step 3: Solve for yy: Substitute x=10x = 10 into the equation y=6.5x60y = 6.5x - 60: y=6.5(10)60=6560=5y = 6.5(10) - 60 = 65 - 60 = 5

Solution: x=10,y=5x = 10, \quad y = 5


Problem #18: Loan Problem

Let the two loans be:

  • Loan 1 at 2.8% interest = xx
  • Loan 2 at 4.5% interest = 31,000x31,000 - x

The total interest earned after one year is $1024.40. Using the simple interest formula I=P×r×tI = P \times r \times t where t=1t = 1 year, we can write the equation:

0.028x+0.045(31,000x)=1024.400.028x + 0.045(31,000 - x) = 1024.40

Step 1: Expand the equation: 0.028x+0.045(31,000)0.045x=1024.400.028x + 0.045(31,000) - 0.045x = 1024.40 0.028x+13950.045x=1024.400.028x + 1395 - 0.045x = 1024.40

Step 2: Combine like terms: 0.017x+1395=1024.40-0.017x + 1395 = 1024.40 0.017x=1024.401395-0.017x = 1024.40 - 1395 0.017x=370.60-0.017x = -370.60 x=370.600.017=21,800x = \frac{-370.60}{-0.017} = 21,800

Step 3: Calculate the other loan: Loan 2=31,00021,800=9,200Loan \ 2 = 31,000 - 21,800 = 9,200

Solution:

  • Loan 1: $21,800 at 2.8%
  • Loan 2: $9,200 at 4.5%

Problem #76:

Simplify:

(10xy4x2y23y3)(9x2y2+4y37xy)(10xy - 4x^2y^2 - 3y^3) - (-9x^2y^2 + 4y^3 - 7xy)

Step 1: Distribute the negative sign: 10xy4x2y23y3+9x2y24y3+7xy10xy - 4x^2y^2 - 3y^3 + 9x^2y^2 - 4y^3 + 7xy

Step 2: Combine like terms:

  • 10xy+7xy=17xy10xy + 7xy = 17xy
  • 4x2y2+9x2y2=5x2y2-4x^2y^2 + 9x^2y^2 = 5x^2y^2
  • 3y34y3=7y3-3y^3 - 4y^3 = -7y^3

Solution: 17xy+5x2y27y317xy + 5x^2y^2 - 7y^3


Problem #80:

Multiply:

(2xy)(2x+y)(4x2y2)(2x - y)(2x + y)(4x^2 - y^2)

Step 1: Multiply the first two factors using the difference of squares: (2xy)(2x+y)=4x2y2(2x - y)(2x + y) = 4x^2 - y^2

Step 2: Now multiply the result with the third factor: (4x2y2)(4x2y2)=(4x2y2)2(4x^2 - y^2)(4x^2 - y^2) = (4x^2 - y^2)^2

Solution: (4x2y2)2=16x48x2y2+y4(4x^2 - y^2)^2 = 16x^4 - 8x^2y^2 + y^4


Let me know if you'd like more details or explanations on any of these problems! Here are some related questions you could explore:

  1. How can we verify the solution of a system of linear equations graphically?
  2. What other methods could be used to solve the loan interest problem?
  3. Why is the difference of squares useful in simplifying certain types of polynomials?
  4. How would the loan problem change if compounded interest were involved instead of simple interest?
  5. What are the real-life applications of systems of linear equations?

Tip: When solving a system of equations, it's often useful to double-check your solution by plugging the values of the variables back into the original equations.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Systems of Equations
Interest Calculation
Simple Interest
Polynomial Operations
Difference of Squares

Formulas

Systems of linear equations solution methods (substitution, elimination)
Simple Interest Formula: I = P × r × t
Polynomial subtraction
Difference of squares: (a - b)(a + b) = a^2 - b^2

Theorems

Properties of linear equations
Simple Interest calculation
Polynomial rules
Difference of squares theorem

Suitable Grade Level

Grades 9-11