Problem 31 - Precalculus Mathematics for Calculus

Question

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered pair form given in Example 6. $$\left\\{\begin{array}{l}\frac{1}{2} x+\frac{1}{3} y=2 \\\\\frac{1}{5} x-\frac{2}{3} y=8\end{array}\right.$$

Step-by-Step Solution

Answer

The system has a single solution: \((10, -9)\).

Step 1: Clear the Fractions

To easily work with the system, eliminate the fractions by finding the least common multiple (LCM) of the denominators in each equation. For the first equation, the denominators are 2 and 3 with LCM 6. Multiply the entire first equation by 6: \(3x + 2y = 12\). For the second equation, the denominators are 5 and 3 with LCM 15. Multiply the entire second equation by 15: \(3x -10y = 120\).

Step 2: Eliminate One Variable

Subtract the first equation \(3x + 2y = 12\) from the second equation \(3x - 10y = 120\) to eliminate \(x\).Subtracting gives:\((3x - 10y) - (3x + 2y) = 120 - 12\), resulting in \(-12y = 108\).

Step 3: Solve for y

From \(-12y = 108\), solve for \(y\) by dividing both sides by -12:\(y = -9\).

Step 4: Substitute y back to find x

Use \(y = -9\) in one of the original equations to find \(x\). Substitute into the equation \(3x + 2y = 12\):\(3x + 2(-9) = 12\).Simplify:\(3x - 18 = 12\)Add 18 to both sides:\(3x = 30\).Divide by 3:\(x = 10\).

Step 5: Solutions in Ordered Pair

The solution to the system is the ordered pair \((x, y)\).Thus, the solution is \((10, -9)\).

Key Concepts

Fractions

Fractions in equations can be tricky, but understanding how to work with them is crucial. Fractions represent a part of a whole, and in algebra, they are often seen when coefficients of variables are not whole numbers. When working with systems of equations that involve fractions, the first step is often to eliminate them. This makes the equations simpler and easier to solve.

• To clear fractions, convert the equation to one with whole numbers by multiplying every term by the least common multiple (LCM) of all the denominators.
• This multiplication cancels out the denominators, turning fractional coefficients into whole numbers.

In the given system, the first equation has fractions with denominators 2 and 3, while the second has denominations 5 and 3. By converting fractions into whole numbers, solving the system becomes more straightforward. Always check your multiplications to ensure you transform every term correctly.

Least Common Multiple

Finding the least common multiple (LCM) is a vital skill in dealing with equations that have fractions. The LCM of two or more numbers is the smallest number that is evenly divisible by each of them. This concept is especially useful when you need to eliminate fractions from an algebraic equation.

• For example, in the first equation, the LCM of 2 and 3 is 6.
• This means that multiplying each term by 6 will clear the fractions.

To determine the LCM, you can list the multiples of each denominator and find the smallest common multiple. Alternatively, use the prime factorization method to get the LCM by multiplying the highest power of each prime number seen in the factorizations. Applying this concept helps simplify the equation, making it easier to solve for the variables involved.

Variables Elimination

Eliminating variables is a common strategy for solving systems of equations. Once the equations are simplified, the next step is to eliminate one variable to make solving easier. This usually involves adding or subtracting the equations.

• You balance the system in a way that one of the variables cancels out.
• This transforms the system into a single equation with one variable.

In the given problem, the system is set up so that by subtracting the transformed first equation from the second, the variable \(x\) is eliminated. This strategy leads to a simpler equation involving only one variable, \(y\). Solving for this variable then becomes a straightforward task. Finally, the value of the eliminated variable is substituted back into one of the equations to find the complete solution, yielding the ordered pair that represents the solution to the system.