Problem 71 - Precalculus

Question

Explain the mistake that is made. Solve the system of equations by substitution. $$ \begin{array}{r} x+3 y=-4 \\ -x+2 y=-6 \end{array} $$ Solution: Solve Equation (1) for \(x\) Substitute \(x=-3 y-4\) into Equation ( 2 ) Solve for \(y\) $$\begin{aligned} x=-3 y &-4 \\ -(-3 y-4)+2 y &=-6 \\ 3 y-4+2 y &=-6 \\ 5 y &=-2 \\ y &=-\frac{2}{5} \end{aligned}$$ Substitute \(y=-\frac{2}{5}\) into Equation (1) $$\begin{aligned} &x+3\left(-\frac{2}{5}\right)=-4\\\ &x=-\frac{14}{5} \end{aligned}$$ Solve for \(x\) The answer \(\left(-\frac{2}{5},-\frac{14}{5}\right)\) is incorrect. What mistake was made?

Step-by-Step Solution

Answer

The mistake was in expanding - (-3y - 4) incorrectly. The correct solution is \((x, y) = (2, -2)\).

Step 1: Identify the Mistake

The mistake occurred in the step where the expression for \(x\) from the first equation \(x = -3y - 4\) was substituted into the second equation. The substitution was done correctly, however, the equation \(-(-3y - 4) + 2y = -6\) was not expanded correctly which led to the wrong calculation.

Step 2: Correct the Expansion

Let's correctly expand the equation \(-(-3y - 4) + 2y = -6\). The correctly expanded form should be \(3y + 4 + 2y = -6\) instead of the incorrect \(3y - 4 + 2y = -6\) used initially.

Step 3: Solve the Corrected Equation for y

Combining like terms in the equation \(3y + 4 + 2y = -6\) gives us \(5y + 4 = -6\). Subtract 4 from both sides, giving \(5y = -10\). Divide by 5 to solve for \(y\), resulting in \(y = -2\).

Step 4: Substitute Correct y back into First Equation

Substitute \(y = -2\) into the first equation \(x + 3y = -4\). This gives us \(x + 3(-2) = -4\). Simplify to get \(x - 6 = -4\).

Step 5: Solve for x

Add 6 to both sides of the equation to isolate \(x\): \(x - 6 + 6 = -4 + 6\), therefore, \(x = 2\).

Key Concepts

Substitution Method

The substitution method is a technique used to solve a system of equations by expressing one variable in terms of another. This method involves a few straightforward steps:

• Solve one of the equations for one of the variables.
• Substitute this expression into the other equation.
• Solve the resulting single-variable equation.
• Use this solution to find the value of the other variable.

In the provided exercise, the first equation was solved for the variable \(x\) as \(x = -3y - 4\). This expression was then substituted into the second equation, allowing the system to be reduced into a single equation with one variable. When using the substitution method, it’s crucial to perform each arithmetic step carefully to avoid mistakes, especially during substitution and simplification.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to solve for variables. It is vital in solving systems of equations through methods like substitution.

In the correction step, algebraic manipulation was essential in rewriting the expanded form of the substituted equation correctly. Starting from \(-(-3y - 4) + 2y = -6\), correctly expanding this to \(3y + 4 + 2y = -6\) played a key role in solving the system accurately.

• Always expand parentheses fully and apply operations outside of them correctly.
• Combine like terms to simplify the equation.
• Perform consistent arithmetic operations on both sides of the equation to maintain equality.

Small arithmetic missteps in algebraic manipulation, such as in this exercise, can lead to incorrect solutions, highlighting the need for precision.

Solving Linear Equations

Solving linear equations involves finding the value of the unknown that satisfies the equation. This process is central in finding solutions for systems of linear equations. Once the substitution and algebraic manipulation are complete, you isolate the variable.

• First, gather all terms involving the variable on one side of the equation.
• Move constant terms to the other side.
• Finally, divide by the coefficient of the variable to solve it.

In the exercise, correctly solving the equation after manipulating it led to \(y = -2\). This value was then substituted back into the first equation to solve for \(x\), resulting in \(x = 2\). The importance of accuracy in each step cannot be overstressed, as in the case of verifying that the values satisfy both original equations.