Problem 41 - 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) $$ \left(\begin{array}{lr} 5(x+1)-(y+3)= & -6 \\ 2(x-2)+3(y-1)= & 0 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = -1\) and \(y = 3\).

Step 1: Simplify Each Equation

For each equation, distribute the numbers inside the brackets.For the first equation: \[5(x+1) - (y+3) = -6 \]Distribute:\[5x + 5 - y - 3 = -6 \]Simplify:\[5x - y + 2 = -6 \]For the second equation:\[2(x-2) + 3(y-1) = 0 \]Distribute:\[2x - 4 + 3y - 3 = 0 \]Simplify:\[2x + 3y - 7 = 0 \]

Step 2: Arrange Equations in Standard Form

Rewrite each equation in the standard form of a linear equation: AX + BY = C.Rewritten first equation:\[5x - y = -8\]Rewritten second equation:\[2x + 3y = 7\]

Step 3: Choose the Elimination Method

Observe the coefficients of the variables to decide which method to use. We choose elimination because the coefficients make this method straightforward after adjustment.

Step 4: Eliminate One Variable

To eliminate one of the variables, let us eliminate \(y\) by making the coefficients of \(y\) in both equations equal.Multiply the first equation by 3 to get the coefficients of \(y\) to match:\[3(5x - y) = 3(-8) \]This results in:\[15x - 3y = -24 \]Subtract this new equation from the second equation:\[2x + 3y = 7 \]Add:\[(2x + 3y) + (15x - 3y) = 7 + (-24) \]Simplified:\[17x = -17 \]

Step 5: Solve for x

From the previous equation, solve for \(x\).\[17x = -17 \]Divide both sides by 17:\[x = -1 \]

Step 6: Substitute to Find y

Use the value of \(x\) in one of the original equations to find \(y\). Using the first modified equation:\[5x - y = -8 \]Substitute \(x = -1\):\[5(-1) - y = -8 \]Simplify:\[-5 - y = -8 \]Add 5 to both sides:\[-y = -3 \]Multiply by -1:\[y = 3\]

Step 7: Write the Solution

Combine the values of \(x\) and \(y\) to express the solution to the system of equations:\[x = -1, y = 3\]

Key Concepts

Elimination Method

The elimination method is an effective way to solve systems of linear equations. It involves combining two equations to eliminate one of the variables, hence the name "elimination." This method can be particularly useful when the coefficients of one variable are easily manipulated to be equal.

Here's how it works:

• You first need to arrange the equations in the standard linear form, which is typically \( AX + BY = C \).
• Next, decide which variable to eliminate. In this case, we chose to eliminate \( y \).
• Manipulate equations, such as multiplying one or both of them, to ensure that the coefficients of one variable become opposites in each equation. For instance, if one equation has \( +3y \) and the other \( -3y \), adding them will eliminate \( y \).
• After elimination, you'll have a single-variable equation which you can solve easily.

Once you've found the value of the single variable, go back to one of the original equations, substitute this value, and solve for the second variable. The elimination method is handy when it makes the computation straightforward and quick.

Substitution Method

The substitution method offers another approach to solving systems of equations and is especially useful when one of the equations is already solved for one of the variables, or can be solved easily.

Here's a step-by-step on how you can use the substitution method:

• First, solve one of the equations for one variable in terms of the other. Suppose you have an equation in the form of \( y = 5x - 2 \).
• Substitute the expression found into the other equation. This reduces the system to a single equation with one variable.
• Solve this single equation for the remaining variable. Once you find this value, use it to find the other variable by substituting back into the initial equation.
• Check your solutions by plugging them back into the original equations to ensure they satisfy both equations.

This method is often straightforward and reduces the complexity of handling two equations at once. It gives a clear pathway by breaking down a system into simpler parts.

Linear Equations

Linear equations are foundational in algebra and describe a line on a graph. They are written in the form \( AX + BY = C \), where \( A \), \( B \), and \( C \) are constants.

Key properties and characteristics include:

• Each variable is raised to the power of one, which is why they form straight lines when graphed.
• The solutions to linear equations involve finding the values of \( x \) and \( y \) that make both sides of the equation equal.
• In systems of linear equations, you're looking for a point where the graphs of all the equations intersect. This point represents the solution to the system.
• There are several methods to solve them, such as graphing, substitution, and elimination – each offering different strengths.

Linear equations form the basis for more complex algebra and even more advanced areas like calculus. Understanding them helps build a foundation for further mathematical learning.