Problem 21 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{l} 3 x+5 y=22 \\ 4 x-7 y=-39 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = -1 \) and \( y = 5 \).

Step 1: Solve one equation for one variable

Let's solve the first equation, \( 3x + 5y = 22 \), for \( x \).Rearrange the terms to isolate \( x \):\[ 3x = 22 - 5y \] Divide every term by 3:\[ x = \frac{22 - 5y}{3} \]

Step 2: Substitute into the second equation

Now substitute \( x = \frac{22 - 5y}{3} \) into the second equation, \( 4x - 7y = -39 \).Replace \( x \) with \( \frac{22 - 5y}{3} \) :\[ 4 \left(\frac{22 - 5y}{3}\right) - 7y = -39 \] Simplify: \[ \frac{88 - 20y}{3} - 7y = -39 \]

Step 3: Eliminate the fraction

Multiply every term by 3 to eliminate the fraction:\[ 88 - 20y - 21y = -117 \] Combine like terms:\[ 88 - 41y = -117 \]

Step 4: Solve for \( y \)

Isolate \( y \) by moving 88 to the other side:\[ -41y = -117 - 88 \] Calculate:\[ -41y = -205 \] Divide by -41 to solve for \( y \):\[ y = 5 \]

Step 5: Substitute back to find \( x \)

Substitute \( y = 5 \) back into the expression for \( x \):\( x = \frac{22 - 5y}{3} \).\[ x = \frac{22 - 5(5)}{3} \]\[ x = \frac{22 - 25}{3} \]\[ x = \frac{-3}{3} \]\[ x = -1 \]

Step 6: Verify the solution

Check the solution \( (x, y) = (-1, 5) \) in the original system.For \( 3x + 5y = 22 \):\( 3(-1) + 5(5) = -3 + 25 = 22 \), which is true.For \( 4x - 7y = -39 \):\( 4(-1) - 7(5) = -4 - 35 = -39 \), which is also true.Thus, the solution is correct.

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. Solving systems of equations means finding the values of these variables that satisfy all equations simultaneously. In this example, we have two linear equations, each involving variables \( x \) and \( y \):

• \( 3x + 5y = 22 \)
• \( 4x - 7y = -39 \)

These equations form a system because they relate the same two variables, \( x \) and \( y \), within different contexts or situations. By solving the system, we find the intersection point on a graph where both lines represented by the equations meet. This intersection corresponds to the solution of the system: the values of \( x \) and \( y \) that make both equations true.
The substitution method is particularly handy for systems where one equation can be easily solved for a single variable. It involves substituting this expression into the other equation, making it easier to find the variable values.

Solving Equations

Solving equations involves finding the values of variables that make the equations true. In our system of equations, we solve by first isolating one variable called substitution. Here, we select the first equation \( 3x + 5y = 22 \) and solve for \( x \) in terms of \( y \):
\[ x = \frac{22 - 5y}{3} \]This expression means that \( x \) changes depending on the value of \( y \). Substituting this expression into the second equation \( 4x - 7y = -39 \) allows the elimination of \( x \), leaving an equation with only \( y \):
\[ 4\left(\frac{22 - 5y}{3}\right) - 7y = -39 \]By simplifying and solving this equation, we discover the value of \( y = 5 \).
The next step is to substitute \( y = 5 \) back into the expression \( x = \frac{22 - 5y}{3} \) to solve for \( x \), which gives \( x = -1 \).
This process of substitution and solving is a core technique in algebra, allowing us to strategically approach complex problems.

Algebra Techniques

The substitution method employs various essential algebra techniques that simplify solving systems of equations. First is isolating a variable, which involves rearranging an equation to express one variable explicitly in terms of others. In our example:

• We rearrange \( 3x + 5y = 22 \) to solve for \( x \), yielding \( x = \frac{22 - 5y}{3} \).

Substitution is the next technique, allowing us to substitute this expression into the second equation to reduce it from two variables to a single one.
Further algebraic manipulation involves eliminating fractions by multiplying through by the denominator, collecting like terms, and isolating the variable:

• In the step \( \frac{88 - 20y}{3} - 7y = -39 \), multiplying through by 3 eliminates the fraction.
• Then combining like terms simplifies \( 88 - 41y = -117 \).
• Finally, solving for \( y \) and verifying its correctness are integral practices in ensuring the accuracy of our solution.

These techniques provide a structured pathway through complex algebraic problems, making even challenging systems more manageable by breaking them down into steps.