Math Problem Statement

Solve the system of equations: (1/2)x - (1/3)y = -5 and 2x - (1/5)y = 2/5.

Solution

We are given the following system of equations:

12x13y=5\frac{1}{2}x - \frac{1}{3}y = -5 2x15y=252x - \frac{1}{5}y = \frac{2}{5}

Let's solve this step-by-step using substitution or elimination. First, we can eliminate the fractions by multiplying each equation by the least common denominator (LCD) of the fractions in each.

Step 1: Eliminate fractions in the first equation

The first equation is: 12x13y=5\frac{1}{2}x - \frac{1}{3}y = -5 The LCD of 2 and 3 is 6. Multiply the entire equation by 6: 6×(12x13y)=6×(5)6 \times \left(\frac{1}{2}x - \frac{1}{3}y\right) = 6 \times (-5) This simplifies to: 3x2y=30(Equation 1)3x - 2y = -30 \quad \text{(Equation 1)}

Step 2: Eliminate fractions in the second equation

The second equation is: 2x15y=252x - \frac{1}{5}y = \frac{2}{5} Multiply the entire equation by 5 to eliminate the fractions: 5×(2x15y)=5×255 \times \left(2x - \frac{1}{5}y\right) = 5 \times \frac{2}{5} This simplifies to: 10xy=2(Equation 2)10x - y = 2 \quad \text{(Equation 2)}

Step 3: Solve the system of equations

We now have the system: 3x2y=30(Equation 1)3x - 2y = -30 \quad \text{(Equation 1)} 10xy=2(Equation 2)10x - y = 2 \quad \text{(Equation 2)}

We can use substitution or elimination to solve this system. Let's use elimination by first multiplying Equation 2 by 2 so that the coefficients of yy match:

2(10xy)=2(2)2(10x - y) = 2(2) 20x2y=4(Equation 3)20x - 2y = 4 \quad \text{(Equation 3)}

Now subtract Equation 1 from Equation 3:

(20x2y)(3x2y)=4(30)(20x - 2y) - (3x - 2y) = 4 - (-30) 20x3x=4+3020x - 3x = 4 + 30 17x=3417x = 34 x=2x = 2

Step 4: Substitute x=2x = 2 into one of the original equations

Substitute x=2x = 2 into Equation 2: 10(2)y=210(2) - y = 2 20y=220 - y = 2 y=18y = 18

Final Solution:

The solution to the system is: x=2,y=18x = 2, \quad y = 18

Would you like more details on any part of the process or have any questions?


Here are some related questions for further exploration:

  1. How do you solve a system of equations using the substitution method?
  2. What is the process of solving a system of equations with fractions?
  3. How does elimination work for solving systems of equations?
  4. What happens when you have no solution or infinitely many solutions to a system?
  5. Can systems of equations be solved graphically? How?

Tip: When dealing with fractions in equations, multiplying through by the least common denominator is a quick way to simplify the system.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Systems
Fractions

Formulas

Elimination Method
Substitution Method

Theorems

Elimination Theorem for Linear Systems

Suitable Grade Level

Grades 9-12