Problem 21 - Algebra A Combined Function

Question

Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or decimals.$$ \left\\{\begin{array}{l} 4 x-3 y=7 \\ 7 x+5 y=2 \end{array}\right. $$

Step-by-Step Solution

Answer

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

Step 1: Align and Simplify the Equations

Write down both equations as given: \( 4x - 3y = 7 \) and \( 7x + 5y = 2 \). Since there are no fractions or decimals, proceed to find coefficients that make it easy to eliminate a variable by addition.

Step 2: Determine Elimination Coefficients

Examine the coefficients of \(y\) in both equations: \(-3\) and \(5\). The least common multiple (LCM) of 3 and 5 is 15, so we'll aim to adjust the coefficients to eliminate \(y\).

Step 3: Multiply Each Equation to Align Coefficients

Multiply the first equation \(4x - 3y = 7\) by 5 and the second equation \(7x + 5y = 2\) by 3, resulting in the system: \[(5)(4x - 3y) = 5(7) \quad \Rightarrow \quad 20x - 15y = 35 \]\[(3)(7x + 5y) = 3(2) \quad \Rightarrow \quad 21x + 15y = 6\]

Step 4: Add Equations to Eliminate y

Add the two equations from Step 3: \[20x - 15y + 21x + 15y = 35 + 6 \]The \(y\) terms cancel out, giving us:\[ 41x = 41\]

Step 5: Solve for x

Divide both sides by 41: \[ x = \frac{41}{41} = 1\].

Step 6: Substitute x into One Equation

Use \(4x - 3y = 7\) to find \(y\). Substitute \(x = 1\):\[4(1) - 3y = 7\]This simplifies to: \[4 - 3y = 7\].

Step 7: Solve for y

Rearrange the equation from Step 6:\[-3y = 7 - 4 \quad \Rightarrow \quad -3y = 3\] Divide both sides by -3:\[y = \frac{3}{-3} = -1\]

Step 8: Conclude with the Solution

The solution to the system of equations is \(x = 1\) and \(y = -1\), which satisfies both original equations.

Key Concepts

System of Equations

A system of equations consists of two or more equations that are solved together since they share common variables. In our exercise, we have two equations:

• \(4x - 3y = 7\)
• \(7x + 5y = 2\)

These equations make up a system because both equations need to be satisfied at the same time by the same variable values \(x\) and \(y\). When solving such a system, we're finding the values of \(x\) and \(y\) that make both equations true. The solution is a point where the two lines represented by these equations intersect on a graph.To solve the system, we can use several methods. Here, we are using the addition method, also known as the elimination method, which simplifies the process by removing one of the variables.

Linear Equations

Linear equations describe relationships between variables that form a straight line when graphed. The general form of a linear equation is \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.In our system:

• The first equation \(4x - 3y = 7\) is linear, with coefficients 4 and -3, and a constant 7.
• The second equation \(7x + 5y = 2\) is also linear, with coefficients 7 and 5, and a constant 2.

These linear equations intersect at a single solution point when the system is consistent and independent. The key feature of linear equations is that they graph as straight lines. The solution to the system is the intersection of these two lines, meaning the point where both lines meet on the graph.

Variable Elimination

Variable elimination, through the addition method, helps us find the values of variables by strategically removing one variable. This method involves:

• Aligning the two equations to make variable elimination simple.
• Finding coefficients so that adding or subtracting the equations cancels out a variable.

In our exercise, we chose to eliminate \(y\) by finding the least common multiple (LCM) of their coefficients, which were -3 and 5. The LCM is 15, so we multiplied the first equation by 5 and the second by 3 to equalize the coefficients of \(y\). By doing this, we obtained:

• \(20x - 15y = 35\)
• \(21x + 15y = 6\)

Adding these modified equations removed \(y\), leaving us with only \(x\) to solve for. This step significantly simplifies the system, making it easier to isolate and solve for one variable at a time.

Solution to Equations

Finding the solution to equations means determining the values of the variables that satisfy all equations in the system. For our system, after eliminating \(y\), we reduced it to a single equation:\[ 41x = 41 \]By dividing both sides by 41, we find that \(x = 1\). However, we must still solve for \(y\). Using the value of \(x\), we substituted it back into one of the original equations (\(4x - 3y = 7\)) to find \(y\):

• \(4(1) - 3y = 7\)
• \(4 - 3y = 7\)

Solving this, we find \(y = -1\). Thus, the solution to the system of equations is \(x = 1\) and \(y = -1\). These values make both original equations true, verifying that they are the correct solution. This process highlights the effectiveness of the addition method in solving systems of linear equations.