Problem 33 - Intermediate Algebra

Question

For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$ \left(\begin{array}{l} -2 x+5 y=-16 \\ x=\frac{3}{4} y+1 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( (x, y) = (-2, -4) \).

Step 1: Identify the method

Since the second equation is already solved for \( x \), substitution appears to be the appropriate method. We can substitute the expression for \( x \) from the second equation into the first equation.

Step 2: Substitute the expression for x

Replace \( x \) in the first equation with \( \frac{3}{4}y + 1 \) from the second equation: \(-2(\frac{3}{4}y + 1) + 5y = -16\).

Step 3: Distribute and simplify

Distribute \(-2\) across \(\left(\frac{3}{4}y + 1\right)\), yielding \(-\frac{3}{2}y - 2\). The equation becomes: \(-\frac{3}{2}y - 2 + 5y = -16\).

Step 4: Combine like terms

Combine \(-\frac{3}{2}y\) and \(5y\): \( \frac{7}{2}y - 2 = -16 \).

Step 5: Solve for y

Add \(2\) to both sides to isolate the \( y \) term: \( \frac{7}{2}y = -14 \).Now, multiply both sides by \( \frac{2}{7} \) to solve for \( y \): \( y = -4 \).

Step 6: Solve for x using y

Substitute \( y = -4 \) back into the second original equation: \(x = \frac{3}{4}(-4) + 1 = -3 + 1 = -2\).

Step 7: Write the solution

The solution to the system of equations is \( (x, y) = (-2, -4) \).

Key Concepts

Substitution Method

The substitution method is a handy tool for solving systems of equations, especially when one of the equations is already solved for one variable. In this method, you replace one variable with an equivalent expression from another equation. For our specific problem, the second equation already expresses \( x \) in terms of \( y \), making substitution the ideal choice. When substituting, be sure to replace the entire variable with the expression, ensuring all instances in the equation are accounted for. This creates a single-variable equation that is generally easier to solve.
For example, in our case: \(-2x + 5y = -16\) and \( x = \frac{3}{4} y + 1 \). By substituting \( x\) in the first equation with \( \frac{3}{4}y + 1 \), the equation transforms into a one-variable equation that we can solve to find the value of \( y\). Once you have found \( y\), use it to calculate the value of \( x\) by substituting back into the initial expression for \( x\). This completes the process efficiently, finding a solution to the system of equations.

Elimination Method

The elimination method, another technique for solving systems of equations, involves manipulating the equations to eliminate one of the variables. This is done so that you can solve for the remaining variable quickly. The elimination method is particularly powerful when both equations are in standard form and easy to align or adjust for elimination.
To use this method, you might start by adding or subtracting entire equations to cancel a variable. Multiplying one or both equations by appropriate values can also achieve alignment for elimination. For example, if solving a different system, you might first adjust coefficients so that adding or subtracting equations removes one variable. Once one variable is eliminated, the resulting simple equation lets you solve for the other variable.
While our exercise applied substitution, the elimination method works best when substitution isn't straightforward, and equations are naturally aligned for cancellation.

Algebraic Solutions

Algebraic solutions refer to solving problems using algebraic manipulations to find the exact values of variables. In solving systems of equations, both substitution and elimination methods are algebraic techniques. The strength of algebraic solutions lies in arriving at precise and exact answers rather than approximations.
When solving systems algebraically, always check each step for accuracy. This ensures solutions are correct and consistent. Solving systems algebraically involves combining like terms, distributing constants, and rearranging equations to isolate variables. These techniques aid in methodically reaching the solution, as shown step by step in our exercise.
This thorough method results in explicit values for variables, as we found \( x = -2 \) and \( y = -4 \) in the exercise above. By examining each step critically, students can grasp the logic behind algebraic manipulations, resulting in a clearer understanding of systems of equations.