Problem 26 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{l} 7 x-2 y=1 \\ 4 x+5 y=2 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution to the system is \( x = \frac{63}{301}, y = \frac{10}{43} \).

Step 1: Solve one equation for one variable

First, choose one of the equations and solve it for one of the variables. Let's solve the first equation for "x": \[ 7x - 2y = 1 \] Add 2y to both sides: \[ 7x = 2y + 1 \] Divide the entire equation by 7 to solve for "x": \[ x = \frac{2y + 1}{7} \].

Step 2: Substitute the expression in the second equation

Substitute the expression for "x" into the second equation: \[ 4x + 5y = 2 \] Replace "x" with \( \frac{2y + 1}{7} \): \[ 4\left( \frac{2y + 1}{7} \right) + 5y = 2 \] Simplify the left side: \[ \frac{8y + 4}{7} + 5y = 2 \].

Step 3: Clear the fraction

Multiply the entire equation by 7 to eliminate the fraction: \[ 8y + 4 + 35y = 14 \] Combine like terms: \[ 43y + 4 = 14 \].

Step 4: Solve for y

Subtract 4 from both sides: \[ 43y = 10 \] Divide by 43 to solve for "y": \[ y = \frac{10}{43} \].

Step 5: Substitute back to find x

Now substitute \( y = \frac{10}{43} \) back into the expression for "x": \[ x = \frac{2\left( \frac{10}{43} \right) + 1}{7} \] Calculate \( 2 \times \frac{10}{43} = \frac{20}{43} \): \[ x = \frac{\frac{20}{43} + 1}{7} = \frac{20}{43} + \frac{43}{43} \] Combine the fractions: \[ x = \frac{63}{43} \frac{1}{7} = \frac{63}{301} \].

Step 6: Verify the solution

Verify the solution by substituting \( x = \frac{63}{301} \) and \( y = \frac{10}{43} \) back into the original equations. Check if both are satisfied:For the first equation: \[ 7\left( \frac{63}{301} \right) - 2\left( \frac{10}{43} \right) = 1 \] and for the second equation: \[ 4\left( \frac{63}{301} \right) + 5\left( \frac{10}{43} \right) = 2 \]. Both should be true if the solution is correct. However, as checking here is error-prone by hand, double-check with computations.

Key Concepts

System of Equations

A system of equations is simply a set of two or more equations that share the same variables. They are often used to find values of these variables that satisfy all equations within the system at the same time.
In our exercise, the given system consists of two linear equations:

• \(7x - 2y = 1\)
• \(4x + 5y = 2\)

This means you need to find numbers for \(x\) and \(y\) that make both equations true simultaneously. There are several methods to solve a system of equations, such as substitution, elimination, and using matrices. Here, we're focusing on the substitution method, which is particularly good when you can easily solve one of the equations for a single variable.

Algebraic Solution

An algebraic solution involves manipulating equations using algebraic operations to find the value of variables. To solve a system algebraically via substitution, you'll typically follow these steps:
First, pick one of the equations and solve it for one of the variables, either \(x\) or \(y\). This can involve adding, subtracting, multiplying, or dividing both sides of the equation to isolate the variable of choice.
For instance, in the equation \(7x - 2y = 1\), we solve for \(x\):

• Add \(2y\) to both sides: \(7x = 2y + 1\)
• Divide by 7: \(x = \frac{2y + 1}{7}\)

Once a variable is isolated, substitute its expression into the other equation. This substitution effectively reduces the system of two equations in two variables to a single equation in one variable. Solving this equation gives a direct value for one of the variables.

Variable Isolation

Variable isolation is a crucial part of solving equations, particularly when using the substitution method. It involves rearranging an equation so one variable stands alone on one side of the equation, expressed in terms of other variables. This helps us easily substitute into other equations within the system.
In our given system, we isolated \(x\) in the first equation to get \(x = \frac{2y + 1}{7}\). By doing this, \(x\) is expressed in terms of \(y\), allowing us to replace \(x\) in the second equation \(4x + 5y = 2\) without changing the system's balance.
This leads to simplifying equations step by step until you find the specific values for your variables. In this case, after substituting and simplifying, you'll eventually isolate \(y\), giving you \(y = \frac{10}{43}\). You can then use this value to find \(x\) which helps in completely solving the system.