Problem 45 - A Graphical Approach to Precalculus with Limits

Question

Solve each system by elimination. $$\begin{array}{l}\frac{2 x-1}{3}+\frac{y+2}{4}=4 \\\\\frac{x+3}{2}-\frac{x-y}{3}=3\end{array}$$

Step-by-Step Solution

Answer

The solution is \((x, y) = (5, 2)\).

Step 1: Clear Fractions in the First Equation

Multiply each term in the first equation by the least common multiple of the denominators, which is 12:\[4(2x - 1) + 3(y + 2) = 48\]This simplifies to:\[8x - 4 + 3y + 6 = 48\]Which simplifies further to:\[8x + 3y = 46\]

Step 2: Clear Fractions in the Second Equation

Multiply each term in the second equation by the least common multiple of the denominators, which is 6:\[3(x + 3) - 2(x - y) = 18\]Distribute the terms:\[3x + 9 - 2x + 2y = 18\]Combine like terms:\[x + 2y + 9 = 18\]Subtract 9 from both sides:\[x + 2y = 9\]

Step 3: Align the System for Elimination

Now, write the two equations for elimination:\[\begin{align*}8x + 3y &= 46 \x + 2y &= 9\end{align*}\]

Step 4: Eliminate x by Substitution

Multiply the second equation by 8 to eliminate \(x\):\[8(x + 2y) = 8(9)\]Which becomes:\[8x + 16y = 72\]

Step 5: Subtract the Equations

Subtract the first equation from the modified second equation:\[\begin{align*}(8x + 16y) - (8x + 3y) &= 72 - 46 \16y - 3y &= 26\end{align*}\]Simplify to find \(y\):\[13y = 26\]Divide by 13:\[y = 2\]

Step 6: Substitute y back to find x

Substitute \(y = 2\) back into the simplified second equation:\[x + 2(2) = 9\]Simplify:\[x + 4 = 9\]Subtract 4 from both sides:\[x = 5\]

Step 7: Concluding Step: Verify the Solution

Substitute \(x = 5\) and \(y = 2\) into both original equations to verify the solution:First equation:\[\frac{2(5) - 1}{3} + \frac{2 + 2}{4} = 4 \Rightarrow \frac{9}{3} + \frac{4}{4} = 3 + 1 = 4\]Second equation:\[\frac{5+3}{2} - \frac{5-2}{3} = 3 \Rightarrow 4 - 1 = 3\]Since both verifications hold, the solution \((x, y) = (5, 2)\) is correct.

Key Concepts

System of Equations

A system of equations consists of two or more equations with the same set of variables. The main objective in solving a system of equations is to find the values of these variables that satisfy all the equations simultaneously. In the given exercise, we are working with a system of two equations:

• \( \frac{2x-1}{3}+\frac{y+2}{4}=4 \)
• \( \frac{x+3}{2}-\frac{x-y}{3}=3 \)

When tackling such problems, the elimination method is very useful. This method involves manipulating the equations to eliminate one of the variables, making it easier to solve for the remaining variable. Once one variable is solved, it is substituted back into one of the original equations to find the other variable. This process is iterative, refining the solution with each step until all variables are determined. By effectively using the elimination method, you can simplify and solve complex systems of equations with ease.

Fraction Equations

Fraction equations are those in which the variables are within fractions, as seen in our system. To solve these equations, a common step is to clear the fractions by finding the least common multiple (LCM) of the denominators. This allows us to remove the fractions, transforming the equation into an easier format to handle.

In our exercise, for the first equation, the LCM of 3 and 4 is 12. By multiplying through by 12, the equation no longer has fractions:

• \( 4(2x - 1) + 3(y + 2) = 48 \)

Similarly, for the second equation, the LCM of 2 and 3 is 6, so multiplying by 6 clears those fractions:

• \( 3(x + 3) - 2(x - y) = 18 \)

By handling the fractions upfront with algebraic manipulations, equations become linear and significantly easier to solve.

Algebraic Manipulation

Algebraic manipulation is a critical skill in solving equations, especially systems involving fractions. It involves rearranging and simplifying equations to reveal the values of unknowns. In this exercise, we've used several techniques:

• Clearing fractions by multiplying each term by the LCM of the denominators.
• Combining like terms, as seen when converting \( 8x - 4 + 3y + 6 = 48 \) to \( 8x + 3y = 46 \).
• Aligning terms for elimination by creating equivalent forms of equations, such as multiplying the second equation by 8 to match coefficients for elimination.

After manipulating the equations, subtraction helped eliminate one variable, simplifying the system further. Algebraic manipulation in solving such problems requires accuracy and attention to detail, but with practice, it becomes a mastered tool in your problem-solving arsenal.