Problem 22 - A Graphical Approach to College Algebra

Question

Solve each system by substitution. $$\begin{aligned}&2 x+6 y=-18\\\&5 y=-29+3 x\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{66}{7}\), \(y = \frac{1}{7}\).

Step 1: Solve One Equation for a Variable

We will first solve the second equation for one of the variables. Let's solve the second equation, \(5y = -29 + 3x\), for \(y\). We'll isolate \(y\) by adding 29 to both sides and then dividing by 5. This gives us:\[y = \frac{3x + 29}{5}\]

Step 2: Substitute Expression in the Other Equation

Now that we have \(y\) in terms of \(x\), we'll substitute this expression into the first equation, \(2x + 6y = -18\). Substituting gives us:\[2x + 6\left(\frac{3x + 29}{5}\right) = -18\]

Step 3: Solve for x

Simplify the equation to find \(x\).First, distribute the 6:\[2x + \frac{18x + 174}{5} = -18\]To eliminate the fraction, multiply every term by 5:\[10x + 18x + 174 = -90\]Combine like terms:\[28x + 174 = -90\]Subtract 174 from both sides:\[28x = -264\]Finally, divide by 28:\[x = -\frac{264}{28} = -\frac{66}{7}\]

Step 4: Substitute x Back to Find y

Using the value of \(x\) we found, substitute back into the expression for \(y\):\[y = \frac{3\left(-\frac{66}{7}\right) + 29}{5}\]\[y = \frac{-\frac{198}{7} + \frac{203}{7}}{5}\]\[y = \frac{\frac{5}{7}}{5}\]\[y = \frac{1}{7}\]

Step 5: Conclusion

We found that the solution to the system of equations is \(x = -\frac{66}{7}\) and \(y = \frac{1}{7}\). These values satisfy both equations.

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations, where we express one variable in terms of another and substitute it into the other equation. This reduces the system to a single equation, allowing us to solve for the unknowns step by step. In the given problem, the method begins by transforming the second equation into a simpler form. We started with the equation:

• \(5y = -29 + 3x\)

We isolated \(y\) by rearranging the terms:

• \(y = \frac{3x + 29}{5}\)

This is the expression for \(y\) in terms of \(x\). By substituting this expression into the other equation, we convert it into a one-variable equation. This allows us to solve for \(x\) first. Once \(x\) is known, we can easily find \(y\) by plugging the \(x\) value back into any of the original equations.

Linear Equations

Linear equations are fundamental algebraic expressions where each term is either a constant or the product of a constant and a single variable. The instructors in our exercise exemplified this with two linear equations:

• \(2x + 6y = -18\)
• \(5y = -29 + 3x\)

Both of these have terms with variables raised to the first power, making them linear. Linear equations like these often represent lines on a graph. When using the substitution method, we attempt to find the point where these lines intersect. This intersection point represents the solution to the system. The beauty of linear equations is their predictability and uniformity, offering a straight path for problem-solving using techniques like substitution or elimination.

Algebraic Expressions

Algebraic expressions are combinations of numbers, operations, and variables. They form the basis of equations and inequalities in mathematics. In our exercise, an algebraic expression such as \(5y = -29 + 3x\) neatly encapsulates relationships between variables \(x\) and \(y\). By manipulating these expressions, we solve for the unknowns within them. We turned the equation into the expression:

• \(y = \frac{3x + 29}{5}\).

This step demonstrates a key algebraic skill: the ability to isolate and solve for one variable within an expression. Once modified, expressions become much easier to handle, acting as keys to unlocking the values of unknowns in systems of equations. Understanding how to work through algebraic expressions by performing operations like addition, multiplication, or division is crucial for effectively utilizing both the substitution method and other techniques in algebra.