Problem 91 - Elementary Algebra

Question

In the following exercises, solve the systems of equations by substitution. $$ \left\\{\begin{array}{l} y=-2 x-1 \\ y=-\frac{1}{3} x+4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = -3\) and \(y = 5\).

Step 1 - Set the equations equal to each other

Since both equations equal to y, we can set them equal to each other: \(-2x - 1 = -\frac{1}{3}x + 4\).

Step 2 - Eliminate the fraction

Multiply every term in the equation by 3 to eliminate the fraction: \(3(-2x - 1) = 3(-\frac{1}{3}x) + 3(4)\) which simplifies to \(-6x - 3 = -x + 12\).

Step 3 - Solve for x

Add x to both sides to isolate terms containing x on one side:\(-6x + x - 3 = 12\), which simplifies to \(-5x - 3 = 12\).Then add 3 to both sides:\(-5x = 15\).Finally, divide by -5:\(x = -3\).

Step 4 - Substitute x back into one of the original equations to solve for y

Use the first equation \(y = -2x - 1\).Substitute \(x = -3\) into the equation:\(y = -2(-3) - 1\).This simplifies to:\(y = 6 - 1 = 5\).

Step 5 - Verify the solution

Substitute both \(x\) and \(y\) values into the second equation to verify correctness:\(y = -\frac{1}{3}x + 4\).Substitute \(x = -3\) and \(y = 5\):\(5 = -\frac{1}{3}(-3) + 4\).Simplifies to:\(5 = 1 + 4 = 5\).Therefore, the solution \((x, y) = (-3, 5)\) is correct.

Key Concepts

Substitution Method

The substitution method is a way to solve systems of equations. It involves solving one of the equations for one variable in terms of the other variable and then substituting this expression into the other equation.
This substitution allows us to solve for one variable first and then use that value to find the other variable.
In our case, we have two equations:
1. \(y = -2x - 1\)
2. \(y = -\frac{1}{3}x + 4\)

Since both equations are already solved for \(y\), we can set them equal to each other to eliminate \(y\), providing us a start for solving the system.

Eliminating Fractions

Eliminating fractions simplifies equations and makes algebraic manipulations easier.
In our problem, the second equation has a fraction: \(-\frac{1}{3}x\).

To clear the fraction, we multiply every term by 3. This will give:
\[ 3(-2x - 1) = 3\left(-\frac{1}{3}x + 4\right) \] Simplifying this we get:
\[ -6x - 3 = -x + 12 \] Now the equation is free of fractions, making it more straightforward to solve.

Algebraic Verification

After finding the potential solution for \(x\) and \(y\), we need to verify it by substituting these values back into the original equations to ensure both are satisfied.
In our case, we solved:
\(x = -3\)
and \(y = 5\).

To verify, replace \(x\) and \(y\) in the original equations:
For the first equation: \(y = -2x - 1\), substituting \(x = -3\) and \(y = 5\) shows:
\[ 5 = -2(-3) - 1 \] Simplifies to: \[ 5 = 6 - 1 = 5 \]
For the second equation: \(y = -\frac{1}{3}x + 4\), substituting \(x = -3\) and \(y = 5\) shows:
\[ 5 = -\frac{1}{3}(-3) + 4 \] Simplifies to: \[ 5 = 1 + 4 = 5 \]
Therefore, the solution \((x, y) = (-3, 5)\) is verified as correct.

Linear Equations

Linear equations are algebraic equations where each term is either a constant or the product of a constant and a single variable.
The standard form of a linear equation in two variables is \(Ax + By = C\), where \(A\), \(B\), and \(C\) are real numbers.
In our problem:
The first equation, \(y = -2x - 1\), can be written in standard form as \(2x + y = -1\).

The second equation, \(y = -\frac{1}{3}x + 4\), simplifies in standard form to \(x + 3y = 12\).
Linear equations graph as straight lines, and the solution to a system of linear equations is the point(s) where the lines intersect.
Here, the solution \((x, y) = (-3, 5)\) represents the intersection point of the two lines described by our equations.